Past Exam Questions

(i) The number of red or silver cars manufactured in Korea is \(34 + 30 = 64\). The probability is \(\frac{64}{250} = 0.256\).

(ii) The number of cars not manufactured in Japan is \(250 - (26 + 22 + 12) = 190\). The probability is \(\frac{190}{250} = 0.76\).

(iii) Calculate \(P(X)\), \(P(Y)\), and \(P(X \cap Y)\):

\(P(X) = \frac{80}{250} = \frac{8}{25}\)

\(P(Y) = \frac{100}{250} = \frac{2}{5}\)

\(P(X \cap Y) = \frac{32}{250} = \frac{16}{125}\)

Check independence: \(P(X) \times P(Y) = \frac{8}{25} \times \frac{2}{5} = \frac{16}{125}\)

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

(i) To complete the table, we use the given information:

• Total number of adults = 30
• Right-handed adults = 25
• Adults who wear spectacles = 8
• Right-handed and do not wear spectacles = 19

From this, we can fill in the table:

• Right-handed and wears spectacles = 25 - 19 = 6
• Not right-handed = 30 - 25 = 5
• Not right-handed and wears spectacles = 8 - 6 = 2
• Not right-handed and does not wear spectacles = 5 - 2 = 3

The completed table is:

(ii) To determine if events X and Y are independent, we check if:

\(P(X \cap Y) = P(X) \times P(Y)\)

Calculate the probabilities:

• \(P(X) = \frac{25}{30}\)
• \(P(Y) = \frac{8}{30}\)
• \(P(X \cap Y) = \frac{6}{30}\)

Check independence:

\(P(X) \times P(Y) = \frac{25}{30} \times \frac{8}{30} = \frac{200}{900} = \frac{2}{9}\)

\(P(X \cap Y) = \frac{6}{30} = \frac{1}{5}\)

Since \(P(X \cap Y) \neq P(X) \times P(Y)\), events X and Y are not independent.

(i) To find the probability of event R, we list the pairs where one score is exactly 3 greater than the other: \((1, 4), (2, 5), (3, 6), (4, 7), (5, 8)\). Each pair can occur in two ways (e.g., \((1, 4)\) and \((4, 1)\)). Thus, there are \(5 \times 2 = 10\) favorable outcomes. The total number of outcomes when the die is thrown twice is \(8 \times 8 = 64\). Therefore, \(P(R) = \frac{10}{64}\).

(ii) For event S, we need the product of the scores to be more than 19. The pairs satisfying this condition are \((3, 8), (3, 7), (4, 8), (4, 7), (4, 6), (4, 5), (5, 8), (5, 7), (5, 6), (6, 8), (6, 7), (7, 8)\). Additionally, \((5, 5), (6, 6), (7, 7), (8, 8)\) also satisfy the condition. The total number of favorable outcomes is \(24 + 4 = 28\). Thus, \(P(S) = \frac{28}{64}\).

(iii) To determine if R and S are independent, we check if \(P(R \cap S) = P(R) \times P(S)\). From the mark scheme, \(P(R \cap S) = \frac{4}{64}\). Calculating \(P(R) \times P(S) = \frac{10}{64} \times \frac{28}{64} = \frac{280}{4096}\), which simplifies to \(\frac{35}{512}\), not equal to \(\frac{4}{64}\). Therefore, events R and S are not independent.

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

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

The possible outcomes for the sum to be 4 are (1,3), (2,2), and (3,1). Thus, \(P(S) = \frac{3}{16}\).

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

The product is odd if both numbers are odd. The possible outcomes are (1,1), (1,3), (3,1), and (3,3). Thus, \(P(T) = \frac{4}{16}\).

Now, calculate \(P(S \cap T)\):

The outcomes that satisfy both events are (1,3) and (3,1). Thus, \(P(S \cap T) = \frac{2}{16}\).

Check independence:

\(P(S) \times P(T) = \frac{3}{16} \times \frac{4}{16} = \frac{12}{256} = \frac{3}{64}\).

