To find the probability that the two marbles come from bag B given that one is white and one is red, we use conditional probability:

\(P(B \mid WR \text{ or } RW) = \frac{P(W \& R \text{ from bag B})}{P(W \text{ and } R)}\)

First, calculate \(P(W \& R \text{ from bag B})\):

\(P(W \& R \text{ from bag B}) = \frac{2}{3} \times \frac{6}{15} \times \frac{7}{14} + \frac{2}{3} \times \frac{7}{15} \times \frac{6}{14} = \frac{4}{15}\)

Next, calculate \(P(W \text{ and } R)\):

\(P(W \text{ and } R) = P(W \& R \text{ from bag A}) + P(W \& R \text{ from bag B})\)

\(P(W \& R \text{ from bag A}) = \frac{2}{6} \times \frac{8}{15} \times \frac{4}{14} + \frac{2}{6} \times \frac{4}{15} \times \frac{8}{14} = \frac{116}{315}\)

\(P(W \text{ and } R) = \frac{116}{315} + \frac{4}{15} = \frac{232}{630}\)

Finally, calculate the conditional probability:

\(P(B \mid WR \text{ or } RW) = \frac{\frac{4}{15}}{\frac{232}{630}} = \frac{168}{232} = 0.724\)

(a) The probabilities for the tree diagram are:

• From Box A to Box B (Red to Red): \(\frac{x+1}{x+10}\)
• From Box A to Box B (Red to Blue): \(\frac{9}{x+10}\)
• From Box A to Box B (Blue to Red): \(\frac{x}{x+10}\)
• From Box A to Box B (Blue to Blue): \(\frac{10}{x+10}\)

(b) The probability that both balls chosen are blue is:

\(\frac{4}{10} \times \frac{10}{x+10} = \frac{4}{x+10}\)

(c) Given \(\frac{4}{x+10} = \frac{1}{6}\), solve for \(x\):

\(\frac{4}{x+10} = \frac{1}{6}\)

\(4 \times 6 = x + 10\)

\(x + 10 = 24\)

\(x = 14\)

Now, find the probability that the ball chosen from box A is red given that the ball chosen from box B is red:

\(P(A \text{ Red } | B \text{ Red }) = \frac{P(A \text{ Red } \cap B \text{ Red })}{P(B \text{ Red })}\)

\(P(A \text{ Red } \cap B \text{ Red }) = \frac{6}{10} \times \frac{x+1}{x+10} = \frac{6}{10} \times \frac{15}{24} = \frac{3}{8}\)

\(P(B \text{ Red }) = \frac{6}{10} \times \frac{15}{24} + \frac{4}{10} \times \frac{14}{24} = \frac{8}{73}\)

\(P(A \text{ Red } | B \text{ Red }) = \frac{3/8}{8/73} = \frac{45}{73} \approx 0.616\)

To find the probability that a student is male given that they play the drums, we use conditional probability:

\(P(M \mid D) = \frac{P(M \cap D)}{P(D)}\)

From the table, the number of male students who play the drums is 11. The total number of students who play the drums is 11 (male) + 20 (female) = 31.

Thus, \(P(M \cap D) = \frac{11}{180}\) and \(P(D) = \frac{31}{180}\).

Substituting these into the formula gives:

\(P(M \mid D) = \frac{\frac{11}{180}}{\frac{31}{180}} = \frac{11}{31}\)

Let \(C\), \(T\), and \(M\) represent the events of having chocolate, tea, and milk, respectively. Let \(B\) represent the event of having a biscuit.

The probability that Suki does not have a biscuit, \(P(B')\), is calculated as follows:

\(P(B') = P(C \cap B') + P(T \cap B') + P(M \cap B')\)

\(P(C \cap B') = 0.45 \times (1 - 0.3) = 0.45 \times 0.7 = 0.315\)

\(P(T \cap B') = 0.35 \times (1 - 0.6) = 0.35 \times 0.4 = 0.14\)

\(P(M \cap B') = 0.2 \times 1 = 0.2\)

Thus, \(P(B') = 0.315 + 0.14 + 0.2 = 0.655\)

We need to find \(P(T | B')\), the probability that Suki has tea given that she does not have a biscuit:

\(P(T | B') = \frac{P(T \cap B')}{P(B')} = \frac{0.14}{0.655}\)

Calculating this gives:

\(P(T | B') = \frac{0.14}{0.655} \approx 0.214\)

Therefore, the probability that Suki has tea given that she does not have a biscuit is \(0.214\) or \(\frac{28}{131}\).

(i) The number of households in Shan with poor broadband service is 40. The total number of households is 800. Therefore, the probability is given by:

\(P(\text{Shan and Poor}) = \frac{40}{800} = \frac{1}{20}\)

(ii) The number of households in Shan with good broadband service is 177. The total number of households in Shan is \(223 + 177 + 40 = 440\). Therefore, the probability is given by:

\(P(\text{Good} | \text{Shan}) = \frac{177}{440}\)

(a) Let the probability that Alexa arrives late be \(P(\text{late})\). We know that \(P(\text{not late}) = 0.48\).

Using the total probability theorem, \(P(\text{not late}) = 0.4 \times 0.45 + 0.35 \times 0.3 + 0.25 \times (1-x) = 0.48\).

Calculating, \(0.18 + 0.105 + 0.25(1-x) = 0.48\).

\(0.285 + 0.25 - 0.25x = 0.48\).

\(0.535 - 0.25x = 0.48\).

\(0.25x = 0.535 - 0.48\).

\(0.25x = 0.055\).

\(x = \frac{0.055}{0.25} = 0.22\).

(b) To find \(P(\text{train} | \text{late})\), use Bayes' theorem:

\(P(\text{train} | \text{late}) = \frac{P(\text{train} \cap \text{late})}{P(\text{late})}\).

\(P(\text{train} \cap \text{late}) = 0.35 \times 0.7 = 0.245\).

\(P(\text{late}) = 1 - P(\text{not late}) = 0.52\).

Thus, \(P(\text{train} | \text{late}) = \frac{0.245}{0.52} = \frac{49}{104}\) or 0.471.

(a) The tree diagram is constructed as follows:

• First branch: Pass first written test (W1P) with probability 0.8, Fail first written test (W1F) with probability 0.2.
• Second branch (if W1F): Pass second written test (W2P) with probability 0.6, Fail second written test (W2F) with probability 0.4.
• Third branch (if W1P or W2P): Pass practical test (PP) with probability 0.3, Fail practical test (PF) with probability 0.7.

(b) The probability that a student will succeed in gaining a place is calculated by considering the successful paths:

\(P(W1P \cap PP) + P(W1F \cap W2P \cap PP) = (0.8 \times 0.3) + (0.2 \times 0.6 \times 0.3) = 0.24 + 0.036 = 0.276\)

(c) The probability that a student passes the written test at the first attempt given that they succeed in gaining a place is:

\(P(W1P \mid \text{getting place}) = \frac{P(W1P \cap PP)}{P(\text{getting place})} = \frac{0.24}{0.276} = \frac{20}{23} \approx 0.87\)

(a) To find the probability that Georgie wears a hat, we calculate the total probability of wearing a hat with each scarf:

\(P(H) = 0.2 \times 1 + 0.45 \times 0.4 + 0.35 \times 0.3\)

\(= 0.2 + 0.18 + 0.105\)

\(= 0.485\)

Thus, the probability that Georgie wears a hat is \(\frac{97}{200}\) or 0.485.

(b) To find the probability that Georgie wears a yellow scarf given that she does not wear a hat, we use conditional probability:

\(P(Y | \overline{H}) = \frac{P(Y \cap \overline{H})}{P(\overline{H})}\)

\(P(Y \cap \overline{H}) = 0.35 \times 0.7 = 0.245\)

\(P(\overline{H}) = 1 - P(H) = 1 - 0.485 = 0.515\)

\(P(Y | \overline{H}) = \frac{0.245}{0.515} \approx 0.476\)

Thus, the probability that Georgie wears a yellow scarf given that she does not wear a hat is \(\frac{49}{103}\) or 0.476.

(a) The tree diagram is completed with the following probabilities:

• 1 April Fine to 2 April Fine: 0.75
• 1 April Fine to 2 April Rainy: 0.25
• 1 April Rainy to 2 April Fine: 0.4
• 1 April Rainy to 2 April Rainy: 0.6

(b) The probability that 2 April is fine is calculated as follows:

\(P(\text{2 April Fine}) = 0.8 \times 0.75 + 0.2 \times 0.4 = 0.6 + 0.08 = 0.68\)

Thus, \(P(\text{2 April Fine}) = \frac{17}{25}\).

(c) To find \(P(X \cap Y)\), we calculate:

\(P(X \cap Y) = 0.8 \times 0.25 + 0.8 \times 0.6 = 0.2 + 0.12 = 0.27\)

(d) To find the probability that 1 April is fine given that 3 April is rainy, we use:

\(P(Y) = 0.2 \times 0.4 \times 0.25 + 0.2 \times 0.6 \times 0.6 = 0.362\)

\(P(X|Y) = \frac{P(X \cap Y)}{P(Y)} = \frac{0.27}{0.362} = 0.746\)

(a) The tree diagram is constructed as follows:

• First level: Modes of transport with probabilities: Car (0.2), Bus (0.45), Walk (0.35).
• Second level: Arrival times for each mode: Early (E) or Not Early (NE).
• Probabilities for Car: E (0.6), NE (0.4).
• Probabilities for Bus: E (0.1), NE (0.9).
• Probabilities for Walk: E (1), NE (0).

(b) To find the probability that Juan goes to college by car given that he arrives early, we use Bayes' theorem:

\(P(C|E) = \frac{P(C \cap E)}{P(E)}\)

Where:

• \(P(C \cap E) = 0.2 \times 0.6 = 0.12\)
• \(P(E) = (0.2 \times 0.6) + (0.45 \times 0.1) + (0.35 \times 1) = 0.12 + 0.045 + 0.35 = 0.515\)

Thus,

\(P(C|E) = \frac{0.12}{0.515} = \frac{12}{515} \approx 0.233\)

To find the probability that a randomly chosen student is at Devar college given that they prefer soccer, we use conditional probability.

