Past Exam Questions

(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)

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