Past Exam Questions

(a) The mean of the 40 values of \(x\) is given by:

\(\frac{\Sigma x}{40} = 34\)

We know \(\Sigma(x-k) = 520\), so:

\(\frac{\Sigma x - 40k}{40} = \frac{520}{40}\)

\(34 - k = 13\)

\(k = 34 - 13 = 21\)

(b) The variance is calculated using:

\(\text{Var} = \frac{\Sigma(x-k)^2}{40} - \left(\frac{\Sigma(x-k)}{40}\right)^2\)

Substitute the given values:

\(\text{Var} = \frac{9640}{40} - \left(\frac{520}{40}\right)^2\)

\(\text{Var} = 241 - 13^2\)

\(\text{Var} = 241 - 169 = 72\)

We start with the equation \(\Sigma (x - 50) = 144\).

This can be expanded to \(\Sigma x - 50n = 144\).

We are also given \(\Sigma x = 944\).

Substitute \(\Sigma x = 944\) into the expanded equation:

\(944 - 50n = 144\).

Rearrange to solve for \(n\):

\(944 - 144 = 50n\)

\(800 = 50n\)

\(n = \frac{800}{50}\)

\(n = 16\)

(iii) To find \(\Sigma(x - 60)^2\), calculate each \((x - 60)^2\) for the given times:

\((-15)^2, (-13)^2, (-7)^2, (-4)^2, (-4)^2, 1^2, 4^2, 6^2, 9^2, 13^2, 23^2, 15^2, 18^2\).

Summing these gives:

\(225 + 169 + 49 + 16 + 16 + 1 + 16 + 36 + 81 + 169 + 529 + 225 + 324 = 1856\).

Thus, \(\Sigma(x - 60)^2 = 1856\).

(iv) The variance is calculated using the formula:

\(\text{Var} = \frac{\Sigma(x - 60)^2}{n} - \left(\frac{\Sigma(x - 60)}{n}\right)^2\).

Given \(\Sigma(x - 60) = 46\) and \(n = 13\), substitute the values:

\(\text{Var} = \frac{1856}{13} - \left(\frac{46}{13}\right)^2\).

Calculate each part:

\(\frac{1856}{13} = 142.7692 \ldots\)

\(\left(\frac{46}{13}\right)^2 = 12.5385 \ldots\)

Thus, \(\text{Var} = 142.7692 - 12.5385 = 130.2307 \ldots \approx 130\).

(i) Calculate \(\Sigma(t - 120)\):

\(\Sigma(t - 120) = (95 - 120) + (126 - 120) + (117 - 120) + (135 - 120) + (120 - 120) + (125 - 120) + (114 - 120) + (119 - 120) + (136 - 120)\)

\(= -25 + 6 - 3 + 15 + 0 + 5 - 6 - 1 + 16 = 7\)

Calculate \(\Sigma(t - 120)^2\):

\(\Sigma(t - 120)^2 = (-25)^2 + 6^2 + (-3)^2 + 15^2 + 0^2 + 5^2 + (-6)^2 + (-1)^2 + 16^2\)

\(= 625 + 36 + 9 + 225 + 0 + 25 + 36 + 1 + 256 = 1213\)

(ii) Calculate the variance of \(t\):

Variance \(= \frac{\Sigma(t - 120)^2}{9} - \left(\frac{\Sigma(t - 120)}{9}\right)^2\)

\(= \frac{1213}{9} - \left(\frac{7}{9}\right)^2\)

\(= 134.2\)

(i) The formula for the standard deviation \(\sigma\) is:

\(\sigma^2 = \frac{\Sigma (x-c)^2}{n} - \left(\frac{\Sigma (x-c)}{n}\right)^2\)

Given \(\sigma = 3.2\), \(n = 40\), and \(\Sigma (x-c)^2 = 3099.2\), we have:

\(3.2^2 = \frac{3099.2}{40} - \left(\frac{\Sigma (x-c)}{40}\right)^2\)

\(10.24 = 77.48 - \left(\frac{\Sigma (x-c)}{40}\right)^2\)

\(\left(\frac{\Sigma (x-c)}{40}\right)^2 = 67.24\)

\(\Sigma (x-c) = 40 \times \sqrt{67.24} = 328\)

(ii) Given \(c = 50\), the mean \(\bar{x}\) is:

\(\Sigma x - 40c = 328\)

\(\Sigma x = 328 + 40 \times 50 = 2328\)

\(\text{Mean} = \frac{2328}{40} = 58.2\)

First, calculate \(\Sigma(x - 10)\):

\(\Sigma(x - 10) = \Sigma x - 12 \times 10 = 186 - 120 = 66\)

Next, use the formula for variance to find \(\Sigma(x - 10)^2\):

Given the standard deviation \(\sigma = 4.5\), the variance is \(\sigma^2 = 4.5^2 = 20.25\).

