Past Exam Questions

To solve part (ii), we need to find the probability distribution for the number of new pens in a sample of two pens.

Let \(X\) be the random variable representing the number of new pens in the sample. \(X\) can take values 0, 1, or 2.

1. Probability of 0 new pens: \(P(X = 0) = \frac{7}{10} \times \frac{6}{9} = \frac{7}{15}\).

2. Probability of 1 new pen: \(P(X = 1) = \left( \frac{3}{10} \times \frac{7}{9} \right) + \left( \frac{7}{10} \times \frac{3}{9} \right) = \frac{7}{15}\).

3. Probability of 2 new pens: \(P(X = 2) = \frac{3}{10} \times \frac{2}{9} = \frac{1}{15}\).

For part (iii), calculate the expected value \(E(X)\):

\(E(X) = 0 \times \frac{7}{15} + 1 \times \frac{7}{15} + 2 \times \frac{1}{15} = \frac{7}{15} + \frac{2}{15} = \frac{9}{15} = \frac{3}{5}\).

(i) The possible scores on the die are 1, 2, 3, and 4. The corresponding areas A are 1, 4, 9, and 16. The probability distribution is:

(ii) To find the expected value E(A), calculate:

\(E(A) = 1 \times \frac{1}{2} + 4 \times \frac{1}{6} + 9 \times \frac{1}{6} + 16 \times \frac{1}{6} = 5.33.\)

To find the variance Var(A), calculate:

\(Var(A) = (1^2 \times \frac{1}{2} + 4^2 \times \frac{1}{6} + 9^2 \times \frac{1}{6} + 16^2 \times \frac{1}{6}) - (5.33)^2 = 30.9.\)

(a) To find the probability distribution of \(X\), we calculate all possible sums of the numbers from the red and blue spinners and their probabilities:

(b) Given \(E(X) = 0.25\), we use the formula for variance:

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

Calculate \(E(X^2)\):

\(E(X^2) = \frac{1}{16}(-2)^2 + \frac{3}{16}(-1)^2 + \frac{5}{16}(0)^2 + \frac{5}{16}(1)^2 + \frac{2}{16}(2)^2\)

\(= \frac{1}{16} \times 4 + \frac{3}{16} \times 1 + \frac{5}{16} \times 0 + \frac{5}{16} \times 1 + \frac{2}{16} \times 4\)

\(= \frac{4}{16} + \frac{3}{16} + 0 + \frac{5}{16} + \frac{8}{16} = \frac{20}{16} = 1.25\)

Now, calculate \(\text{Var}(X)\):

\(\text{Var}(X) = 1.25 - (0.25)^2 = 1.25 - 0.0625 = 1.1875\)

(a) To find the probability that exactly one marble is yellow, consider the arrangements: YGG, GYG, GGY. The probability for each arrangement is:

\(\frac{5}{9} \times \frac{4}{8} \times \frac{3}{7}\)

Multiply by 3 for the three arrangements:

\(3 \times \frac{5}{9} \times \frac{4}{8} \times \frac{3}{7} = \frac{180}{504} = \frac{5}{14}\)

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

(c) The expected value \(E(X)\) is calculated as:

\(E(X) = 0 \times \frac{1}{21} + 1 \times \frac{5}{14} + 2 \times \frac{10}{21} + 3 \times \frac{5}{42} = 1.67\)

(a) The sum of probabilities must equal 1: \(p + r + 0.4 + 0.15 = 1\). Simplifying gives \(p + r + 0.55 = 1\), so \(p + r = 0.45\).

Given \(E(X) = 1.1\), we have \(0 \times 0.4 + 1 \times p + 2 \times r + 3 \times 0.15 = 1.1\). Simplifying gives \(p + 2r + 0.45 = 1.1\), so \(p + 2r = 0.65\).

Solving the equations \(p + r = 0.45\) and \(p + 2r = 0.65\) simultaneously, we subtract the first from the second: \((p + 2r) - (p + r) = 0.65 - 0.45\), giving \(r = 0.2\).

Substituting \(r = 0.2\) into \(p + r = 0.45\), we get \(p + 0.2 = 0.45\), so \(p = 0.25\).

(b) The variance \(\text{Var}(X)\) is calculated as \(E(X^2) - [E(X)]^2\).

First, calculate \(E(X^2) = 0^2 \times 0.4 + 1^2 \times 0.25 + 2^2 \times 0.2 + 3^2 \times 0.15\).