The formula for conditional probability is:

\(P(D|S) = \frac{P(D \cap S)}{P(S)}\)

Where:

• \(P(D|S)\) is the probability that a student is at Devar college given they prefer soccer.
• \(P(D \cap S)\) is the probability that a student is both at Devar college and prefers soccer.
• \(P(S)\) is the probability that a student prefers soccer.

From the table, the number of students who are at Devar college and prefer soccer is 120.

The total number of students who prefer soccer is 280.

Thus, \(P(D|S) = \frac{120}{280}\).

Simplifying \(\frac{120}{280}\) gives \(\frac{3}{7}\).

Therefore, the probability is \(\frac{3}{7}\).

(a) The tree diagram shows the following probabilities:

• Choosing bag X or Y: each with probability \(\frac{1}{2}\).
• For bag X: Removing a red marble \(\frac{7}{10}\), blue marble \(\frac{3}{10}\).
• For bag Y: Removing a red marble \(\frac{4}{5}\), blue marble \(\frac{1}{5}\).
• After adding a red marble, for bag X: Removing a red marble \(\frac{8}{10}\), blue marble \(\frac{2}{10}\).
• After adding a red marble, for bag Y: Removing a red marble \(\frac{5}{6}\), blue marble \(\frac{1}{6}\).

(b) Probability that both marbles are the same colour:

\(P(\text{both same colour}) = P(BB) + P(RR) = P(XBB) + P(XRR) + P(YRR)\)

\(= \frac{1}{2} \times \frac{3}{10} \times \frac{2}{10} + \frac{1}{2} \times \frac{7}{10} \times \frac{7}{10} + \frac{1}{2} \times \frac{4}{5} \times \frac{4}{5}\)

\(= \frac{6}{200} + \frac{49}{200} + \frac{16}{50}\)

\(= 0.03 + 0.245 + 0.32 = 0.595\)

(c) Probability that bag Y is chosen given different colours:

\(P(\text{bag } Y | \text{different colours}) = \frac{P(\text{bag } Y \cap \text{different colours})}{P(\text{different colours})}\)

\(= \frac{\frac{1}{2} \times \frac{4}{5} \times \frac{1}{6} + \frac{1}{2} \times \frac{1}{5} \times \frac{5}{6}}{1 - 0.595}\)

\(= \frac{\frac{4}{50} + \frac{1}{12}}{0.405}\)

\(= \frac{9}{81} = 0.444\)

(a) The tree diagram is as follows:

- First branch: Pizza (0.35), Burger (0.44), Curry (0.21)

- Second branch for Pizza: Fruit (0.3), No Fruit (0.7)

- Second branch for Burger: Fruit (0.8), No Fruit (0.2)

- Second branch for Curry: Fruit (0), No Fruit (1)

(b) The probability that Rani has some fruit is calculated as:

\(0.35 \times 0.3 + 0.44 \times 0.8 + 0.21 \times 0 = 0.105 + 0.352 + 0 = 0.457\)

(c) The probability that Rani does not have a burger given that she does not have any fruit is:

\(P(\text{not B} | \text{not fruit}) = \frac{P(\text{not B} \cap \text{not fruit})}{P(\text{not fruit})}\)

Calculate \(P(\text{not B} \cap \text{not fruit}) = 0.35 \times 0.7 + 0.21 \times 1 = 0.245 + 0.21 = 0.455\)

Calculate \(P(\text{not fruit}) = 1 - 0.457 = 0.543\)

Thus, \(P(\text{not B} | \text{not fruit}) = \frac{0.455}{0.543} = 0.838\)

(a) For Box A, the probability of choosing a red ball is \(\frac{7}{8}\) and a blue ball is \(\frac{1}{8}\).

For Box B, if a red ball is added, the probability of choosing a red ball is \(\frac{10}{15}\) and a blue ball is \(\frac{5}{15}\). If a blue ball is added, the probability of choosing a red ball is \(\frac{9}{15}\) and a blue ball is \(\frac{6}{15}\).

(b) The probability that the two balls chosen are not the same colour is calculated as:

\(\frac{7}{8} \times \frac{5}{15} + \frac{1}{8} \times \frac{9}{15} = \frac{44}{120} = \frac{11}{30}\)

(c) The probability that the ball chosen from box A is blue given that the ball chosen from box B is blue is:

\(P(A \text{ blue } | B \text{ blue }) = \frac{P(A \text{ blue } \cap B \text{ blue })}{P(B \text{ blue })}\)

\(= \frac{\frac{1}{8} \times \frac{6}{15}}{\frac{7}{8} \times \frac{5}{15} + \frac{1}{8} \times \frac{6}{15}} = \frac{\frac{1}{20}}{\frac{41}{120}} = \frac{6}{41}\)

(i) The total probability that Benju is late for work is given by:

\(0.4x + 0.6 \times 2x = 0.36\)

Simplifying, we have:

\(0.4x + 1.2x = 0.36\)

\(1.6x = 0.36\)

Solving for \(x\), we get:

\(x = \frac{0.36}{1.6} = 0.225\)

(ii) We need to find the probability that Benju chooses the hilly route given that he is not late for work. The probability that he is not late is \(1 - 0.36 = 0.64\).

The probability that he chooses the hilly route and is not late is \(0.4(1-x)\).

The probability that he chooses the busy route and is not late is \(0.6(1-2x)\).

Thus, the probability that he is not late is:

\(0.4(1-x) + 0.6(1-2x) = 0.64\)

Using \(x = 0.225\), the probability that he chooses the hilly route and is not late is:

\(0.4(1-0.225) = 0.4 \times 0.775 = 0.31\)

The probability that he is not late is:

\(0.4 \times 0.775 + 0.6 \times 0.55 = 0.31 + 0.33 = 0.64\)

Therefore, the probability that he chooses the hilly route given that he is not late is:

\(\frac{0.31}{0.64} = \frac{31}{64} \approx 0.484\)

(i) The tree diagram is drawn with the following branches:

• First set of branches for the type of message: text (0.3), email (0.2), social media (0.5).
• Second set of branches for replies:
• Text: reply (0.4), no reply (0.6).
• Email: reply (0.15), no reply (0.85).
• Social media: reply (0.6), no reply (0.4).

(ii) To find the probability that the message was an email given no immediate reply, use Bayes' theorem:

\(P(\text{email} \mid \text{NR}) = \frac{P(\text{email} \cap \text{NR})}{P(\text{NR})}\)

Calculate \(P(\text{email} \cap \text{NR}) = 0.2 \times 0.85 = 0.17\).

Calculate \(P(\text{NR}) = 0.3 \times 0.6 + 0.2 \times 0.85 + 0.5 \times 0.4 = 0.18 + 0.17 + 0.2 = 0.55\).

Thus, \(P(\text{email} \mid \text{NR}) = \frac{0.17}{0.55} = \frac{17}{55} \approx 0.309\)

(i) The probability that Tamar wears red socks is calculated by considering both cases: wearing a blue suit and wearing a grey suit.

Probability of wearing red socks with a blue suit: \(0.6 \times 0.2 = 0.12\).

Probability of wearing red socks with a grey suit: \(0.4 \times 0.32 = 0.128\).

Total probability of wearing red socks: \(0.12 + 0.128 = 0.248\).

Expressed as a fraction: \(\frac{31}{125}\).

(ii) We need to find the probability that Tamar is wearing a grey suit given that he is not wearing red socks.

Probability of not wearing red socks with a blue suit: \(0.6 \times 0.8 = 0.48\).

Probability of not wearing red socks with a grey suit: \(0.4 \times 0.68 = 0.272\).

Total probability of not wearing red socks: \(0.48 + 0.272 = 0.752\).

Using Bayes' theorem, the probability of wearing a grey suit given not wearing red socks is:

\(\frac{0.272}{0.752} = \frac{17}{47}\).

(i) The tree diagram is completed as follows:

• First ball: Probability of red (R) = \(\frac{3}{8}\), Probability of blue (B) = \(\frac{5}{8}\).
• Second ball after red: Probability of red (R) = \(\frac{2}{8}\), Probability of blue (B) = \(\frac{5}{8}\), Probability of yellow (Y) = \(\frac{1}{8}\).
• Second ball after blue: Probability of red (R) = \(\frac{3}{8}\), Probability of blue (B) = \(\frac{4}{8}\), Probability of yellow (Y) = \(\frac{1}{8}\).

(ii) Probability that the two balls taken are the same colour:

\(P(RR) = \frac{3}{8} \times \frac{2}{8} = \frac{3}{32}\)

\(P(BB) = \frac{5}{8} \times \frac{4}{8} = \frac{5}{16}\)

Total probability = \(\frac{3}{32} + \frac{5}{16} = \frac{13}{32}\) or 0.406

(iii) Probability that the first ball taken is red, given that the second ball taken is blue:

\(P(RB) = \frac{3}{8} \times \frac{5}{8} = \frac{15}{64}\)

\(P(B) = \frac{3}{8} \times \frac{5}{8} + \frac{5}{8} \times \frac{4}{8} = \frac{35}{64}\)

\(P(R|B) = \frac{P(RB)}{P(B)} = \frac{15/64}{35/64} = \frac{3}{7}\) or 0.429

First, calculate the total number of girls:

\(15 + 12 + 37 = 64\).

Next, calculate the number of girls not studying Drama:

\(15 + 12 = 27\).

The probability that a randomly chosen student is not studying Drama, given that the student is a girl, is:

\(\frac{27}{64}\).