The formula for variance is:

\(\frac{\Sigma(x - 10)^2}{12} - \left(\frac{\Sigma(x - 10)}{12}\right)^2 = 20.25\)

Substitute \(\Sigma(x - 10) = 66\):

\(\frac{\Sigma(x - 10)^2}{12} - \left(\frac{66}{12}\right)^2 = 20.25\)

\(\frac{\Sigma(x - 10)^2}{12} - 5.5^2 = 20.25\)

\(\frac{\Sigma(x - 10)^2}{12} - 30.25 = 20.25\)

\(\frac{\Sigma(x - 10)^2}{12} = 50.5\)

\(\Sigma(x - 10)^2 = 606\)

The probability of rolling a 1 or 2 is \(\frac{2}{6}\), and the probability of rolling a 3, 4, 5, or 6 is \(\frac{4}{6}\).

For bag A, the probability of picking two white marbles is:

\(\frac{8}{15} \times \frac{7}{14}\)

For bag B, the probability of picking two white marbles is:

\(\frac{6}{15} \times \frac{5}{14}\)

The total probability that both marbles are white is:

\(\frac{2}{6} \times \frac{8}{15} \times \frac{7}{14} + \frac{4}{6} \times \frac{6}{15} \times \frac{5}{14}\)

Calculating each part:

\(\frac{2}{6} \times \frac{8}{15} \times \frac{7}{14} = \frac{56}{630}\)

\(\frac{4}{6} \times \frac{6}{15} \times \frac{5}{14} = \frac{60}{630}\)

Adding these probabilities gives:

\(\frac{56}{630} + \frac{60}{630} = \frac{116}{630} = \frac{58}{315}\)

Thus, the probability that both marbles are white is \(\frac{58}{315}\) or 0.184.

To find the probability that exactly 1 student is studying Music and exactly 2 are boys, we consider two scenarios:

1. 1 boy studying Music, 1 boy not studying Music, and 1 girl not studying Music.
2. 1 girl studying Music, 2 boys not studying Music.

For the first scenario:

\(P(B \text{ M}) \times P(B \text{ NM}) \times P(G \text{ NM}) = \frac{40}{160} \times \frac{56}{159} \times \frac{52}{158} \times 3! = 0.17387\)

For the second scenario:

\(P(G \text{ M}) \times P(B \text{ NM}) \times P(B \text{ NM}) = \frac{12}{160} \times \frac{56}{159} \times \frac{55}{158} \times \frac{3!}{2!} = 0.02759\)

Adding both probabilities gives:

\(P = 0.17387 + 0.02759 = 0.201\)

The possible outcomes for the sum of 9 when the die is thrown twice are: (3, 6), (6, 3), (4, 5), and (5, 4).

The probability of rolling a 3 is 0.1 and the probability of rolling a 6 is 0.3. Thus, the probability of (3, 6) is:

\(P(3, 6) = 0.1 \times 0.3 = 0.03\)

The probability of (6, 3) is the same as (3, 6) because the events are independent:

\(P(6, 3) = 0.3 \times 0.1 = 0.03\)

The probability of rolling a 4 is 0.2 and the probability of rolling a 5 is 0.1. Thus, the probability of (4, 5) is:

\(P(4, 5) = 0.2 \times 0.1 = 0.02\)

The probability of (5, 4) is the same as (4, 5):

\(P(5, 4) = 0.1 \times 0.2 = 0.02\)

Therefore, the total probability that the sum is 9 is:

\(P(\text{sum is 9}) = P(3, 6) + P(6, 3) + P(4, 5) + P(5, 4) = 0.03 + 0.03 + 0.02 + 0.02 = 0.1\)

To find \(r\), note that the only way to get a score of 6 is by landing on 3 on both spinners, so \(P(\text{score is } 6) = r^2 = \frac{1}{36}\). Solving for \(r\), we get \(r = \frac{1}{6}\).

