Past Exam Questions

(i) Since the captain, a midfield player, must be included, we need to choose 1 player from the 3 defence players and 1 player from the 5 forward players. The number of ways to do this is given by:

\(\binom{3}{1} \times \binom{5}{1} = 3 \times 5 = 15\)

(ii) If exactly one forward player must be included, we choose 1 player from the 5 forward players. Then, we choose 2 more players from the remaining 6 players (3 defence and 3 midfield). The number of ways to do this is given by:

\(\binom{5}{1} \times \binom{6}{2} = 5 \times 15 = 75\)

(a) (i) To find the number of different three-course meals, multiply the number of choices for each course. There are 3 choices for the starter, 5 choices for the main course, and 3 choices for the dessert. Thus, the total number of three-course meals is:

\(3 \times 5 \times 3 = 45\)

(a) (ii) If customers choose only two of the three courses, consider the combinations:

• Starter and Main: \(3 \times 5 = 15\)
• Starter and Dessert: \(3 \times 3 = 9\)
• Main and Dessert: \(5 \times 3 = 15\)

Summing these gives:

\(15 + 9 + 15 = 39\)

(b) To divide 14 people into groups of 5, 5, and 4, use combinations:

\(\binom{14}{5} \times \binom{9}{5} \times \binom{4}{4} = 252252\)

(a) To find the number of ways to select the committee with exactly one of Mr Lan or Mrs Lan, we consider two cases:

1. Mr Lan is on the committee, but Mrs Lan is not. We choose 4 more members from the remaining 12 people (6 men and 6 women): \\(\binom{12}{4} = 495 \\\)
2. Mrs Lan is on the committee, but Mr Lan is not. We choose 4 more members from the remaining 12 people (6 men and 6 women): \\(\binom{12}{4} = 495 \\\)

Thus, the total number of ways is \\(495 + 495 = 990 \\\).

(b) If Mrs Lan must be on the committee and there must be more women than men, we consider the following scenarios:

1. 2 women and 2 men in addition to Mrs Lan: \\(\binom{7}{2} \times \binom{6}{2} = 315 \\\)
2. 3 women and 1 man in addition to Mrs Lan: \\(\binom{7}{3} \times \binom{6}{1} = 210 \\\)
3. 4 women in addition to Mrs Lan: \\(\binom{7}{4} = 35 \\\)

Adding these gives the total number of ways: \\(315 + 210 + 35 = 560 \\\).

(a) (i) The number of ways to choose 6 books from 18 is given by the combination formula \(\binom{18}{6}\). Calculating this gives:

\(\binom{18}{6} = \frac{18 \times 17 \times 16 \times 15 \times 14 \times 13}{6 \times 5 \times 4 \times 3 \times 2 \times 1} = 18564\)

(ii) To include the Harry Potter book, choose 5 more books from the remaining 17. This is \(\binom{17}{5}\):

\(\binom{17}{5} = \frac{17 \times 16 \times 15 \times 14 \times 13}{5 \times 4 \times 3 \times 2 \times 1} = 6188\)

(b) (i) The total number of ways to arrange 8 people (5 boys and 3 girls) is \(8!\):

\(8! = 40320\)

(ii) Treat the 5 boys as a single unit. This gives 4 units to arrange (3 girls + 1 unit of boys). The number of ways to arrange these 4 units is \(4!\). Within the boys' unit, the 5 boys can be arranged in \(5!\) ways. Therefore, the total number of arrangements is:

\(5! \times 4! = 120 \times 24 = 2880\)

(i) To choose 3 men from 6 and 2 women from 4, we use combinations:

\(\binom{6}{3} \times \binom{4}{2} = 20 \times 6 = 120\).

(ii) More men than women means either 3 men and 2 women, 4 men and 1 woman, or 5 men and 0 women. Calculate each case:

• 3 men and 2 women: \(\binom{6}{3} \times \binom{4}{2} = 120\)
• 4 men and 1 woman: \(\binom{6}{4} \times \binom{4}{1} = 15 \times 4 = 60\)
• 5 men and 0 women: \(\binom{6}{5} \times \binom{4}{0} = 6 \times 1 = 6\)

Total: \(120 + 60 + 6 = 186\).

(iii) If one particular woman and one particular man cannot be together, calculate the total ways without restriction and subtract the restricted cases:

Total ways for 3 men and 2 women: \(120\).