(i) Let the probability of being a Beginner be 0.5. The probability of being a Beginner is given by:

\(0.7x + 0.45(1-x) = 0.5\)

Solving for \(x\):

\(0.7x + 0.45 - 0.45x = 0.5\)

\(0.25x = 0.05\)

\(x = 0.2\)

(ii) The probability of being an Advanced swimmer is also 0.5. The probability of being an Advanced swimmer and male is:

\(0.3x = 0.3 \times 0.2 = 0.06\)

The probability of being an Advanced swimmer is:

\(0.5 = 0.3 \times 0.2 + 0.55 \times 0.8\)

The probability that a randomly chosen Advanced swimmer is male is:

\(\frac{0.06}{0.5} = 0.12 = \frac{3}{25}\)

Let the events be defined as follows:

• ext{SSS}: All three vehicles go straight on.
• ext{LLL}: All three vehicles turn left.
• ext{RRR}: All three vehicles turn right.

The probability that all three vehicles choose the same direction is given by:

\(P( ext{same direction}) = P( ext{SSS}) + P( ext{LLL}) + P( ext{RRR})\)

Calculating each probability:

\(P( ext{SSS}) = 0.3^3\)

\(P( ext{LLL}) = 0.55^3\)

\(P( ext{RRR}) = 0.15^3\)

Thus,

\(P( ext{same direction}) = 0.3^3 + 0.55^3 + 0.15^3\)

We need to find the probability that all three vehicles go straight on given that they all choose the same direction:

\(P( ext{SSS} \,|\, ext{all same direction}) = \frac{P( ext{SSS and same direction})}{P( ext{same direction})}\)

Since ext{SSS} is a subset of ext{same direction},

\(P( ext{SSS and same direction}) = P( ext{SSS}) = 0.3^3\)

Therefore,

\(P( ext{SSS} \,|\, ext{all same direction}) = \frac{0.3^3}{0.3^3 + 0.55^3 + 0.15^3}\)

Calculating the values:

\(0.3^3 = 0.027\)

\(0.55^3 = 0.166375\)

\(0.15^3 = 0.003375\)

Thus,

\(P( ext{same direction}) = 0.027 + 0.166375 + 0.003375 = 0.197\)

Finally,

\(P( ext{SSS} \,|\, ext{all same direction}) = \frac{0.027}{0.197} \approx 0.137\)

(i) The tree diagram is as follows:

First attempt: Pass (0.85), Fail (0.15)

If Fail: Retake Pass (0.65), Retake Fail (0.35)

(ii) We need to find the probability that a student fails the first attempt given that they gain the certificate.

Let \(F\) be the event of failing the first attempt and \(P\) be the event of passing overall.

The probability of passing overall, \(P(P)\), is given by:

\(P(P) = P(\text{Pass first attempt}) + P(\text{Fail first attempt and Pass retake})\)

\(P(P) = 0.85 + (0.15 \times 0.65) = 0.85 + 0.0975 = 0.9475\)

The probability of failing the first attempt and passing the retake, \(P(F \cap P)\), is:

\(P(F \cap P) = 0.15 \times 0.65 = 0.0975\)

Using conditional probability:

\(P(F | P) = \frac{P(F \cap P)}{P(P)} = \frac{0.0975}{0.9475} = \frac{39}{379} \approx 0.103\)

To find \(P(B \,|\, A')\), we use the formula for conditional probability:

\(P(B \,|\, A') = \frac{P(B \cap A')}{P(A')}\)

First, calculate \(P(A')\):

The total number of outcomes when throwing two dice is 36. The event \(A\) (sum less than 6) includes the outcomes: (1,1), (1,2), (1,3), (1,4), (2,1), (2,2), (2,3), (3,1), (3,2), (4,1). There are 10 outcomes, so \(P(A) = \frac{10}{36}\).

Thus, \(P(A') = 1 - P(A) = 1 - \frac{10}{36} = \frac{26}{36}\).

Next, calculate \(P(B \cap A')\):

The event \(B\) (difference at most 2) includes outcomes where the difference between the numbers is 0, 1, or 2. The outcomes that satisfy \(B \cap A'\) are those that are not in \(A\) but satisfy \(B\). There are 16 such outcomes.

Thus, \(P(B \cap A') = \frac{16}{36}\).

Now, substitute these into the conditional probability formula:

\(P(B \,|\, A') = \frac{\frac{16}{36}}{\frac{26}{36}} = \frac{16}{26} = \frac{8}{13}\)

(i) The tree diagram is as follows:

• First branch: Economy (E) with probability 0.82, First class (F) with probability 0.18.
• Second branch from Economy: Good Night's Sleep (GNS) with probability \(x\), Not GNS with probability \(1-x\).
• Second branch from First class: GNS with probability 0.9, Not GNS with probability 0.1.

(ii) The probability of getting a good night's sleep is given by:

\(0.82x + 0.18 imes 0.9 = 0.285\)

Solving for \(x\):

\(0.82x + 0.162 = 0.285\)

\(0.82x = 0.123\)

\(x = \frac{0.123}{0.82} = 0.15\)

(iii) We need to find \(P(E | \text{not GNS})\):

\(P(E | \text{not GNS}) = \frac{P(E \cap \text{not GNS})}{P(\text{not GNS})}\)

\(P(E \cap \text{not GNS}) = 0.82 imes (1 - 0.15) = 0.82 imes 0.85 = 0.697\)

\(P(\text{not GNS}) = 1 - 0.285 = 0.715\)

\(P(E | \text{not GNS}) = \frac{0.697}{0.715} = 0.975\)

(i) To find the probability that the jar is small, we calculate the probability of a jar being small for both Café Premium and Café Standard and sum them:

\(P(S) = 0.65 \times 0.6 + 0.35 \times 0.75\)

\(= 0.39 + 0.2625 = 0.6525\)

Therefore, the probability that the jar is small is approximately 0.653.

(ii) To find the probability that the jar is Café Standard given that it is large, we use conditional probability:

\(P(\text{Std} \mid L) = \frac{P(\text{Std} \cap L)}{P(L)}\)

Where:

\(P(\text{Std} \cap L) = 0.35 \times 0.25 = 0.0875\)

And:

\(P(L) = 1 - P(S) = 1 - 0.6525 = 0.3475\)

Thus:

\(P(\text{Std} \mid L) = \frac{0.0875}{0.3475} \approx 0.252\)

Therefore, the probability that the jar is Café Standard given that it is large is approximately 0.252.

(i) Let \(P(BH)\) be the probability that Khalid hurts himself while riding his bicycle, and \(P(SH)\) be the probability that he hurts himself while riding his skateboard. We have:

\(P(BH) = 0.6 \times 0.05 = 0.03\)

\(P(SH) = 0.4 \times 0.75 = 0.3\)

The probability that Khalid hurts himself on any particular day is:

\(P(H) = P(BH) + P(SH) = 0.03 + 0.3 = 0.33\)

(ii) We need to find \(P(S|H)\), the probability that Khalid is riding on his skateboard given that he hurts himself. Using the formula for conditional probability:

\(P(S|H) = \frac{P(S \cap H)}{P(H)}\)

\(P(S \cap H) = P(SH) = 0.3\)

\(P(H) = 0.33\)

Thus, \(P(S|H) = \frac{0.3}{0.33} = \frac{10}{11} \approx 0.909\)

(i) The tree diagram is drawn with the following branches:

• First match: Win (\(\frac{3}{5}\)), Draw (\(\frac{1}{5}\)), Lose (\(\frac{1}{5}\))
• Second match if Win first: Win (\(\frac{7}{10}\)), Draw (\(\frac{2}{10}\)), Lose (\(\frac{1}{10}\))
• Second match if Draw first: Win (\(\frac{1}{3}\)), Draw (\(\frac{1}{3}\)), Lose (\(\frac{1}{3}\))
• Second match if Lose first: Win (\(\frac{3}{10}\)), Draw (\(\frac{1}{20}\)), Lose (\(\frac{13}{20}\))

(ii) To find the probability that they lose the first match given they win the second match, use Bayes' theorem:

\(P(L_1 \mid W_2) = \frac{P(L_1 \cap W_2)}{P(W_2)}\)

Calculate \(P(L_1 \cap W_2)\):

\(P(L_1 \cap W_2) = \frac{1}{5} \times \frac{3}{10} = \frac{3}{50}\)

Calculate \(P(W_2)\):

\(P(W_2) = \frac{3}{5} \times \frac{7}{10} + \frac{1}{5} \times \frac{1}{3} + \frac{1}{5} \times \frac{3}{10}\)

\(P(W_2) = \frac{21}{50} + \frac{1}{15} + \frac{3}{50}\)

\(P(W_2) = \frac{41}{75}\)

Thus,

\(P(L_1 \mid W_2) = \frac{3/50}{41/75} = \frac{9}{82}\)

Let \(W\), \(C\), and \(B\) represent the events that Anya walks, cycles, and takes the bus, respectively. Let \(L\) represent the event that Anya is late.

The probability that Anya cycles and is late is \(P(C \cap L) = P(C) \times P(L \mid C) = 0.65 \times 0.1 = 0.065\).

The total probability that Anya is late, \(P(L)\), is given by:

\(P(L) = P(W \cap L) + P(C \cap L) + P(B \cap L)\)

\(= (0.3 \times 0.15) + (0.65 \times 0.1) + (0.05 \times 0.6)\)

\(= 0.045 + 0.065 + 0.03 = 0.14\)

The probability that Anya cycles given that she is late is:

\(P(C \mid L) = \frac{P(C \cap L)}{P(L)} = \frac{0.065}{0.14} = 0.464\)

Thus, the probability that Anya cycles given that she is late is \(0.464\) or \(\frac{13}{28}\).

(i) The probability of taking a blue pen from the left pocket is \(\frac{1}{4}\). After moving this pen to the right pocket, the probability of taking a blue pen from the right pocket is \(\frac{2}{5}\). Therefore, the probability is \(\frac{1}{4} \times \frac{2}{5} = \frac{1}{10}\).

(ii) To have 1 blue pen in the left pocket after operation T, either a red pen is moved from left to right and a red pen is moved back, or a blue pen is moved from left to right and a blue pen is moved back. The probability of moving a red pen from left to right is \(\frac{3}{4}\) and then moving a red pen back is \(\frac{4}{5}\). The probability of moving a blue pen from left to right and then moving a blue pen back is \(\frac{1}{10}\). Thus, \(P(X = 1) = \frac{3}{4} \times \frac{4}{5} + \frac{1}{10} = \frac{14}{20} = \frac{7}{10}\).