Since \(\frac{3}{64} \neq \frac{2}{16}\), events S and T are not independent.

For exclusivity, events are exclusive if \(P(S \cap T) = 0\).

Since \(P(S \cap T) = \frac{2}{16} \neq 0\), events S and T are not exclusive.

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 \(P(A)\):

There are 2 numbers divisible by 3 (3 and 6) and 4 numbers not divisible by 3 (1, 2, 4, 5) on a die. The probability that one die shows a number divisible by 3 and the other does not is:

\(P(A) = \frac{1}{3} \times \frac{2}{3} + \frac{2}{3} \times \frac{1}{3} = \frac{4}{9}\)

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

The product is even if at least one number is even. There are 3 even numbers (2, 4, 6) on a die. The probability that the product is even is:

\(P(B) = 1 - \left(\frac{1}{2} \times \frac{1}{2}\right) = \frac{3}{4}\)

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

\(P(A \cap B) = \frac{12}{36} = \frac{1}{3}\)

Check independence:

\(P(A) \times P(B) = \frac{4}{9} \times \frac{3}{4} = \frac{1}{3}\)

Since \(P(A \cap B) = P(A) \times P(B)\), events A and B are independent.

For mutual exclusivity, events are mutually exclusive if \(P(A \cap B) = 0\). Since \(P(A \cap B) = \frac{1}{3} \neq 0\), events A and B are not mutually exclusive.

The probability of event A, where the white marble is in either box L or M, is given by:

\(P(A) = \frac{1}{2}\)

The probability of event B, where the red marble is to the left of both the green marble and the yellow marble, is given by:

\(P(B) = \frac{8}{24} = \frac{1}{3}\)

The probability of both events A and B occurring together is:

\(P(A \cap B) = \frac{1}{6}\)

To check for independence, we verify if:

\(P(A) \times P(B) = \frac{1}{2} \times \frac{1}{3} = \frac{1}{6}\)

Since \(P(A \cap B) = P(A) \times P(B)\), events A and B are independent.

(iii) To determine if events \(R\) and \(S\) are independent, we check if \(P(R \cap S) = P(R) \times P(S)\).

Given: \(P(R) = 0.5\), \(P(S) = 0.4\), and \(P(R \cap S) = \frac{120}{720} = \frac{1}{6}\).

Calculate \(P(R) \times P(S) = 0.5 \times 0.4 = 0.2\).

Since \(P(R \cap S) = \frac{1}{6} \neq 0.2\), events \(R\) and \(S\) are not independent.

(iv) Events \(R\) and \(S\) are exclusive if \(P(R \cap S) = 0\). However, \(P(R \cap S) = \frac{1}{6} \neq 0\), indicating there is an overlap between \(R\) and \(S\) (e.g., the combination 3, 4, 4).

Therefore, events \(R\) and \(S\) are not exclusive.

The probability of event Q, getting a total of 5, can occur with the following pairs: (1,4), (2,3), (3,2), (4,1). Thus,

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

The probability of event S, getting one low score and one high score, can occur with the following pairs: (1,4), (1,5), (1,6), (2,4), (2,5), (2,6), (3,4), (3,5), (3,6), (4,1), (5,1), (6,1), (4,2), (5,2), (6,2), (4,3), (5,3), (6,3). Thus,

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

The probability of both events Q and S occurring, i.e., the intersection \(Q \cap S\), is given by the pairs: (1,4), (2,3), (3,2), (4,1). Thus,

\(P(Q \cap S) = \frac{4}{36} = \frac{1}{9}\).

To check for independence, we verify if \(P(Q \cap S) = P(Q) \times P(S)\).

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

Since \(P(Q \cap S) = \frac{1}{9}\) and \(P(Q) \times P(S) = \frac{1}{18}\), the events are independent.

(i) The total number of countries is 170. The number of countries with a medium GDP is \(20 + 42 + 12 = 74\). Therefore, the probability is \(\frac{74}{170} = \frac{37}{85} \approx 0.435\).