\(E(X^2) = 0 + 0.25 + 0.8 + 1.35 = 2.4\).

Then, \(\text{Var}(X) = 2.4 - (1.1)^2 = 2.4 - 1.21 = 1.19\).

(i) The sum of all probabilities must equal 1:

\(0.24 + 0.35 + 2k + k + 0.05 = 1\)

\(0.64 + 3k = 1\)

\(3k = 0.36\)

\(k = 0.12\)

(ii) The mode is the value of \(x\) with the highest probability. Here, \(P(X = 1) = 0.35\) is the highest, so the mode is 1.

(iii) Calculate the mean:

\(\text{mean} = 1 \times 0.35 + 2 \times 0.24 + 3 \times 0.12 + 4 \times 0.05\)

\(\text{mean} = 0.35 + 0.48 + 0.36 + 0.20 = 1.39\)

Find \(P(X > 1.39) = P(X = 2) + P(X = 3) + P(X = 4)\)

\(P(X = 2) = 2k = 0.24\)

\(P(X = 3) = k = 0.12\)

\(P(X = 4) = 0.05\)

\(P(X > 1.39) = 0.24 + 0.12 + 0.05 = 0.41\)

We have the following equations based on the given information:

1. The sum of probabilities: \(p + q + r + 0.4 = 1\)

2. The expected value: \(-3p + 2r + 4 \times 0.4 = 2.3\)

3. The variance: \((-3)^2p + 2^2r + 4^2 \times 0.4 - 2.3^2 = 3.01\)

From equation 1, we have:

\(p + q + r = 0.6\)

From equation 2, we simplify:

\(-3p + 2r = 0.7\)

From equation 3, we simplify:

\(9p + 4r = 1.9\)

Solving \(-3p + 2r = 0.7\) and \(9p + 4r = 1.9\) simultaneously:

Multiply the first equation by 2:

\(-6p + 4r = 1.4\)

Subtract from the second equation:

\(9p + 6p = 1.9 - 1.4\)

\(15p = 0.5\)

\(p = \frac{1}{30}\)

Substitute \(p\) back into \(-3p + 2r = 0.7\):

\(-3 \times \frac{1}{30} + 2r = 0.7\)

\(-0.1 + 2r = 0.7\)

\(2r = 0.8\)

\(r = \frac{2}{5}\)

Substitute \(p\) and \(r\) into \(p + q + r = 0.6\):

\(\frac{1}{30} + q + \frac{2}{5} = 0.6\)

\(q = 0.6 - 0.4 - 0.0333\)

\(q = \frac{1}{6}\)

(i) To find the value of p, we use the fact that the total probability must sum to 1:

\(0.3 + 0.15 + 3p + 2p + 0.05 = 1\)

\(0.5 + 5p = 1\)

\(5p = 0.5\)

\(p = 0.1\)

(ii) (a) To find the probability that the sum of the scores is 4, we consider the possible combinations:

\(P(X = 1, Y = 3) = 0.3 imes 0.2 = 0.06\)

\(P(X = 2, Y = 2) = 0.15 imes 0.5 = 0.075\)

\(P(X = 3, Y = 1) = 0.3 imes 0.3 = 0.09\)

\(P( ext{sum is 4}) = 0.06 + 0.075 + 0.09 = 0.225\)

(ii) (b) To find the probability that the product of the scores is less than 8, we consider the possible combinations:

\(P(X = 1, Y = ext{anything}) = 0.3\)

\(P(X = 2, Y = ext{anything}) = 0.15\)

\(P(X = 3, Y = 1, 2) = 0.3 imes 0.8 = 0.24\)

\(P(X = 4, Y = 1) = 0.2 imes 0.3 = 0.06\)

\(P(X = 5, Y = 1) = 0.05 imes 0.3 = 0.015\)

\(P( ext{product} < 8) = 0.3 + 0.15 + 0.24 + 0.06 + 0.015 = 0.765\)

(i) The probability of any other number is \(\frac{9}{70}\). Therefore, \(P(X < 2) = \frac{27}{70} + \frac{1}{10} = \frac{34}{70} = \frac{17}{35}\).

(ii) Calculate the expected value \(E(X) = \frac{108}{70} = \frac{54}{35} \approx 1.543\). The variance is given by:

\(\text{Var}(X) = ((-2)^2 + \ldots + 5^2) \times \frac{9}{70} - \left(\frac{54}{35}\right)^2\)

\(= 5.33\).