For a score of 5, the possible outcomes are (2, 3) and (3, 2). Thus, \(P(\text{score is } 5) = P(2, 3) + P(3, 2) = qr + rq = 2qr = \frac{1}{9}\). Substituting \(r = \frac{1}{6}\), we have:

\(2q \times \frac{1}{6} = \frac{1}{9}\)

\(\frac{q}{3} = \frac{1}{9}\)

\(q = \frac{1}{3}\)

Since \(p + q + r = 1\), we have:

\(p + \frac{1}{3} + \frac{1}{6} = 1\)

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

\(p = \frac{1}{2}\)

The total number of balloons is 40 (10 pink + 9 yellow + 12 green + 9 white).

We need to find the probability of selecting exactly 3 green balloons out of 7 selected balloons.

Using combinations, the number of ways to choose 3 green balloons from 12 is given by:

\(\binom{12}{3}\)

The number of ways to choose the remaining 4 balloons from the 28 non-green balloons is:

\(\binom{28}{4}\)

The total number of ways to choose 7 balloons from 40 is:

\(\binom{40}{7}\)

The probability is then given by:

\(\frac{\binom{12}{3} \times \binom{28}{4}}{\binom{40}{7}}\)

Calculating each part:

\(\binom{12}{3} = 220\)

\(\binom{28}{4} = 20475\)

\(\binom{40}{7} = 18643560\)

Thus, the probability is:

\(\frac{220 \times 20475}{18643560} = 0.242\)

(i) The probability of getting a green robot from one packet is \(\frac{1}{4}\) since there are four equally likely colours.

(ii) The probability that Nick gets his first green robot on the fifth packet means he does not get a green robot in the first four packets and then gets a green robot on the fifth. The probability of not getting a green robot in one packet is \(\frac{3}{4}\). Therefore, the probability is \(\left( \frac{3}{4} \right)^4 \times \frac{1}{4} = \frac{81}{1024} = 0.0791\).

(iii) The probability that the first four packets Amos opens all contain different coloured robots is calculated by considering the permutations of the four colours. The probability is \(\frac{1}{4} \times \frac{1}{4} \times \frac{1}{4} \times \frac{1}{4} \times 4! = \frac{3}{32}\).

There are 3 even numbers (2, 4, 6) and 2 odd numbers (1, 7) in the box.

The total number of ways to choose 3 discs from 5 is given by \(\binom{5}{3} = 10\).

To find the number of favorable outcomes (2 even and 1 odd), we can choose 2 even numbers from 3 and 1 odd number from 2:

\(\binom{3}{2} \times \binom{2}{1} = 3 \times 2 = 6\).

Thus, the probability is:

\(\frac{6}{10} = 0.6\).

The tree diagram is constructed as follows:

1. First choice: Sharik has a probability of \(\frac{1}{3}\) of choosing the correct answer (C) and \(\frac{2}{3}\) of choosing a wrong answer (W).
2. Second choice (if first was wrong): From the remaining two answers, the probability of choosing the correct answer is \(\frac{1}{2}\) and the wrong answer is \(\frac{1}{2}\).
3. Third choice (if second was also wrong): The remaining answer must be correct, so the probability is 1.

The tree diagram shows these probabilities and the paths leading to the correct answer.

(i) To find the probability that there is a winner after exactly two sets, we consider two scenarios: Roger wins both sets (RR) or Andy wins both sets (AA).

The probability that Roger wins both sets (RR) is:

\(P(RR) = 0.6 \times 0.7 = 0.42\)

The probability that Andy wins both sets (AA) is:

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

Thus, the probability that there is a winner after exactly two sets is:

\(P(2 \text{ sets in match}) = P(RR) + P(AA) = 0.42 + 0.3 = 0.72\)

(ii) To find the probability that Andy wins the match given that there is a winner after exactly two sets, we use conditional probability:

\(P(A \text{ wins and 2 sets}) = P(AA) = 0.3\)

\(P(\text{2 sets}) = 0.72\)

Thus, the probability is:

\(\frac{P(AA)}{P(\text{2 sets})} = \frac{0.3}{0.72} = \frac{5}{12}\)

To find the probability that Nur chooses a play-house, we calculate the probability for each playground and sum them up.