(ii) The number of countries that do not have a medium GDP is \(170 - 74 = 96\). The number of countries with a low birth rate and not a medium GDP is \(3 + 35 = 38\). Therefore, the probability is \(\frac{38}{96} = \frac{19}{49} \approx 0.396\).

(iii) The probability of a country having both a high GDP and a high birth rate is 0, as there are no such countries. Therefore, the events are exclusive.

(iv) The probability of choosing a country with a medium GDP is \(\frac{74}{170} = \frac{37}{85}\). The probability of choosing a country with a medium birth rate is \(\frac{42}{96} = \frac{7}{16}\). The combined probability is \(\frac{37}{85} \times \frac{7}{16} = \frac{287}{1360} \approx 0.211\).

(i) Complete the table:

(ii) The probability of choosing the pair of low-heeled shoes with designer labels is \(\frac{1}{20}\).

(iii) The probability of choosing a pair of sports shoes is \(\frac{10}{20} = \frac{1}{2}\).

(iv) The probability of choosing a pair of high-heeled shoes given that it has a designer label is \(\frac{2}{8} = \frac{1}{4}\).

(v) To check independence, calculate:

\(P(D) = \frac{8}{20} = 0.4\)

\(P(S) = \frac{10}{20} = 0.5\)

\(P(D \cap S) = \frac{5}{20} = 0.25\)

Since \(P(D) \times P(S) = 0.4 \times 0.5 = 0.2 \neq 0.25 = P(D \cap S)\), the events are not independent.

(i) Let \(F\) be the event that a person is female, and \(W\) be the event that a person watches 'Kops are Kids'.

\(P(F) = \frac{12}{30} = 0.4\)

\(P(W) = \frac{16}{30} = 0.533\)

\(P(F \cap W') = \frac{5}{30} = 0.167\)

Using the formula for the union of two events:

\(P(F \cup W) = P(F) + P(W) - P(F \cap W')\)

\(P(F \cup W) = \frac{12}{30} + \frac{16}{30} - \frac{5}{30} = \frac{5}{6} = 0.833\)

(ii) Let \(M\) be the event that a person is male.

\(P(M) = \frac{18}{30} = 0.6\)

\(P(W) = \frac{16}{30} = 0.533\)

\(P(M \cap W) = \frac{13}{30} = 0.433\)

Check independence by comparing \(P(M \cap W)\) and \(P(M) \times P(W)\):

\(P(M) \times P(W) = \frac{18}{30} \times \frac{16}{30} = \frac{8}{25} = 0.32\)

Since \(P(M \cap W) \neq P(M) \times P(W)\), the events are not independent.

(i) Calculate the probability for each scenario where exactly two balls have the same number:

- Probability of selecting 2 from A, 2 from B, and any from C: \(\frac{1}{4} \times \frac{1}{5} \times \frac{1}{7} = \frac{1}{140}\)

- Probability of selecting 8 from A, 8 from B, and any from C: \(\frac{1}{4} \times \frac{2}{5} \times \frac{3}{7} = \frac{3}{70}\)

- Probability of selecting 8 from A, any from B, and 8 from C: \(\frac{1}{4} \times \frac{3}{5} \times \frac{4}{7} = \frac{3}{35}\)

- Probability of selecting any from A, 8 from B, and 8 from C: \(\frac{3}{4} \times \frac{2}{5} \times \frac{4}{7} = \frac{6}{35}\)

Summing these probabilities gives \(\frac{47}{140}\).

(ii) Given that exactly two of the selected balls have the same number, the probability that they are both numbered 2 is:

\(\frac{\frac{1}{28}}{\frac{47}{140}} = \frac{5}{47}\)

(iii) Calculate \(P(X) = \frac{47}{140}\) and \(P(Y) = \frac{1}{4}\). Check independence:

\(P(X \cap Y) = \frac{1}{28}\)

Since \(P(X \cap Y) \neq P(X) \times P(Y)\), events X and Y are not independent.