Restricted case (both are in the committee): Choose 2 more men from the remaining 5 and 1 more woman from the remaining 3: \(\binom{5}{2} \times \binom{3}{1} = 10 \times 3 = 30\).

Subtract restricted from total: \(120 - 30 = 90\).

(i) To find the number of ways to punch exactly 2 holes in a column with 8 positions, we use the combination formula: \(\binom{8}{2} = 28\).

(ii) The number of different patterns of holes in a column is the sum of combinations for 1, 2, 3, and 4 holes: \(\binom{8}{1} + \binom{8}{2} + \binom{8}{3} + \binom{8}{4} = 8 + 28 + 56 + 70 = 162\).

(iii) Since each column can have 162 different patterns, and there are 4 columns, the total number of different possible key cards is \(162^4 = 688,747,536\).

(a) To choose a team of 5 people with exactly one adult, select 1 adult from 5 and 4 others from the remaining 7 (3 boys and 4 girls):

\(\binom{5}{1} \times \binom{7}{4} = 5 \times 35 = 175\)

(b) Consider the cases where the team includes at least 2 boys and at least 1 girl:

1. 2 boys, 1 girl, 2 adults: \(\binom{3}{2} \times \binom{4}{1} \times \binom{5}{2} = 3 \times 4 \times 10 = 120\)
2. 2 boys, 2 girls, 1 adult: \(\binom{3}{2} \times \binom{4}{2} \times \binom{5}{1} = 3 \times 6 \times 5 = 90\)
3. 2 boys, 3 girls: \(\binom{3}{2} \times \binom{4}{3} = 3 \times 4 = 12\)
4. 3 boys, 1 girl, 1 adult: \(\binom{3}{3} \times \binom{4}{1} \times \binom{5}{1} = 1 \times 4 \times 5 = 20\)
5. 3 boys, 2 girls: \(\binom{3}{3} \times \binom{4}{2} = 1 \times 6 = 6\)

Summing these cases gives the total number of ways:

\(120 + 90 + 12 + 20 + 6 = 248\)

Since Mr and Mrs Khan and Abad must be included, we have already chosen 3 people. Therefore, we need to choose 17 more people from the remaining 23 members of the club.

The number of ways to choose 17 people from 23 is given by the combination formula:

\(\binom{23}{17} = \frac{23!}{17!(23-17)!} = \frac{23!}{17!6!}\)

Calculating this gives:

\(\binom{23}{17} = 100947\)

Thus, there are 100947 ways to choose the 20 people.

(a) To find \(k\), use the fact that the sum of probabilities must equal 1:

\(2k + 6k + 12k + 20k = 1\)

\(40k = 1\)

\(k = \frac{1}{40}\)

Now, calculate the probabilities:

(b) Calculate \(E(X)\):

\(E(X) = \frac{1 \times 2 + 2 \times 6 + 3 \times 12 + 4 \times 20}{40}\)

\(E(X) = \frac{2 + 12 + 36 + 80}{40}\)

\(E(X) = \frac{130}{40} = 3.25\)

Calculate \(\text{Var}(X)\):

\(\text{Var}(X) = \frac{1^2 \times 2 + 2^2 \times 6 + 3^2 \times 12 + 4^2 \times 20}{40} - (E(X))^2\)

\(\text{Var}(X) = \frac{2 + 24 + 108 + 320}{40} - (3.25)^2\)

\(\text{Var}(X) = \frac{454}{40} - 10.5625\)

\(\text{Var}(X) = 11.35 - 10.5625 = 0.7875\)

(a) To find the probability distribution of \(X\), we consider all possible sums of the numbers from the two spinners:

• Spinner 1: 0, 1, 2, 2
• Spinner 2: -1, 0, 1

Possible sums \(X\) are: -1, 0, 1, 2, 3.