(iii) Given that a blue pen is taken from the right pocket, the probability that a blue pen was initially taken from the left pocket is \(\frac{P(B \cap B)}{P(B)} = \frac{1/10}{3/4 \times 1/5 + 1/4 \times 2/5} = \frac{1/10}{1/5} = \frac{2}{5}\).

(i) The tree diagram is constructed as follows:

• First branch: Tea (T) with probability 0.65, Coffee (C) with probability 0.28, Hot Chocolate (HC) with probability 0.07.
• Second branch for Tea: Milk (M) with probability 0.8, No Milk (NM) with probability 0.2.
• Second branch for Coffee: Milk (M) with probability 0.5, No Milk (NM) with probability 0.5.
• Second branch for Hot Chocolate: Milk (M) with probability 1, No Milk (NM) with probability 0.

(ii) To find the probability that Ayman’s breakfast drink is coffee, given that his drink has milk in it, we use Bayes' theorem:

\(P(C \mid \text{milk}) = \frac{P(C \cap \text{milk})}{P(\text{milk})}\)

Calculate \(P(C \cap \text{milk})\):

\(P(C \cap \text{milk}) = 0.28 \times 0.5 = 0.14\)

Calculate \(P(\text{milk})\):

\(P(\text{milk}) = (0.65 \times 0.8) + (0.28 \times 0.5) + (0.07 \times 1) = 0.52 + 0.14 + 0.07 = 0.73\)

Substitute these into the formula:

\(P(C \mid \text{milk}) = \frac{0.14}{0.73} \approx 0.192\)

(i) To find the probability that Sam gets a cup of coffee, we use the law of total probability:

\(P( ext{cup of coffee}) = P( ext{on time}) \times P( ext{coffee | on time}) + P( ext{not on time}) \times P( ext{coffee | not on time})\)

\(= 0.6 \times 0.9 + 0.4 \times 0.3\)

\(= 0.54 + 0.12 = 0.66\)

(ii) To find the probability that the bus is not on time given that Sam does not get a cup of coffee, we use Bayes' theorem:

\(P( ext{not on time | no cup}) = \frac{P( ext{not on time} \cap ext{no cup})}{P( ext{no cup})}\)

First, calculate \(P( ext{not on time} \cap ext{no cup})\):

\(P( ext{not on time} \cap ext{no cup}) = P( ext{not on time}) \times P( ext{no cup | not on time})\)

\(= 0.4 \times 0.7 = 0.28\)

Next, calculate \(P( ext{no cup})\):

\(P( ext{no cup}) = 1 - P( ext{cup of coffee}) = 1 - 0.66 = 0.34\)

Finally, substitute these into Bayes' theorem:

\(P( ext{not on time | no cup}) = \frac{0.28}{0.34} \approx 0.824\)

Let:

• \(A\) be the event that a person goes abroad.
• \(C\) be the event that a person goes camping.

We need to find \(P(A \mid C)\), the probability that a person goes abroad given that they go camping.

Using the formula for conditional probability:

\(P(A \mid C) = \frac{P(A \cap C)}{P(C)}\)

Calculate \(P(A \cap C)\):

\(P(A \cap C) = P(A) \times P(C \mid A) = 0.35 \times 0.15 = 0.0525\)

Calculate \(P(C)\):

\(P(C) = P(A \cap C) + P(H \cap C)\)

Where \(H\) is the event of taking a holiday in their own country.

\(P(H \cap C) = P(H) \times P(C \mid H) = 0.65 \times 0.4 = 0.26\)

Thus,

\(P(C) = 0.0525 + 0.26 = 0.3125\)

Finally, calculate \(P(A \mid C)\):

\(P(A \mid C) = \frac{0.0525}{0.3125} = 0.168\)

(i) To complete the table, we use the given probabilities and the fact that the total probability is 1.

The probability that the meal is served on time is \(\frac{1}{5}\), so the probability that it is not served on time is \(1 - \frac{1}{5} = \frac{4}{5}\).

The probability that the kitchen is left in a mess is \(\frac{3}{5}\), so the probability that it is not left in a mess is \(1 - \frac{3}{5} = \frac{2}{5}\).

We know that the probability that the meal is not served on time and the kitchen is not left in a mess is \(\frac{3}{10}\).

Using these, we can fill in the table:

(ii) To find the probability that the meal is not served on time given that the kitchen is left in a mess, we use conditional probability:

\(P(\text{Not on time} | \text{Kitchen mess}) = \frac{P(\text{Not on time and Kitchen mess})}{P(\text{Kitchen mess})}\)

From the table, \(P(\text{Not on time and Kitchen mess}) = \frac{1}{2}\) and \(P(\text{Kitchen mess}) = \frac{3}{5}\).

Thus, \(P(\text{Not on time} | \text{Kitchen mess}) = \frac{\frac{1}{2}}{\frac{3}{5}} = \frac{1}{2} \times \frac{5}{3} = \frac{5}{6}\)

To solve this problem, we need to find the conditional probability \(P(X > 0 \mid X \neq 2)\).

First, calculate the total number of outcomes when two spinners are spun. Each spinner has 5 sides, so there are \(5 \times 5 = 25\) possible outcomes.

Next, determine the number of outcomes where the score is not equal to 2. The score is 2 when the numbers are (1,3), (3,1), (2,4), (4,2), (3,5), and (5,3). There are 6 such outcomes.

Thus, the number of outcomes where the score is not 2 is \(25 - 6 = 19\).

Now, find the number of outcomes where the score is greater than 0 and not equal to 2. This includes all outcomes except those where the numbers are equal (score = 0) and those where the score is 2. There are 5 outcomes where the numbers are equal: (1,1), (2,2), (3,3), (4,4), (5,5).

Therefore, the number of outcomes where the score is greater than 0 and not equal to 2 is \(25 - 5 - 6 = 14\).

The conditional probability is then given by:

\(P(X > 0 \mid X \neq 2) = \frac{\text{Number of outcomes } (X > 0 \cap X \neq 2)}{\text{Number of outcomes } X \neq 2} = \frac{14}{19}\)

Thus, the probability is \(\frac{14}{19}\) or approximately 0.737.

(i) To find the probability \(x\), we use the total probability that Nikita’s mother likes the present:

\(0.3 \times 0.72 + 0.7 \times x = 0.783\)

Simplifying, we get:

\(0.216 + 0.7x = 0.783\)

\(0.7x = 0.783 - 0.216\)

\(0.7x = 0.567\)

\(x = \frac{0.567}{0.7} = 0.81\)

(ii) To find the probability that the present is a scarf given that Nikita’s mother does not like it, we use conditional probability:

\(P(S \mid NL) = \frac{P(S \cap NL)}{P(NL)}\)

\(P(S \cap NL) = 0.3 \times (1 - 0.72) = 0.3 \times 0.28 = 0.084\)

\(P(NL) = 1 - P(L) = 1 - 0.783 = 0.217\)

\(P(S \mid NL) = \frac{0.084}{0.217} \approx 0.387\)

Thus, the probability that the present is a scarf given that Nikita’s mother does not like it is \(0.387\) or \(\frac{12}{31}\).

(i) We know the total probability of viewing photos fewer than 3 times is 0.801. This can be expressed as:

\((1-x) \times 0.9 + x \times 0.24 = 0.801\)

Solving for \(x\):

\(0.9 - 0.9x + 0.24x = 0.801\)

\(0.9 - 0.66x = 0.801\)

\(0.66x = 0.9 - 0.801\)

\(0.66x = 0.099\)

\(x = \frac{0.099}{0.66} = 0.15\)

(ii) We need to find \(P(\text{at least 100 photos} \mid \text{at least 3 views})\).

Using Bayes' theorem:

\(P(\text{at least 100 photos} \mid \text{at least 3 views}) = \frac{P(\text{at least 100 photos} \cap \text{at least 3 views})}{P(\text{at least 3 views})}\)

\(P(\text{at least 100 photos} \cap \text{at least 3 views}) = 0.85 \times 0.1\)

\(P(\text{at least 3 views}) = 1 - 0.801 = 0.199\)

\(P(\text{at least 100 photos} \mid \text{at least 3 views}) = \frac{0.85 \times 0.1}{0.85 \times 0.1 + 0.15 \times 0.76}\)

\(= \frac{0.085}{0.085 + 0.114}\)

\(= \frac{0.085}{0.199} \approx 0.427\)

(i) The probability that the person chosen is from country X is given by the ratio of the population of country X to the total population:

\(P(X) = \frac{20}{28} = \frac{5}{7} \approx 0.714\)

(ii) The probability that the person chosen has fair hair is calculated by considering both countries:

In country X, 25% have fair hair, so \(0.25 \times 20 = 5\) million people.

In country Y, 60% have fair hair, so \(0.60 \times 8 = 4.8\) million people.

Total with fair hair = \(5 + 4.8 = 9.8\) million people.

