Past Exam Questions

(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 \(\frac{25}{78}\).

To determine if events A and B are independent, we need to check if \(P(A \cap B) = P(A) \times P(B)\).

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

The possible outcomes for the sum of two dice being less than 6 are: (1,1), (1,2), (1,3), (1,4), (2,1), (2,2), (2,3), (3,1), (3,2), (4,1). There are 10 outcomes.

Thus, \(P(A) = \frac{10}{36} = \frac{5}{18} \approx 0.278\).

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

The possible outcomes for the difference between the numbers being at most 2 are: (1,1), (1,2), (1,3), (2,1), (2,2), (2,3), (2,4), (3,1), (3,2), (3,3), (3,4), (3,5), (4,2), (4,3), (4,4), (4,5), (4,6), (5,3), (5,4), (5,5), (5,6), (6,4), (6,5), (6,6). There are 24 outcomes.

Thus, \(P(B) = \frac{24}{36} = \frac{2}{3} \approx 0.667\).

Now, calculate \(P(A \cap B)\):

The outcomes that satisfy both conditions are: (1,1), (1,2), (1,3), (2,1), (2,2), (3,1), (3,2), (4,1). There are 8 outcomes.

Thus, \(P(A \cap B) = \frac{8}{36} = \frac{2}{9} \approx 0.222\).

Check independence:

\(P(A) \times P(B) = \frac{5}{18} \times \frac{2}{3} = \frac{5}{27} \approx 0.185\).

Since \(P(A \cap B) \neq P(A) \times P(B)\), the events are not independent.

To determine if events X and Y are independent, we need to check if \(P(X \cap Y) = P(X) \times P(Y)\).

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

The possible outcomes for the total score to be 4 are (1,3), (2,2), and (3,1). Thus, there are 3 favorable outcomes.

Since there are 36 possible outcomes when a die is thrown twice, \(P(X) = \frac{3}{36}\).

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

The first score can be 2 or 5. For each of these, there are 6 possible outcomes for the second die. Thus, there are 12 favorable outcomes.

So, \(P(Y) = \frac{12}{36}\).

Now, calculate \(P(X \cap Y)\):

The outcomes that satisfy both X and Y are (2,2). Thus, there is 1 favorable outcome.

So, \(P(X \cap Y) = \frac{1}{36}\).

Check for independence:

\(P(X) \times P(Y) = \frac{3}{36} \times \frac{12}{36} = \frac{1}{36}\).

Since \(P(X \cap Y) = P(X) \times P(Y)\), events X and Y are independent.

(i) The total number of students studying Music is \(40 + 12 = 52\). The total number of students is 160. Therefore, the probability that a randomly chosen student is studying Music is \(\frac{52}{160} = \frac{13}{40}\) or 0.325.

(ii) To determine independence, we check if \(P(\text{boy}) \times P(\text{Music}) = P(\text{boy and Music})\).

\(P(\text{boy}) = \frac{96}{160}\) since there are \(24 + 40 + 32 = 96\) boys.

\(P(\text{Music}) = \frac{52}{160}\) as calculated earlier.

\(P(\text{boy and Music}) = \frac{40}{160}\) since there are 40 boys studying Music.

Now, \(\frac{96}{160} \times \frac{52}{160} \neq \frac{40}{160}\), so the events are not independent.

(i) Let \(M\) be the event that a student is male, \(F\) be the event that a student is female, and \(H\) be the event that a student likes their curry hot.

We have \(P(M) = \frac{3}{4}\) and \(P(F) = \frac{1}{4}\).

The probability that a male student likes their curry hot is \(P(H|M) = \frac{3}{5}\), so \(P(M \cap H) = P(M) \cdot P(H|M) = \frac{3}{4} \cdot \frac{3}{5} = \frac{9}{20}\).

The probability that a female student likes their curry hot is \(P(H|F) = \frac{4}{5}\), so \(P(F \cap H) = P(F) \cdot P(H|F) = \frac{1}{4} \cdot \frac{4}{5} = \frac{1}{5} = \frac{4}{20}\).

The probability that a student likes their curry hot is \(P(H) = P(M \cap H) + P(F \cap H) = \frac{9}{20} + \frac{4}{20} = \frac{13}{20}\).

The probability that the student is either female, or likes their curry hot, or both is \(P(F \cup H) = P(F) + P(H) - P(F \cap H) = \frac{1}{4} + \frac{13}{20} - \frac{4}{20} = \frac{5}{20} + \frac{13}{20} - \frac{4}{20} = \frac{14}{20} = \frac{7}{10}\).

(ii) To determine independence, check if \(P(M \cap H) = P(M) \cdot P(H)\).

We have \(P(M \cap H) = \frac{9}{20}\) and \(P(M) \cdot P(H) = \frac{3}{4} \cdot \frac{13}{20} = \frac{39}{80}\).

Since \(\frac{9}{20} \neq \frac{39}{80}\), the events are not independent.

(i) The number of silver estate cars is 10. The total number of cars is 160. The probability is \(\frac{10}{160} = \frac{1}{16} = 0.0625\).

(ii) The total number of hatchback cars is \(40 + 26 + 24 = 90\). The probability is \(\frac{90}{160} = \frac{9}{16} = 0.5625\).

(iii) The number of red hatchback cars is 40. The probability of a car being a hatchback is \(\frac{90}{160}\). The probability of a car being red given it is a hatchback is \(\frac{40}{90} = \frac{4}{9}\).

(iv) For independence, \(P(\text{red}) \times P(\text{hatchback}) = \frac{72}{160} \times \frac{90}{160}\) should equal \(\frac{40}{160}\). Calculating, \(\frac{72}{160} \times \frac{90}{160} \neq \frac{40}{160}\), so the events are not independent.

First, calculate the probability of event R (product is 12). The possible pairs are (3, 4), (4, 3), (2, 6), and (6, 2). Thus,

\(P(R) = \frac{4}{36} = \frac{1}{9}\).

Next, calculate the probability of event T (one number is odd, one is even). There are 18 such pairs out of 36 total possibilities, so

\(P(T) = \frac{18}{36} = \frac{1}{2}\).

Now, calculate the probability of both events occurring together, \(P(R \cap T)\). The pairs (3, 4) and (4, 3) satisfy both conditions, so

\(P(R \cap T) = \frac{2}{36} = \frac{1}{18}\).

Check for independence by verifying if \(P(R) \times P(T) = P(R \cap T)\):

\(\frac{1}{9} \times \frac{1}{2} = \frac{1}{18}\).

Since \(P(R) \times P(T) = P(R \cap T)\), the events are independent.