(i) (a) To get a final score of 12, the second throw must be 6 and the first throw must also be 6, since \(6 + 6 = 12\). The probability of this is \(\frac{1}{6} \times \frac{1}{6} = \frac{1}{36}\).

(b) To get a final score of 5, the possible outcomes are (1,4), (2,3), (3,2), (4,1) where the second throw is not 5. The probability is \(\frac{4}{36} = \frac{1}{9}\).

(ii) Calculate probabilities:

\(P(A) = \frac{1}{6}\)

\(P(B) = \frac{5}{36}\) for pairs (1,5), (2,4), (3,3), (4,2), (5,1)

\(P(C) = 1 - P(O, O) = \frac{3}{4}\)

Check independence:

\(P(A \cap B) = \frac{1}{36}\) which is not equal to \(P(A) \times P(B)\)

\(P(A \cap C) = \frac{3}{36}\) which is not equal to \(P(A) \times P(C)\)

\(P(B \cap C) = \frac{2}{36}\) which is not equal to \(P(B) \times P(C)\)

None of the events are independent.

(i) To find \(P(Q)\), we need the pairs \((a, b)\) such that \(a \times b = 24\). The possible pairs are \((2, 12), (3, 8), (4, 6), (6, 4), (8, 3), (12, 2)\). There are 6 favorable outcomes. The total number of outcomes when two dice are thrown is \(12 \times 12 = 144\). Thus, \(P(Q) = \frac{6}{144} = \frac{1}{24}\).

(ii) For \(P(R)\), both numbers must be greater than 8. The numbers greater than 8 are 9, 10, 11, and 12. Thus, there are 4 choices for each die, giving \(4 \times 4 = 16\) favorable outcomes. Therefore, \(P(R) = \frac{16}{144} = \frac{1}{9}\).

(iii) Events \(Q\) and \(R\) are exclusive if they cannot occur simultaneously. Since \(R\) requires both numbers to be greater than 8, and none of the pairs for \(Q\) satisfy this, \(P(R \cap Q) = 0\). Thus, \(R\) and \(Q\) are exclusive.

(iv) Events \(Q\) and \(R\) are independent if \(P(R \cap Q) = P(R) \times P(Q)\). Since \(P(R \cap Q) = 0\) and \(P(R) \times P(Q) \neq 0\), they are not independent.

(i) To find the probability of event A, we list the pairs where the scores differ by 3 or more: (1,4), (1,5), (1,6), (2,5), (2,6), (3,6) and their reverses: (4,1), (5,1), (6,1), (5,2), (6,2), (6,3). There are 12 such pairs out of 36 possible outcomes.

Thus, \(P(A) = \frac{12}{36} = \frac{1}{3}\).

(ii) For event B, we need the product of the scores to be greater than 8. The pairs are: (2,5), (2,6), (3,4), (3,5), (3,6), (4,3), (4,4), (4,5), (4,6), (5,2), (5,3), (5,4), (5,5), (5,6), (6,2), (6,3), (6,4), (6,5), (6,6). There are 20 such pairs.

Thus, \(P(B) = \frac{20}{36} = \frac{5}{9}\).

(iii) Events A and B are not mutually exclusive because \(P(A \cap B) \neq 0\). Specifically, there are 6 outcomes that satisfy both conditions: (2,5), (2,6), (3,6), (5,2), (6,2), (6,3). Therefore, \(P(A \cap B) = \frac{6}{36}\), so they are not mutually exclusive.

(i) To find \(P(M)\), calculate the total number of males: \(206 + 412 = 618\). The total number of people is \(206 + 412 + 358 + 305 = 1281\). Thus, \(P(M) = \frac{618}{1281} \approx 0.482\).

(ii) To find \(P(M \text{ and } E)\), use the number of employed males: \(412\). Therefore, \(P(M \text{ and } E) = \frac{412}{1281} \approx 0.322\).

(iii) To check if \(M\) and \(E\) are independent, calculate \(P(E)\): \(P(E) = \frac{717}{1281}\) (since \(412 + 305 = 717\)). Check if \(P(M) \times P(E) = P(M \text{ and } E)\). \(\frac{618}{1281} \times \frac{717}{1281} \neq \frac{412}{1281}\), so they are not independent.