\(P(F) = \frac{9.8}{28} = \frac{7}{20} = 0.35\)

(iii) The probability that the person is from country X given they have fair hair is:

\(P(X|F) = \frac{P(X \cap F)}{P(F)}\)

\(P(X \cap F) = \frac{5}{28}\) (people with fair hair from X)

\(P(X|F) = \frac{\frac{5}{28}}{\frac{7}{20}} = \frac{5}{28} \times \frac{20}{7} = \frac{25}{49} \approx 0.51\)

To find the probability that the coin shows a head given that Jodie’s score is 8, we use conditional probability:

\(P(H | 8) = \frac{P(H \cap 8)}{P(8)}\)

First, calculate \(P(8)\):

\(P(8) = P(H \text{ and } 4+4) + P(T \text{ and } 2 \times 4) + P(T \text{ and } 4 \times 2)\)

\(= \frac{1}{3} \times \frac{1}{16} + \frac{2}{3} \times \frac{1}{16} + \frac{2}{3} \times \frac{1}{16}\)

\(= \frac{1}{48} + \frac{2}{48} + \frac{2}{48}\)

\(= \frac{5}{48}\)

Next, calculate \(P(H \cap 8)\):

\(P(H \cap 8) = P(H \text{ and } 4+4) = \frac{1}{3} \times \frac{1}{16} = \frac{1}{48}\)

Now, substitute these into the conditional probability formula:

\(P(H | 8) = \frac{\frac{1}{48}}{\frac{5}{48}} = \frac{1}{5}\)

(i) The probability that Ben becomes the champion after playing exactly 2 games is when Ben wins both games. The probability of Ben winning a game is 0.7. Therefore, the probability is:

\(P(B \text{ champ}) = 0.7 \times 0.7 = 0.49\)

(ii) The probability that Ben becomes the champion can occur in the following sequences: WW, WLW, LWW.

\(P(B \text{ champ}) = P(WW) + P(WLW) + P(LWW)\)

\(= (0.7 \times 0.7) + (0.7 \times 0.3 \times 0.7) + (0.3 \times 0.7 \times 0.7)\)

\(= 0.49 + 0.147 + 0.147\)

\(= 0.784\)

(iii) Given that Tom becomes the champion, we need to find the probability that he won the 2nd game. This can occur in the sequences: TT or TBT.

\(P(T2 \cap T) = (0.3 \times 0.3) + (0.7 \times 0.3 \times 0.3) = 0.216\)

The probability that Tom becomes the champion is \(1 - 0.784 = 0.216\).

Thus, the probability is:

\(P(T2 | T) = \frac{P(T2 \cap T)}{P(T)} = \frac{0.216}{0.216} = 0.708\)

To find the probability that Nur chose playground Y given that she chose a climbing frame, we use Bayes' theorem:

\(P(Y | C) = \frac{P(Y \cap C)}{P(C)}\)

First, calculate \(P(Y \cap C)\):

Nur chooses playground Y with probability \(\frac{1}{4}\) and then chooses a climbing frame with probability \(\frac{1}{12}\) (since there is 1 climbing frame out of 12 total pieces of equipment in playground Y).

\(P(Y \cap C) = \frac{1}{4} \times \frac{1}{12} = \frac{1}{48}\)

Next, calculate \(P(C)\), the total probability of choosing a climbing frame:

\(P(C) = P(X \cap C) + P(Y \cap C) + P(Z \cap C)\)

\(P(X \cap C) = 0\) since there are no climbing frames in playground X.

\(P(Y \cap C) = \frac{1}{48}\) as calculated above.

\(P(Z \cap C) = \frac{1}{2} \times \frac{4}{16} = \frac{1}{8}\) since there are 4 climbing frames out of 16 total pieces of equipment in playground Z.

\(P(C) = 0 + \frac{1}{48} + \frac{1}{8} = \frac{1}{48} + \frac{6}{48} = \frac{7}{48}\)

Finally, substitute back into Bayes' theorem:

\(P(Y | C) = \frac{\frac{1}{48}}{\frac{7}{48}} = \frac{1}{7}\)

Let the event that the first digit is 5 be denoted as \(5_1\) and the event that the second digit is not 5 be denoted as \(\overline{5}_2\).

The probability that the first digit is 5 and the second digit is not 5 is:

\(P(5_1 \cap \overline{5}_2) = \frac{3}{9} \times \frac{6}{8} = \frac{1}{4}\)

The probability that the second digit is not 5 is:

\(P(\overline{5}_2) = \frac{1}{4} + \frac{6}{9} \times \frac{5}{8} = \frac{48}{72} = 0.6666\)

The conditional probability that the first digit is 5 given that the second digit is not 5 is:

\(P(5_1 | \overline{5}_2) = \frac{\frac{1}{4}}{\frac{48}{72}} = \frac{3}{8} = 0.375\)

(i) The probability that Mohit goes to the cinema is 0.35. The probability that he spends less than $50 at the cinema is 1 - 0.8 = 0.2. Therefore, the probability that Mohit will go to the cinema and spend less than $50 is:

\(P(C \cap <50) = 0.35 \times 0.2 = 0.07\)

(ii) We need to find \(P(C \mid <50)\), the probability that Mohit went to the cinema given that he spent less than $50. This is given by:

\(P(C \mid <50) = \frac{P(C \cap <50)}{P(<50)}\)

We already found \(P(C \cap <50) = 0.07\).

Now, calculate \(P(<50)\):

\(P(<50) = P(S \cap <50) + P(C \cap <50) + P(H \cap <50)\)

\(P(S \cap <50) = 0.25 \times (1 - 0.7) = 0.25 \times 0.3 = 0.075\)

\(P(C \cap <50) = 0.07\)

\(P(H \cap <50) = 0.4 \times 1 = 0.4\)

Thus, \(P(<50) = 0.075 + 0.07 + 0.4 = 0.545\)

Therefore, \(P(C \mid <50) = \frac{0.07}{0.545} = 0.128\) (or \(\frac{14}{109}\))

Let \(M\) be the event that Mumbok is chosen, and \(18{-}60\) be the event that a person aged between 18 and 60 is chosen.

The probability of choosing Mumbok and a person aged 18 to 60 is:

\(P(M \cap 18{-}60) = 0.6 \times \frac{55}{90} = 0.367 \text{ or } \frac{11}{30}\)

The probability of choosing a person aged 18 to 60 from either town is:

\(P(18{-}60) = 0.6 \times \frac{55}{90} + 0.4 \times \frac{95}{160} = \frac{29}{48} \text{ or } 0.604\)

Using Bayes' theorem, the probability that the town chosen was Mumbok given that the person is aged 18 to 60 is:

\(P(M \mid 18{-}60) = \frac{P(M \cap 18{-}60)}{P(18{-}60)} = \frac{\frac{11}{30}}{\frac{29}{48}} = \frac{88}{145} \text{ or } 0.607\)

Let \(W_1\) be the event that John wins his first game and \(W_2\) be the event that he wins his second game.

The probability that John wins his second game, \(P(W_2)\), can be calculated as:

\(P(W_2) = P(W_1 \cap W_2) + P(L_1 \cap W_2)\)

\(= 0.3 \times 0.6 + 0.7 \times 0.15\)

\(= 0.18 + 0.105\)

\(= 0.285\)

We need to find \(P(W_1 | W_2)\), the probability that John won his first game given that he won his second game. This is given by:

\(P(W_1 | W_2) = \frac{P(W_1 \cap W_2)}{P(W_2)}\)

\(= \frac{0.18}{0.285}\)

\(= 0.632\)

Thus, the probability that John won his first game given that he won his second game is \(0.632\) or \(\frac{12}{19}\).

(a) The probability that Tom removes a red disc on his first turn can be calculated by considering two scenarios: Sam removes a red disc first, then Tom removes a red disc, or Sam removes a white disc first, then Tom removes a red disc.

Probability of Sam removing a red disc first: \(\frac{3}{8}\)

Probability of Tom removing a red disc after Sam: \(\frac{2}{7}\)

Probability of Sam removing a white disc first: \(\frac{5}{8}\)

Probability of Tom removing a red disc after Sam: \(\frac{3}{7}\)

Total probability: \(\frac{3}{8} \times \frac{2}{7} + \frac{5}{8} \times \frac{3}{7} = \frac{21}{56} = \frac{3}{8}\)

(b) Tom wins on his second turn if the sequence of draws is either RRWR, WRRR, or WRWR.

Probability of RRWR: \(\frac{3}{8} \times \frac{2}{7} \times \frac{5}{6} \times \frac{1}{5} = \frac{1}{56}\)

Probability of WRRR: \(\frac{5}{8} \times \frac{3}{7} \times \frac{2}{6} \times \frac{1}{5} = \frac{1}{56}\)

Probability of WRWR: \(\frac{5}{8} \times \frac{3}{7} \times \frac{2}{6} \times \frac{3}{5} = \frac{1}{14}\)

Total probability: \(\frac{1}{56} + \frac{1}{56} + \frac{1}{14} = \frac{3}{28}\)

(c) Given that Tom wins on his second turn, we need to find the probability that Sam removes a red disc on his first turn.

Probability of Sam removing a red disc first: \(\frac{3}{8}\)

Probability of Tom winning on his second turn given Sam removes a red disc first: \(\frac{1}{56}\)

Total probability of Tom winning on his second turn: \(\frac{3}{28}\)

Conditional probability: \(\frac{\frac{1}{56}}{\frac{3}{28}} = \frac{1}{6}\)

(i) The probability of taking a toffee from Susan's bag is \(\frac{5}{12}\). After placing it in Ahmad's bag, the probability of taking a boiled sweet from Ahmad's bag is \(\frac{2}{10}\). Therefore, the probability that the two sweets taken are a toffee from Susan’s bag and a boiled sweet from Ahmad’s bag is:

\(P(T, B) = \frac{5}{12} \times \frac{2}{10} = \frac{1}{12}\)

(ii) Let \(C_S\) be the event that a chocolate is taken from Susan's bag, and \(C_A\) be the event that a chocolate is taken from Ahmad's bag. We need to find \(P(C_S | C_A)\).

First, calculate \(P(C_S \cap C_A)\):

\(P(C_S \cap C_A) = \frac{7}{12} \times \frac{4}{10} = \frac{28}{120}\)

Next, calculate \(P(C_A)\):

\(P(C_A) = \frac{7}{12} \times \frac{4}{10} + \frac{5}{12} \times \frac{3}{10} = \frac{28}{120} + \frac{15}{120} = \frac{43}{120}\)

Finally, calculate \(P(C_S | C_A)\):

\(P(C_S | C_A) = \frac{P(C_S \cap C_A)}{P(C_A)} = \frac{28/120}{43/120} = \frac{28}{43}\)

(i) The total number of balls in box B after transferring one ball from box A is \(5 + x + 1 = x + 6\). Therefore, the probability of choosing a yellow ball from box B is \(\frac{x}{x+6}\).