Calculate probabilities:

• \(P(X = -1) = \frac{1}{12}\) (only one combination: 0 + (-1))
• \(P(X = 0) = \frac{2}{12}\) (combinations: 0 + 0, 1 + (-1))
• \(P(X = 1) = \frac{4}{12}\) (combinations: 0 + 1, 1 + 0, 2 + (-1), 2 + (-1))
• \(P(X = 2) = \frac{3}{12}\) (combinations: 1 + 1, 2 + 0, 2 + 0)
• \(P(X = 3) = \frac{2}{12}\) (combinations: 2 + 1, 2 + 1)

(b) To find \(\text{Var}(X)\), first calculate \(E(X)\):

\(E(X) = \sum x p(x) = -1 \times \frac{1}{12} + 0 \times \frac{2}{12} + 1 \times \frac{4}{12} + 2 \times \frac{3}{12} + 3 \times \frac{2}{12} = \frac{15}{12}\)

Then, calculate \(E(X^2)\):

\(E(X^2) = (-1)^2 \times \frac{1}{12} + 0^2 \times \frac{2}{12} + 1^2 \times \frac{4}{12} + 2^2 \times \frac{3}{12} + 3^2 \times \frac{2}{12} = \frac{35}{12}\)

Finally, \(\text{Var}(X) = E(X^2) - (E(X))^2 = \frac{35}{12} - \left( \frac{15}{12} \right)^2 = \frac{65}{48} = 1.35\).

(a) To find the probability distribution of X, we consider all possible outcomes from spinning the two spinners:

(b) To find E(X), we calculate:

\(E(X) = \frac{-1 \times 1 + 0 \times 2 + 1 \times 1 + 2 \times 3 + 3 \times 2}{9} = \frac{-1 + 0 + 1 + 6 + 6}{9} = \frac{12}{9} = \frac{4}{3}\)

To find Var(X), we calculate:

\(\text{Var}(X) = \frac{(-1)^2 \times 1 + (0)^2 \times 2 + (1)^2 \times 1 + (2)^2 \times 3 + (3)^2 \times 2}{9} - (E(X))^2\)

\(= \frac{1 + 0 + 1 + 12 + 18}{9} - \left(\frac{4}{3}\right)^2 = \frac{32}{9} - \frac{16}{9} = \frac{16}{9}\)

(a) To find \(P(X = 3)\), we calculate the probability of opening two non-carrot tins followed by a carrot tin. The probability of the first tin not being carrots is \(\frac{4}{7}\) (since there are 4 non-carrot tins out of 7 total). The probability of the second tin not being carrots is \(\frac{3}{6}\) (since there are now 3 non-carrot tins out of 6 remaining). The probability of the third tin being carrots is \(\frac{3}{5}\) (since there are 3 carrot tins out of 5 remaining). Therefore, \(P(X = 3) = \frac{4}{7} \times \frac{3}{6} \times \frac{3}{5} = \frac{6}{35}\).

(b) The probability distribution table for \(X\) is:

(c) To find \(\text{Var}(X)\), we first calculate \(E(X)\):

\(E(X) = 1 \times \frac{15}{35} + 2 \times \frac{10}{35} + 3 \times \frac{6}{35} + 4 \times \frac{3}{35} + 5 \times \frac{1}{35} = \frac{70}{35} = 2\)

Then, \(\text{Var}(X)\) is calculated as:

\(\text{Var}(X) = \frac{1^2 \times 15 + 2^2 \times 10 + 3^2 \times 6 + 4^2 \times 3 + 5^2 \times 1}{35} - 2^2\)

\(= \frac{15 + 40 + 54 + 48 + 25}{35} - 4 = \frac{182}{35} - 4 = \frac{6}{5}\)

(a) The probability distribution table for X is:

To find k, use the fact that the sum of probabilities is 1:

\(4k + 6k + 6k + 4k = 1\)

\(20k = 1\)

\(k = \frac{1}{20}\)

(b) Calculate the expected value \(E(X)\):

\(E(X) = 1 \times \frac{4}{20} + 2 \times \frac{6}{20} + 3 \times \frac{6}{20} + 4 \times \frac{4}{20}\)

\(= \frac{4}{20} + \frac{12}{20} + \frac{18}{20} + \frac{16}{20}\)

\(= \frac{50}{20} = 2.5\)

Calculate the variance \(\text{Var}(X)\):

\(\text{Var}(X) = 1^2 \times \frac{4}{20} + 2^2 \times \frac{6}{20} + 3^2 \times \frac{6}{20} + 4^2 \times \frac{4}{20} - (2.5)^2\)

\(= \left(\frac{4}{20} + \frac{24}{20} + \frac{54}{20} + \frac{64}{20}\right) - 6.25\)

\(= \frac{146}{20} - 6.25\)

\(= 7.3 - 6.25\)

\(= 1.05\)

(a) To find the probability of obtaining exactly 2 heads and 1 tail, consider the scenarios: HHT, HTH, and THH.