For Playground X, the probability of choosing a play-house is:

\(P(X \text{ and } P) = \frac{1}{4} \times \frac{4}{9} = \frac{1}{9}\)

For Playground Y, the probability of choosing a play-house is:

\(P(Y \text{ and } P) = \frac{1}{4} \times \frac{2}{12} = \frac{1}{24}\)

For Playground Z, the probability of choosing a play-house is:

\(P(Z \text{ and } P) = \frac{1}{2} \times \frac{1}{16} = \frac{1}{32}\)

Summing these probabilities gives:

\(P(P) = \frac{1}{9} + \frac{1}{24} + \frac{1}{32} = \frac{53}{288} \approx 0.184\)

(i) To find the probability that the two digits chosen are equal, consider the digits 1, 3, and 5 in the number 113 333 555. The probabilities are:

\(P(1, 1) = \frac{2}{9} \times \frac{1}{8}\)

\(P(3, 3) = \frac{4}{9} \times \frac{3}{8}\)

\(P(5, 5) = \frac{3}{9} \times \frac{2}{8}\)

Summing these probabilities gives:

\(\frac{2}{9} \times \frac{1}{8} + \frac{4}{9} \times \frac{3}{8} + \frac{3}{9} \times \frac{2}{8} = \frac{5}{18}\)

(ii) To find the probability that one digit is a 5 and one digit is not a 5, consider:

\(P(5, \overline{5}) + P(\overline{5}, 5)\)

\(= \frac{3}{9} \times \frac{6}{8} + \frac{6}{9} \times \frac{3}{8} = \frac{36}{72} = \frac{1}{2}\)

(a) The probability that it does not rain on Sunday is \(0.6\). If it does not rain on any day, the probability for each subsequent day is \(0.8\). Therefore, the probability that it does not rain on any of the 4 days is:

\(P(\text{no rain}) = 0.6 \times 0.8^3 = \frac{192}{625}\)

(b) The probability that it does not rain on Sunday is \(0.6\). The probability that it rains on Tuesday, given it did not rain on Monday, is \(0.2\). Therefore, the probability that the first day it rains is Tuesday is:

\(0.6 \times 0.8 \times 0.2 = \frac{12}{125}\)

(c) The probability that it rains on exactly one of the 4 days can be calculated by considering the scenarios where it rains on one specific day and not on the others. The scenarios are:

• Rains on Sunday: \(0.4 \times 0.3 \times 0.8 \times 0.8 = \frac{48}{625}\)
• Rains on Monday: \(0.6 \times 0.2 \times 0.3 \times 0.8 = \frac{18}{625}\)
• Rains on Tuesday: \(0.6 \times 0.8 \times 0.2 \times 0.3 = \frac{18}{625}\)
• Rains on Wednesday: \(0.6 \times 0.8 \times 0.8 \times 0.2 = \frac{48}{625}\)

Adding these probabilities gives:

\(\frac{48}{625} + \frac{18}{625} + \frac{18}{625} + \frac{48}{625} = \frac{132}{625}\)

The probability of picking a robot card from Jack's pack is \(\frac{10}{15} = \frac{2}{3}\).

Emma has 7 robot cards, so the total number of cards in Emma's pack is \(x + 4\) because \(x - 3\) cards are aeroplanes, making \(x - 3 + 7 = x + 4\) cards in total.

The probability of picking a robot card from Emma's pack is \(\frac{7}{x+4}\).

The probability that both cards have pictures of robots is \(\frac{2}{3} \times \frac{7}{x+4} = \frac{7}{18}\).

Setting up the equation: \(\frac{2}{3} \times \frac{7}{x+4} = \frac{7}{18}\).

Solving for \(x\):

\(\frac{14}{3(x+4)} = \frac{7}{18}\)

Cross-multiply: \(14 \times 18 = 7 \times 3(x+4)\)

\(252 = 21(x+4)\)

\(252 = 21x + 84\)

\(168 = 21x\)

\(x = 8\)

Let the houses with numbers greater than 14 be 16, 18, 20, 22, and 24. There are 5 such houses.

We need to find the probability that at least 2 newspapers are delivered to these 5 houses.

This can be calculated as \(P(2) + P(3)\) or \(1 - P(0, 1)\).

Calculate \(P(2)\):

\(P(2) = \frac{5}{12} \times \frac{4}{11} \times \frac{7}{10} \times \binom{3}{2}\)

Calculate \(P(3)\):

\(P(3) = \frac{5}{12} \times \frac{4}{11} \times \frac{3}{10}\)

Thus, \(P(\text{at least 2}) = P(2) + P(3) = \frac{4}{11}\).

Alternatively, using combinations:

\(\frac{\binom{5}{3} + \binom{5}{2} \times \binom{7}{1}}{\binom{12}{3}} = \frac{4}{11}\).