(ii) The tree diagram is completed as follows:
From box A to box B:
- White to White: \(\frac{6}{x+6}\)
- White to Yellow: \(\frac{x}{x+6}\)
- Yellow to White: \(\frac{5}{x+6}\)
- Yellow to Yellow: \(\frac{x+1}{x+6}\).

(iii) Given that the probability of choosing a white ball from box B after a white ball from box A is \(\frac{1}{3}\), we have:
\(\frac{6}{x+6} = \frac{1}{3}\)
Solving for \(x\):
\(6 = \frac{x+6}{3}\)
\(18 = x + 6\)
\(x = 12\).

(iv) To find the conditional probability that the ball chosen from box A was yellow given that the ball chosen from box B is yellow, we use:
\(P(A_Y | B_Y) = \frac{P(A_Y \cap B_Y)}{P(B_Y)}\)
\(P(B_Y) = \frac{8}{10} \times \frac{12}{18} + \frac{2}{10} \times \frac{13}{18} = \frac{61}{90}\)
\(P(A_Y \cap B_Y) = \frac{2}{10} \times \frac{13}{18} = \frac{13}{90}\)
\(P(A_Y | B_Y) = \frac{13/90}{61/90} = \frac{13}{61}\).

(i) The probability that Fabio chooses Americano and leaves it to drink later is given by the product of the probability of choosing Americano and the probability of leaving it to drink later. This is calculated as:

\(P(A \text{ Later}) = 0.5 \times 0.2 = 0.1\)

(ii) To find the probability that Fabio chose Latte given that he drinks his coffee immediately, we use Bayes' theorem. Let \(L\) be the event that he chooses Latte, and \(I\) be the event that he drinks immediately. We need to find \(P(L \mid I)\).

First, calculate \(P(I)\), the total probability of drinking immediately:

\(P(I) = (0.5 \times 0.8) + (0.3 \times 0.6) + (0.2 \times 0.1) = 0.4 + 0.18 + 0.02 = 0.6\)

Now, calculate \(P(L \cap I)\), the probability that he chooses Latte and drinks immediately:

\(P(L \cap I) = 0.2 \times 0.1 = 0.02\)

Finally, apply Bayes' theorem:

\(P(L \mid I) = \frac{P(L \cap I)}{P(I)} = \frac{0.02}{0.6} = 0.0333\)

Let \(E\), \(L\), and \(OT\) represent the events that Ana is early, late, and on time, respectively. Let \(B\) represent the event that Ana eats a banana, and \(\overline{B}\) represent the event that she does not eat a banana.

We need to find \(P(OT | \overline{B})\).

Using the law of total probability, we have:

\(P(\overline{B}) = P(\overline{B} | E)P(E) + P(\overline{B} | L)P(L) + P(\overline{B} | OT)P(OT)\)

Given:

• \(P(E) = 0.05\)
• \(P(L) = 0.75\)
• \(P(OT) = 1 - P(E) - P(L) = 0.2\)
• \(P(\overline{B} | E) = 1 - 0.7 = 0.3\)
• \(P(\overline{B} | L) = 1\)
• \(P(\overline{B} | OT) = 1 - 0.4 = 0.6\)

Substitute these values into the equation:

\(P(\overline{B}) = 0.3 \times 0.05 + 1 \times 0.75 + 0.6 \times 0.2\)

\(P(\overline{B}) = 0.015 + 0.75 + 0.12 = 0.885\)

Now, use Bayes' theorem to find \(P(OT | \overline{B})\):

\(P(OT | \overline{B}) = \frac{P(\overline{B} | OT)P(OT)}{P(\overline{B})} = \frac{0.6 \times 0.2}{0.885}\)

\(P(OT | \overline{B}) = \frac{0.12}{0.885} \approx 0.136\)

Thus, the probability that Ana is on time given that she does not eat a banana is \(0.136 \left( \frac{8}{59} \right)\).

(i) The probability that the radio is set to station 3 and the presenter is male is given by:

\(0.25p = 0.075\)

Solving for \(p\):

\(p = \frac{0.075}{0.25} = 0.3\)

(ii) To find the probability that the radio was set to station 2 given that Maria hears a male presenter, we use conditional probability:

\(P(2|M) = \frac{P(2 \text{ and } M)}{P(M)}\)

Where:

\(P(2 \text{ and } M) = 0.45 \times 0.85 = 0.3825\)

\(P(M) = 0.3 \times 0.1 + 0.45 \times 0.85 + 0.25 \times 0.3 = 0.4875\)

Thus,

\(P(2|M) = \frac{0.3825}{0.4875} = 0.785\)

Let the event that the pen is in the pencil case be denoted as \(P(C) = 0.7\), and the event that Ted finds the pen given it is in the pencil case be \(P(F|C) = 1\). The event that the pen is not in the pencil case is \(P(C') = 0.3\), and the probability that Ted finds the pen given it is not in the pencil case is \(P(F|C') = 0.2\).

We need to find \(P(C|F)\), the probability that the pen is in the pencil case given that Ted finds it. Using Bayes' theorem:

\(P(C|F) = \frac{P(C \cap F)}{P(F)}\)

Where:

\(P(C \cap F) = P(C) \cdot P(F|C) = 0.7 \times 1 = 0.7\)

\(P(F) = P(C \cap F) + P(C' \cap F) = 0.7 + (0.3 \times 0.2) = 0.7 + 0.06 = 0.76\)

Thus:

\(P(C|F) = \frac{0.7}{0.76} = 0.921\)

(i) The tree diagram is structured as follows:

• First branch: Probability of using a cell phone (Ph) = 0.68, not using a cell phone (NPh) = 0.32.
• Second branch for Ph: Young (Y) = 0.7, Middle-aged (M) = 0.25, Old (O) = 0.05.
• Second branch for NPh: Young (Y) = 0.26, Middle-aged (M) = 0.10, Old (O) = 0.64.

(ii) To find the probability that a passenger aged 45 (middle-aged) uses a cell phone, use conditional probability:

\(P(\text{Ph} \mid \text{M}) = \frac{P(\text{Ph} \cap \text{M})}{P(\text{M})}\)

Where:

• \(P(\text{Ph} \cap \text{M}) = 0.68 \times 0.25 = 0.17\)
• \(P(\text{M}) = (0.68 \times 0.25) + (0.32 \times 0.10) = 0.17 + 0.032 = 0.202\)

Thus,

\(P(\text{Ph} \mid \text{M}) = \frac{0.17}{0.202} \approx 0.842\)

To find the probability that the spinner landed on an even-numbered side given that Raj's score is 12, we use conditional probability:

Let \(E\) be the event that the spinner lands on an even-numbered side. We need to find \(P(E \mid 12)\).

First, calculate \(P(E \text{ and } 12)\):

The probability of landing on an even-numbered side is \(\frac{2}{5}\) (since there are 2 even numbers: 2 and 4).

The possible dice outcomes that multiply to 12 are: (2,6), (6,2), (3,4), (4,3). Each pair has a probability of \(\frac{1}{36}\) (since there are 36 possible outcomes when rolling two dice).

Thus, \(P(E \text{ and } 12) = \frac{2}{5} \times \frac{4}{36} = \frac{8}{180} = \frac{2}{45}\).

Next, calculate \(P(12)\):

For an odd-numbered side, the possible dice outcomes that add to 12 are: (6,6). The probability of this is \(\frac{1}{36}\).

The probability of landing on an odd-numbered side is \(\frac{3}{5}\) (since there are 3 odd numbers: 1, 3, 5).

Thus, \(P(12) = \frac{3}{5} \times \frac{1}{36} + \frac{8}{180} = \frac{11}{180}\).

Finally, use the conditional probability formula:

\(P(E \mid 12) = \frac{P(E \text{ and } 12)}{P(12)} = \frac{\frac{8}{180}}{\frac{11}{180}} = \frac{8}{11}\).

Let \(D\) be the event that the dog chases ducks, and \(G\) be the event that the dog chases geese. Let \(A\) be the event that the dog is attacked, and \(NA\) be the event that the dog is not attacked.

We need to find \(P(G \mid NA)\), which is given by:

\(P(G \mid NA) = \frac{P(G \cap NA)}{P(NA)}\)

First, calculate \(P(G \cap NA)\):

\(P(G \cap NA) = P(G) \times P(NA \mid G) = \frac{2}{5} \times \left(1 - \frac{3}{4}\right) = \frac{2}{5} \times \frac{1}{4} = \frac{1}{10}\)

Next, calculate \(P(NA)\):

\(P(NA) = P(D \cap NA) + P(G \cap NA)\)

\(P(D \cap NA) = P(D) \times P(NA \mid D) = \frac{3}{5} \times \left(1 - \frac{1}{10}\right) = \frac{3}{5} \times \frac{9}{10} = \frac{27}{50}\)

Thus,

\(P(NA) = \frac{27}{50} + \frac{1}{10} = \frac{27}{50} + \frac{5}{50} = \frac{32}{50}\)

Finally, substitute back to find \(P(G \mid NA)\):

\(P(G \mid NA) = \frac{\frac{1}{10}}{\frac{32}{50}} = \frac{1}{10} \times \frac{50}{32} = \frac{5}{32}\)

(i) The probability that the first question is answered correctly is calculated as follows:

\(P(\text{1st correct}) = P(\text{Peter correct}) + P(\text{Peter asks for help and audience correct})\)

\(= 0.7 + 0.2 \times 0.95 = 0.89\)

(ii) To find the probability that the first two questions are both answered correctly, consider the following scenarios:

• Peter answers both questions correctly: \(P(CC) = 0.7 \times 0.7 = 0.49\)
• Peter answers the first correctly, asks for help, and the audience answers correctly: \(P(CHA) = 0.7 \times 0.2 \times 0.95 = 0.133\)
• Peter asks for help on the first, audience answers correctly, and Peter answers the second correctly: \(P(HAC) = 0.2 \times 0.95 \times 0.7 = 0.133\)
• Peter asks for help on the first, audience answers correctly, Peter asks for help on the second, and friend answers correctly: \(P(HAHP) = 0.2 \times 0.95 \times 0.2 \times 0.65 = 0.0247\)

Summing these probabilities gives:

\(P(\text{both correctly answered}) = 0.49 + 0.133 + 0.133 + 0.0247 = 0.781\)

(iii) Given that the first two questions were both answered correctly, the probability that Peter asked the audience is:

\(P(\text{audience} \mid \text{both correct}) = \frac{P(CHA) + P(HAC) + P(HAHP)}{P(\text{both correctly answered})}\)

\(= \frac{0.133 + 0.133 + 0.0247}{0.781} = \frac{0.2907}{0.781} = 0.372\)

(a) The total probability that Kino is late is given by:

\(0.2x + 0.1 \times 2x + 0.7 \times 0.25 = 0.235\)

Simplifying, we have:

\(0.2x + 0.2x + 0.175 = 0.235\)

\(0.4x + 0.175 = 0.235\)

Subtract 0.175 from both sides:

\(0.4x = 0.06\)

Divide by 0.4:

\(x = 0.15\)

(b) We need to find \(P(\text{car} | \text{not late})\).

