Past Exam Questions

The probability of choosing a plum from Jameel's box is \(\frac{5}{8}\) because there are 5 plums out of a total of 8 fruits.

The probability of choosing a plum from Rosa's box is \(\frac{x}{x+6}\) because there are x plums out of a total of \(x + 6\) fruits.

The probability that both fruits chosen are plums is given by:

\(\frac{5}{8} \times \frac{x}{x+6} = \frac{1}{4}\)

To solve for x, multiply both sides by \(8(x+6)\):

\(5x = 2(x+6)\)

Expand and simplify:

\(5x = 2x + 12\)

Subtract \(2x\) from both sides:

\(3x = 12\)

Divide both sides by 3:

\(x = 4\)

(i) Total number of selections = \(\binom{12}{7} = 792\).

Selections with the boy included = \(\binom{11}{6} = 462\).

Probability = \(\frac{462}{792} = \frac{7}{12}\).

(ii) Method 1: Scenarios are:

• 2G + 5B: \(\binom{4}{2} \times \binom{8}{5} = 336\)
• 3G + 4B: \(\binom{4}{3} \times \binom{8}{4} = 280\)
• 4G + 3B: \(\binom{4}{4} \times \binom{8}{3} = 56\)

\(Total = 672.\)

Probability = \(\frac{672}{792} = \frac{28}{33}\).

Let \(C\) represent completing the puzzle and \(C'\) represent not completing it.

The probability that Kenny completes the puzzle on both days is \(P(C, C) = 0.8 \times 0.9 = 0.72\).

The probability that Kenny completes the puzzle on Monday but not on Tuesday is \(P(C, C') = 0.8 \times 0.1 = 0.08\).

The probability that Kenny does not complete the puzzle on Monday but completes it on Tuesday is \(P(C', C) = 0.2 \times 0.6 = 0.12\).

The probability that Kenny completes the puzzle on at least one of the two days is:

\(P(C, C) + P(C, C') + P(C', C) = 0.72 + 0.08 + 0.12 = 0.92\)

Alternatively, the probability that Kenny does not complete the puzzle on both days is \(P(C', C') = 0.2 \times 0.4 = 0.08\).

Thus, the probability that he completes the puzzle on at least one day is:

\(1 - P(C', C') = 1 - 0.08 = 0.92\)

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:

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