For HHT: \(\frac{2}{3} \times \frac{2}{3} \times \frac{1}{5} = \frac{4}{45}\)

For HTH: \(\frac{2}{3} \times \frac{1}{3} \times \frac{4}{5} = \frac{8}{45}\)

For THH: \(\frac{1}{3} \times \frac{2}{3} \times \frac{4}{5} = \frac{8}{45}\)

Total probability = \(\frac{4}{45} + \frac{8}{45} + \frac{8}{45} = \frac{20}{45} = \frac{4}{9}\)

(b) The probability distribution table for \(X\) is:

(c) Given \(\text{E}(X) = \frac{32}{15}\), find \(\text{Var}(X)\):

\(\text{Var}(X) = 0^2 \times \frac{1}{45} + 1^2 \times \frac{8}{45} + 2^2 \times \frac{20}{45} + 3^2 \times \frac{16}{45} - \left( \frac{32}{15} \right)^2\)

\(= \frac{8}{45} + \frac{80}{45} + \frac{144}{45} - \left( \frac{32}{15} \right)^2\)

\(= \frac{136}{225}\) or 0.604

(a) To find the probability that Sadie takes exactly 1 red ball, we calculate the probability of choosing 1 red ball and 2 blue balls. The probability is given by:

\(P(1 \text{ red}) = \frac{5}{8} \times \frac{3}{7} \times \frac{2}{6} = \frac{15}{56}\)

(b) The probability distribution table for X is:

(c) To find \(\text{Var}(X)\), we use the formula:

\(\text{Var}(X) = E(X^2) - (E(X))^2\)

First, calculate \(E(X^2)\):

\(E(X^2) = 0^2 \times \frac{1}{56} + 1^2 \times \frac{15}{56} + 2^2 \times \frac{30}{56} + 3^2 \times \frac{10}{56}\)

\(E(X^2) = \frac{15}{56} + \frac{120}{56} + \frac{90}{56} = \frac{225}{56}\)

Now, substitute into the variance formula:

\(\text{Var}(X) = \frac{225}{56} - \left( \frac{15}{8} \right)^2\)

\(\text{Var}(X) = \frac{225}{56} - \frac{225}{64}\)

\(\text{Var}(X) = 0.502\)

(a) To find the probability distribution of Y, consider all possible pairs of X values:

• \((1,1)\), \((2,2)\), \((3,3)\), \((4,4)\): Y takes the value 1, 2, 3, 4 respectively.
• \((1,2)\), \((2,1)\): Y = 1.
• \((1,3)\), \((3,1)\): Y = 2.
• \((1,4)\), \((4,1)\): Y = 3.
• \((2,3)\), \((3,2)\): Y = 1.
• \((2,4)\), \((4,2)\): Y = 2.
• \((3,4)\), \((4,3)\): Y = 1.

Calculate probabilities:

• \(P(Y=1) = \frac{7}{16}\)
• \(P(Y=2) = \frac{5}{16}\)
• \(P(Y=3) = \frac{3}{16}\)
• \(P(Y=4) = \frac{1}{16}\)

(b) Given Y is even, possible values are 2 and 4. Calculate:

\(P(Y=2 \mid Y \text{ is even}) = \frac{P(Y=2)}{P(Y=2) + P(Y=4)} = \frac{\frac{5}{16}}{\frac{5}{16} + \frac{1}{16}} = \frac{5}{6}\)

(a) To find the probability distribution table for X, we first list all possible outcomes of the sums of the numbers from the two spinners:

• Four-sided spinner outcomes: 1, 2, 2, 3
• Three-sided spinner outcomes: -2, -1, 1

Possible sums (X):

• 1 + (-2) = -1
• 1 + (-1) = 0
• 1 + 1 = 2
• 2 + (-2) = 0
• 2 + (-1) = 1
• 2 + 1 = 3
• 3 + (-2) = 1
• 3 + (-1) = 2
• 3 + 1 = 4

Count the occurrences of each sum and divide by the total number of outcomes (12) to get the probabilities:

• \(P(X = -1) = \frac{1}{12}\)
• \(P(X = 0) = \frac{3}{12}\)
• \(P(X = 1) = \frac{3}{12}\)
• \(P(X = 2) = \frac{2}{12}\)
• \(P(X = 3) = \frac{2}{12}\)
• \(P(X = 4) = \frac{1}{12}\)

(b) To find \(\text{Var}(X)\), we first calculate \(\text{E}(X)\):