\(P(\text{car} | \text{not late}) = \frac{P(\text{car and not late})}{P(\text{not late})}\)

\(P(\text{car and not late}) = 0.1 \times (1 - 2x) = 0.1 \times (1 - 0.3) = 0.1 \times 0.7 = 0.07\)

\(P(\text{not late}) = 1 - 0.235 = 0.765\)

Thus,

\(P(\text{car} | \text{not late}) = \frac{0.07}{0.765} = 0.0915\)

Expressed as a fraction, \(\frac{0.07}{0.765} = \frac{70}{765} = \frac{14}{153}\).

(i) The tree diagram is structured as follows:

• First branch: Toast (T) with probability 0.85, Bread roll (B) with probability 0.15.
• Second branch from Toast: Jam (J) with probability 0.8, No Jam (NJ) with probability 0.2.
• Second branch from Bread roll: Jam (J) with probability 0.4, No Jam (NJ) with probability 0.6.

(ii) To find the probability that Maria had toast given she did not have jam, use Bayes' theorem:

\(P(T \mid NJ) = \frac{P(T \text{ and } NJ)}{P(NJ)}\)

Calculate \(P(T \text{ and } NJ)\):

\(P(T \text{ and } NJ) = 0.85 \times 0.2 = 0.17\)

Calculate \(P(NJ)\):

\(P(NJ) = P(T \text{ and } NJ) + P(B \text{ and } NJ) = 0.85 \times 0.2 + 0.15 \times 0.6 = 0.17 + 0.09 = 0.26\)

Substitute back into Bayes' theorem:

\(P(T \mid NJ) = \frac{0.17}{0.26} = \frac{17}{26} \approx 0.654\)

(i) The probability of selecting 2 pears and 1 orange can be calculated using combinations:

\(\frac{{^4C_2 \times ^7C_1}}{{^{11}C_3}} = \frac{6 \times 7}{165} = 0.255\)

Alternatively, using probabilities:

\(\frac{4}{11} \times \frac{3}{10} \times \frac{7}{9} \times 3 = 0.255\)

(ii) The probability that the third fruit is an orange can be calculated by considering different sequences:

\(P(O) = P(P, P, O) + P(P, O, O) + P(O, P, O) + P(O, O, O)\)

\(= \frac{4}{11} \times \frac{3}{10} \times \frac{7}{9} + \frac{4}{11} \times \frac{7}{10} \times \frac{6}{9} + \frac{7}{11} \times \frac{4}{10} \times \frac{6}{9} + \frac{7}{11} \times \frac{6}{10} \times \frac{5}{9}\)

\(= \frac{14}{165} + \frac{28}{165} + \frac{28}{165} + \frac{7}{33} = \frac{7}{11}\)

(iii) The probability that the first fruit was a pear, given that the third fruit is an orange, is calculated using conditional probability:

\(P(P | O) = \frac{P(P \cap O)}{P(O)}\)

\(= \frac{P(P, P, O) + P(P, O, O)}{P(O)}\)

\(= \frac{\frac{4}{11} \times \frac{3}{10} \times \frac{7}{9} + \frac{4}{11} \times \frac{7}{10} \times \frac{6}{9}}{\frac{7}{11}}\)

\(= \frac{28/110 + 28/70}{7/11} = 0.4\)

(i) The probability of getting 2 heads (HH) is \(\frac{1}{4}\), so \(P(E) = \frac{1}{4}\). The probability of getting 2 tails (TT) is \(\frac{1}{4}\), so \(P(C) = \frac{1}{4}\). The probability of getting 1 head and 1 tail (HT or TH) is \(\frac{1}{2}\), so \(P(JT) = \frac{1}{2}\).

(ii) The tree diagram is as follows:

(iii) The probability that Ravi is frightened is calculated by summing the probabilities of being frightened on each ride:

\(P(F) = \left( \frac{1}{4} \times \frac{6}{10} \right) + \left( \frac{1}{4} \times \frac{7}{10} \right) + \left( \frac{1}{2} \times \frac{8}{10} \right) = \frac{29}{40}\) or 0.725.

(iv) Given that Ravi is not frightened, the probability that he went on the camel ride is:

\(P(C \mid NF) = \frac{P(C \cap NF)}{P(NF)} = \frac{\frac{3}{40}}{1 - \frac{29}{40}} = \frac{3}{11}\) or 0.273.

(i) The tree diagram is drawn with branches for stopping (S) and not stopping (NS) at each set of lights. The probabilities are labeled as follows: first light (S: 0.4, NS: 0.6), second light (S: 0.8, NS: 0.2), third light (S: 0.3, NS: 0.7).

(ii) The probability that Karinne stops at the first two sets but not the third is calculated as:

\(P(S, S, NS) = 0.4 \times 0.8 \times 0.7 = 0.224\) or \(\frac{28}{125}\).

(iii) The probability that Karinne stops at exactly two sets of lights is the sum of the probabilities of the following scenarios:

\(P(S, S, NS) + P(S, NS, S) + P(NS, S, S)\).

Calculating each:

\(P(S, S, NS) = 0.4 \times 0.8 \times 0.7 = 0.224\)

\(P(S, NS, S) = 0.4 \times 0.2 \times 0.3 = 0.024\)

\(P(NS, S, S) = 0.6 \times 0.8 \times 0.3 = 0.144\)

Summing these gives:

\(0.224 + 0.024 + 0.144 = 0.392\) or \(\frac{49}{125}\).

(iv) The probability that Karinne stops at the first set of lights, given that she stops at exactly two sets, is calculated using conditional probability:

\(P(S \text{ at first light} | \text{stops at exactly 2 lights}) = \frac{P(S, NS, S) + P(S, S, NS)}{0.392}\).

\(= \frac{0.024 + 0.224}{0.392} = \frac{0.248}{0.392} = 0.633\) or \(\frac{31}{49}\).

(i) The probability that the person chosen is from country A is given by the ratio of the number of people in country A to the total number of people:

\(P(A) = \frac{3}{15} = 0.2\)

(ii) To find the probability that the person chosen does not have sugar in their tea, we calculate the probabilities for both countries and sum them:

In country A, 70% do not have sugar: \(0.7 \times 3 \text{ million} = 2.1 \text{ million}\)

In country B, 35% do not have sugar: \(0.35 \times 12 \text{ million} = 4.2 \text{ million}\)

Total without sugar: \(2.1 + 4.2 = 6.3 \text{ million}\)

Probability: \(P(\text{not S}) = \frac{6.3}{15} = 0.42\)

(iii) Given that the person does not have sugar, the probability they are from country B is:

\(P(B | \text{not S}) = \frac{0.8 \times 0.35}{0.42} = 0.667\)

(i) The probability of taking a white paper clip from box A is \(\frac{1}{6}\). After transferring this white paper clip to box B, box B will have 7 red and 3 white paper clips. The probability of then taking a red paper clip from box B is \(\frac{7}{10}\). Therefore, the probability of both events occurring is:

\(P(W, R) = \frac{1}{6} \times \frac{7}{10} = \frac{7}{60} \approx 0.117\)

(ii) The probability of taking a red paper clip from box A is \(\frac{5}{6}\). After transferring this red paper clip to box B, box B will have 8 red and 2 white paper clips. The probability of then taking a red paper clip from box B is \(\frac{8}{10}\). Therefore, the probability of both events occurring is:

\(P(R, R) = \frac{5}{6} \times \frac{8}{10} = \frac{40}{60}\)

The total probability of taking a red paper clip from box B is the sum of the probabilities of the two mutually exclusive events (taking a red from A and then a red from B, or taking a white from A and then a red from B):

\(P(\text{red}) = \frac{40}{60} + \frac{7}{60} = \frac{47}{60} \approx 0.783\)

(iii) The probability that the paper clip taken from box A was red, given that the paper clip from box B is red, is calculated using conditional probability:

\(P(R | R) = \frac{P(R \cap R)}{P(R)} = \frac{\frac{40}{60}}{\frac{47}{60}} = \frac{40}{47} \approx 0.851\)

Let \(A\) be the event that Jamie attends the training session, and \(T\) be the event that Jamie is chosen for the team.

Given: \(P(A) = 0.5\), \(P(T|A) = 1\), \(P(T|A^c) = 0.6\).

(i) The probability that Jamie is chosen for the team is:

\(P(T) = P(T|A)P(A) + P(T|A^c)P(A^c) = 1 \times 0.5 + 0.6 \times 0.5 = 0.5 + 0.3 = 0.8\)

(ii) The conditional probability that Jamie attended the training session, given that he was chosen for the team, is:

\(P(A|T) = \frac{P(T|A)P(A)}{P(T)} = \frac{1 \times 0.5}{0.8} = \frac{0.5}{0.8} = 0.625\)

Thus, \(P(A|T) = 0.625\) or \(\frac{5}{8}\).