(iv) Given the person is unemployed, find the probability they are female. Total unemployed is \(206 + 358 = 564\). Thus, \(P(\text{Female} | \text{Unemployed}) = \frac{358}{564} \approx 0.635\).

(a) The tree diagram shows the outcomes of the first and second throws. Each branch represents a possible outcome with a probability of \(\frac{1}{6}\).

(b) To find \(P(A)\), calculate the probability of scoring 5, 6, 7, 8, or 9:

- Score 5: \(\frac{1}{6} \times \frac{1}{6} + \frac{1}{6} = \frac{7}{36}\)

- Score 6: \(\frac{1}{6} \times \frac{1}{6} = \frac{1}{36}\)

- Score 7: \(\frac{1}{6} \times \frac{1}{6} + \frac{1}{6} \times \frac{1}{6} = \frac{2}{36}\)

- Score 8: \(\frac{1}{6} \times \frac{1}{6} = \frac{1}{36}\)

- Score 9: \(\frac{1}{6} \times \frac{1}{6} = \frac{1}{36}\)

Adding these probabilities gives \(P(A) = \frac{12}{36} = \frac{1}{3}\).

(c) To check independence, compare \(P(A \cap B)\) and \(P(A)P(B)\):

- \(P(B) = \frac{1}{3}\), \(P(A \cap B) = \frac{6}{36} = \frac{1}{6}\)

- \(P(A)P(B) = \frac{1}{3} \times \frac{1}{3} = \frac{1}{9}\)

Since \(\frac{1}{6} \neq \frac{1}{9}\), events A and B are not independent.

(d) To find \(P(B | A')\), use the formula:

\(P(B | A') = \frac{P(B \cap A')}{P(A')} = \frac{\frac{6}{36}}{\frac{2}{3}} = \frac{1}{4} = 0.25\)

(i) For events A and B to be independent, \(P(A) \times P(B)\) should equal \(P(A \text{ and } B)\). Calculate \(P(A) \times P(B) = 0.3 \times 0.8 = 0.24\). Since \(0.24 \neq 0.4\), events A and B are not independent.

(ii) For events A and B to be mutually exclusive, \(P(A \text{ and } B)\) should be 0. Since \(P(A \text{ and } B) = 0.4 \neq 0\), events A and B are not mutually exclusive.

(a) The total number of students who play the guitar is \(44 + 38 = 82\).

The probability that a student plays the guitar is \(\frac{82}{180} = \frac{41}{90}\).

(c) Let \(F\) be the event that the student is female, and \(G\) be the event that the student plays the guitar.

\(P(F) = \frac{100}{180} = \frac{5}{9} \approx 0.5556\)

\(P(G) = \frac{82}{180} = \frac{41}{90} \approx 0.4556\)

\(P(F \cap G) = \frac{38}{180} = \frac{19}{90} \approx 0.2111\)

To check for independence, we compare \(P(F \cap G)\) with \(P(F) \times P(G)\).

\(P(F) \times P(G) = \frac{5}{9} \times \frac{41}{90} = \frac{205}{810} = \frac{41}{162} \approx 0.2531\)

Since \(P(F \cap G) \neq P(F) \times P(G)\), the events are not independent.

(i) To find the probability that a student prefers swimming, we sum the number of students who prefer swimming and divide by the total number of students.

The total number of students who prefer swimming is \(104 + 31 = 135\).

The probability is \(\frac{135}{400} = 0.3375\).

(ii) To determine if the events are independent, we check if \(P(M \cap S) = P(M) \times P(S)\).

\(P(M) = \frac{180}{400} = 0.45\)

\(P(S) = \frac{135}{400} = 0.3375\)

\(P(M \cap S) = \frac{31}{400} = 0.0775\)

Calculate \(P(M) \times P(S) = 0.45 \times 0.3375 = 0.151875\)

Since \(0.151875 \neq 0.0775\), the events are not independent.