\(\text{E}(X) = \frac{-1 + 0 + 3 + 4 + 6 + 4}{12} = \frac{16}{12} = \frac{4}{3}\)

Then, calculate \(\text{E}(X^2)\):

\(\text{E}(X^2) = \frac{1 + 0 + 3 + 8 + 18 + 16}{12} = \frac{46}{12}\)

Finally, \(\text{Var}(X) = \text{E}(X^2) - (\text{E}(X))^2\):

\(\text{Var}(X) = \frac{46}{12} - \left(\frac{4}{3}\right)^2 = \frac{37}{18} \approx 2.06\)

(a) To find \(P(X = 3)\), consider the outcomes where the maximum value is 3. The three-sided spinner can land on 3, and the five-sided spinner can land on 3, or the five-sided spinner can land on 3 while the three-sided spinner lands on 1 or 2. The possible outcomes are (3,3), (3,1), (3,2), (1,3), (2,3). Counting these outcomes gives 7 favorable outcomes out of 15 total possible outcomes, so \(P(X = 3) = \frac{7}{15}\).

(b) The probability distribution table for X is:

(c) To find \(E(X)\), use the formula \(E(X) = \sum x_i P(x_i)\):

\(E(X) = 1 \times \frac{2}{15} + 2 \times \frac{6}{15} + 3 \times \frac{7}{15} = \frac{2 + 12 + 21}{15} = \frac{35}{15} = \frac{7}{3}\).

To find \(Var(X)\), use the formula \(Var(X) = E(X^2) - (E(X))^2\):

\(E(X^2) = 1^2 \times \frac{2}{15} + 2^2 \times \frac{6}{15} + 3^2 \times \frac{7}{15} = \frac{2 + 24 + 63}{15} = \frac{89}{15}\).

\(Var(X) = \frac{89}{15} - \left(\frac{7}{3}\right)^2 = \frac{89}{15} - \frac{49}{9} = \frac{22}{45}\).

To find the probability distribution for the number of jellies Jemeel chooses, we calculate the probability for each possible number of jellies (0, 1, 2, 3) out of 3 sweets chosen.

There are a total of 8 sweets (5 jellies and 3 chocolates), and Jemeel chooses 3 sweets. The total number of ways to choose 3 sweets from 8 is \(\binom{8}{3} = 56\).

Probability of choosing 0 jellies:

Choose 3 chocolates from 3: \(\binom{3}{3} = 1\)

Probability: \(\frac{1}{56}\)

Probability of choosing 1 jelly:

Choose 1 jelly from 5 and 2 chocolates from 3: \(\binom{5}{1} \times \binom{3}{2} = 5 \times 3 = 15\)

Probability: \(\frac{15}{56}\)

Probability of choosing 2 jellies:

Choose 2 jellies from 5 and 1 chocolate from 3: \(\binom{5}{2} \times \binom{3}{1} = 10 \times 3 = 30\)

Probability: \(\frac{30}{56}\)

Probability of choosing 3 jellies:

Choose 3 jellies from 5: \(\binom{5}{3} = 10\)

Probability: \(\frac{10}{56}\)

The probability distribution table is:

(a) Given \(P(X = 4) = 3P(X = 2)\), we have:

\(4k(4 + a) = 3 \times 2k(2 + a)\)

\(16k + 4ak = 12k + 6ak\)

\(4k = 2ak\)

\(a = 2\)

Using the sum of probabilities:

\(k(1 + a) + 2k(2 + a) + 3k(3 + a) + 4k(4 + a) = 1\)

\(3k + 8k + 15k + 24k = 1\)

\(50k = 1\)

\(k = \frac{1}{50}\)

(b) The probability distribution table is:

(c) To find \(\text{Var}(X)\), use:

\(\text{Var}(X) = \sum (x^2 \cdot P(X = x)) - (E(X))^2\)

\(\text{Var}(X) = \frac{3}{50} \times 1^2 + \frac{8}{50} \times 2^2 + \frac{15}{50} \times 3^2 + \frac{24}{50} \times 4^2 - 3.2^2\)

\(\text{Var}(X) = \frac{3 + 32 + 135 + 384}{50} - 10.24\)

\(\text{Var}(X) = \frac{554}{50} - 10.24\)

\(\text{Var}(X) = 11.08 - 10.24\)

\(\text{Var}(X) = 0.84\)

\(\text{Var}(X) = \frac{21}{25}\)