(i) To find the probability that a taxi arrives late, we use the law of total probability. Let L be the event that a taxi arrives late. We have:

\(P(L) = P(A) \cdot P(L|A) + P(B) \cdot P(L|B) + P(C) \cdot P(L|C)\)

Substituting the given probabilities:

\(P(L) = 0.5 \times 0.04 + 0.3 \times 0.06 + 0.2 \times 0.17\)

\(P(L) = 0.02 + 0.018 + 0.034 = 0.072\)

Thus, the probability that a taxi arrives late is 0.072.

(ii) To find the conditional probability that Andrea rang company B given that the taxi arrives late, we use Bayes' theorem:

\(P(B|L) = \frac{P(B) \cdot P(L|B)}{P(L)}\)

Substituting the known values:

\(P(B|L) = \frac{0.3 \times 0.06}{0.072}\)

\(P(B|L) = \frac{0.018}{0.072} = 0.25\)

Thus, the conditional probability that she rang company B is 0.25.

(i) The tree diagram is constructed as follows:

• First serve in: Probability = 0.65
• Win: Probability = 0.9
• Lose: Probability = 0.1
• First serve out: Probability = 0.35
• Second serve in: Probability = 0.8
• Win: Probability = 0.6
• Lose: Probability = 0.4
• Second serve out: Probability = 0.2
• Lose: Probability = 1

(ii) The probability that Don loses the point is calculated by considering all paths that lead to losing:

\(P(\text{Lose}) = 0.65 \times 0.1 + 0.35 \times 0.8 \times 0.4 + 0.35 \times 0.2 \times 1\)

\(= 0.065 + 0.112 + 0.07 = 0.247\)

(iii) The conditional probability that Don’s first serve went into the correct area, given that he loses the point, is:

\(P(\text{First in | Lose}) = \frac{0.65 \times 0.1}{0.247}\)

\(= \frac{0.065}{0.247} \approx 0.263\) or \(\frac{5}{19}\)

Let:

• \(M\) = event that a person is male, \(P(M) = 0.54\)

• \(F\) = event that a person is female, \(P(F) = 0.46\)

• \(C\) = event that a person is colour-blind

We know:

• \(P(C|M) = 0.05\)

• \(P(C|F) = 0.02\)

We need to find \(P(M|C)\), the probability that a person is male given that they are colour-blind.

Using Bayes' theorem:

\(P(M|C) = \frac{P(C|M) \cdot P(M)}{P(C)}\)

First, calculate \(P(C)\):

\(P(C) = P(C \cap M) + P(C \cap F)\)

\(P(C \cap M) = P(C|M) \cdot P(M) = 0.05 \times 0.54 = 0.027\)

\(P(C \cap F) = P(C|F) \cdot P(F) = 0.02 \times 0.46 = 0.0092\)

\(P(C) = 0.027 + 0.0092 = 0.0362\)

Now, substitute back into Bayes' theorem:

\(P(M|C) = \frac{0.027}{0.0362} \approx 0.746\)

Thus, the probability that the person is male given that they are colour-blind is approximately 0.746, or \(\frac{135}{181}\).

(a) The tree diagram is as follows:

1st Jump: Success (S) with probability 0.2, Failure (F) with probability 0.8.

2nd Jump: If 1st was S, then S with 0.3, F with 0.7; if 1st was F, then S with 0.1, F with 0.9.

3rd Jump: If 2nd was S, then S with 0.3, F with 0.7; if 2nd was F, then S with 0.1, F with 0.9.

(b) Calculate the probability of exactly one success:

- SFF: \(0.2 \times 0.7 \times 0.9 = 0.126\)

- FSF: \(0.8 \times 0.1 \times 0.7 = 0.056\)

- FFS: \(0.8 \times 0.9 \times 0.1 = 0.072\)

Total probability of exactly one success: \(0.126 + 0.056 + 0.072 = 0.254\)

Probability of at least one success: \(1 - 0.8 \times 0.9 \times 0.9 = 0.352\)

Probability of exactly one success given at least one success: \(\frac{0.254}{0.352} = \frac{127}{176}\)

(c) Calculate the probability for six jumps:

- Only first three are successes: \(0.2 \times 0.3 \times 0.3 \times 0.7 \times 0.9 \times 0.9 = 0.010206\)

- Only last three are successes: \(0.8 \times 0.9 \times 0.9 \times 0.1 \times 0.3 \times 0.3 = 0.005832\)

Total probability: \(0.010206 + 0.005832 = 0.016038\)

(i) To find the probability of choosing a grandparent, calculate the total number of grandparents and divide by the total number of people. There are 6 grandparents (2 in House 1, 3 in House 2, 1 in House 3) and 16 people in total (7 in House 1, 7 in House 2, 2 in House 3). Thus, the probability is \(\frac{6}{16} = \frac{3}{8}\).

(ii) To find the probability that a person chosen is a grandparent, calculate the probability for each house and sum them. The probability for House 1 is \(\frac{1}{3} \times \frac{2}{7}\), for House 2 is \(\frac{1}{3} \times \frac{3}{7}\), and for House 3 is \(\frac{1}{3} \times \frac{1}{2}\). Therefore, the total probability is \(\frac{1}{3} \times \frac{2}{7} + \frac{1}{3} \times \frac{3}{7} + \frac{1}{3} \times \frac{1}{2} = \frac{17}{42}\).

(iii) Given that the person chosen is a grandparent, calculate the probability that there is also a parent in the house. For House 1, the probability is \(\frac{2}{21}\), and for House 2, it is \(\frac{3}{21}\). Thus, the probability is \(\frac{2}{21} + \frac{3}{21} = \frac{5}{21}\). Divide by the probability from part (ii), \(\frac{17}{42}\), to get \(\frac{5}{21} \div \frac{17}{42} = \frac{10}{17}\).

(i) To find the conditional probability that Rachel wins the first game given that she loses the second, we use the formula for conditional probability:

\(P(W_1 | L_2) = \frac{P(W_1 \cap L_2)}{P(L_2)}\)

First, calculate \(P(W_1 \cap L_2)\):

\(P(W_1 \cap L_2) = P(W_1) \times P(L_2 | W_1) = 0.6 \times 0.3 = 0.18\)

Next, calculate \(P(L_2)\):

\(P(L_2) = P(W_1 \cap L_2) + P(L_1 \cap L_2) = (0.6 \times 0.3) + (0.4 \times 0.6) = 0.18 + 0.24 = 0.42\)

Thus, \(P(W_1 | L_2) = \frac{0.18}{0.42} = 0.429\).

(ii) To find the probability that Rachel wins 2 games and loses 1 game out of the first three games, consider the following scenarios:

• \(P(W_1, W_2, L_3) = 0.6 \times 0.7 \times 0.3 = 0.126\)
• \(P(W_1, L_2, W_3) = 0.6 \times 0.3 \times 0.4 = 0.072\)
• \(P(L_1, W_2, W_3) = 0.4 \times 0.4 \times 0.7 = 0.112\)

Summing these probabilities gives:

\(0.126 + 0.072 + 0.112 = 0.31\).

(a) To find the probability that all 3 eggs chosen contain the same colour sweet, consider the combinations:

- All red: \(\frac{3}{12} \times \frac{2}{11} \times \frac{1}{10} = \frac{6}{1320}\)

- All orange: \(\frac{4}{12} \times \frac{3}{11} \times \frac{2}{10} = \frac{24}{1320}\)

- All yellow: \(\frac{5}{12} \times \frac{4}{11} \times \frac{3}{10} = \frac{60}{1320}\)

Total probability: \(\frac{6 + 24 + 60}{1320} = \frac{90}{1320} = 0.0682\)

(b) Given that all three children have the same colour sweet, the probability that all 3 eggs contain a yellow sweet is:

\(\frac{\frac{60}{1320}}{\frac{90}{1320}} = \frac{60}{90} = \frac{2}{3}\) or 0.667

(c) To find the probability that at least one child chooses an egg with an orange sweet, use the complement rule:

Probability of no orange sweets: \(\frac{8}{12} \times \frac{7}{11} \times \frac{6}{10} = \frac{336}{1320} = \frac{14}{55}\)

Probability of at least one orange sweet: \(1 - \frac{14}{55} = \frac{41}{55}\) or 0.745

(a) The probability that Janice moves on to level 2 is calculated as follows:

She can complete level 1 on the first attempt with probability 0.6, or fail the first attempt and complete it on the second attempt with probability \(0.4 \times 0.3 = 0.12\).

Thus, the probability she moves on to level 2 is \(0.6 + 0.12 = 0.72\).

(b) The probability that Janice finishes the game is calculated by considering the successful paths:

• Completes level 1 on the first attempt and level 2 on the first attempt: \(0.6 \times 0.4 = 0.24\)
• Completes level 1 on the first attempt and level 2 on the second attempt: \(0.6 \times 0.6 \times 0.2 = 0.072\)
• Completes level 1 on the second attempt and level 2 on the first attempt: \(0.4 \times 0.3 \times 0.4 = 0.048\)
• Completes level 1 on the second attempt and level 2 on the second attempt: \(0.4 \times 0.3 \times 0.6 \times 0.2 = 0.0144\)

Summing these probabilities gives \(0.24 + 0.072 + 0.048 + 0.0144 = 0.3744\).

(c) The probability that Janice fails exactly one attempt, given that she finishes the game, is:

Calculate the probability of failing exactly one attempt:

• Fails first attempt at level 1 and succeeds second attempt, then succeeds both attempts at level 2: \(0.4 \times 0.3 \times 0.4 = 0.048\)
• Succeeds first attempt at level 1, fails first attempt at level 2, and succeeds second attempt: \(0.6 \times 0.6 \times 0.2 = 0.072\)

The total probability of failing exactly one attempt is \(0.048 + 0.072 = 0.12\).

The required probability is \(\frac{0.12}{0.3744} = 0.321\) or \(\frac{25}{78}\).