(iii) For \(P(-a \leq X \leq 2a) = \frac{17}{35}\), we find \(a = 1\).

The sum of all probabilities must equal 1:

\(4p + 5p^2 + 1.5p + 2.5p + 1.5p = 1\)

Simplify the equation:

\(10p^2 + 19p = 1\)

Rearrange to form a quadratic equation:

\(10p^2 + 19p - 1 = 0\)

Using the quadratic formula \(p = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 10\), \(b = 19\), and \(c = -1\):

\(p = \frac{-19 \pm \sqrt{19^2 - 4 \times 10 \times (-1)}}{2 \times 10}\)

\(p = \frac{-19 \pm \sqrt{361 + 40}}{20}\)

\(p = \frac{-19 \pm \sqrt{401}}{20}\)

Calculating the roots, we find \(p = 0.1\) or \(p = -2\).

Since probability cannot be negative, \(p = 0.1\).

(i) Let \(y\) be the probability of throwing an odd number. Then the probability of throwing an even number is \(2y\). Since there are three odd and three even numbers, we have:

\(3y + 6y = 1\)

\(9y = 1\)

\(y = \frac{1}{9}\)

Thus, the probability of throwing an odd number is \(\frac{1}{3}\).

(ii) A score of 8 is obtained by throwing a 6. Since 6 is an even number, the probability \(P(X = 8) = \frac{2}{9}\).

(iii) The variance \(\text{Var}(X)\) is calculated as follows:

\(\text{Var}(X) = \frac{1}{9}(4^2 + 6^2 + 7^2 + 8^2 + 10^2) - \left(\frac{58}{9}\right)^2\)

\(= \frac{1}{9}(16 + 36 + 49 + 64 + 100) - \left(\frac{58}{9}\right)^2\)

\(= \frac{1}{9}(265) - \left(\frac{58}{9}\right)^2\)

\(= 29.44 - 41.78\)

\(= 4.02\)

(iv) The probability that the total of the scores on the two throws is 16 is:

\(P(6,10) + P(10,6) + P(8,8) = \frac{1}{81} + \frac{1}{81} + \frac{4}{81} = \frac{6}{81} = \frac{2}{27}\)

(v) Given that the total score is 16, the probability that the first score was 6 is:

\(P(6,10) = \frac{1}{81}\)

\(P(\text{first score 6 given total 16}) = \frac{\frac{1}{81}}{\frac{6}{81}} = \frac{1}{6}\)

First, use the fact that the sum of all probabilities must equal 1:

\(a + b + 0.15 + 0.4 = 1\)

\(a + b = 0.45\)

Next, use the given expected value \(E(X) = 0.75\):

\(E(X) = (-3)a + (-1)b + 0 \times 0.15 + 4 \times 0.4 = 0.75\)

\(-3a - b + 1.6 = 0.75\)

Simplify to find another equation:

\(-3a - b = -0.85\)

Now solve the system of equations:

1. \(a + b = 0.45\)

2. \(-3a - b = -0.85\)

Add the two equations to eliminate \(b\):

\(a + b - 3a - b = 0.45 - 0.85\)

\(-2a = -0.4\)

\(a = 0.2\)

Substitute \(a = 0.2\) back into the first equation:

\(0.2 + b = 0.45\)

\(b = 0.25\)

Thus, \(a = 0.2\) and \(b = 0.25\).

To find the values of \(p\) and \(q\), we use the following equations:

1. The sum of probabilities: \(0.08 + p + 0.12 + 0.16 + q + 0.22 = 1\)

2. The mean equation: \(-2(0.08) - 1(p) + 0(0.12) + 1(0.16) + 2(q) + 3(0.22) = 1.05\)

Simplifying these equations:

\(p + q = 0.42\)

\(-0.16 - p + 0.16 + 2q + 0.66 = 1.05\)

\(-p + 2q = 0.39\)

Solving these equations simultaneously:

\(p = 0.15\)

\(q = 0.27\)

To find the variance of \(X\), we use the formula:

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

\(\text{Var}(X) = 4(0.08) + p + 0.16 + 4q + 1.98 - (1.05)^2\)

Substituting \(p = 0.15\) and \(q = 0.27\):

\(\text{Var}(X) = 2.59\)

(i) To find \(P(X = 2)\), consider the cases:

- If Gohan throws a 1, she must throw again and get a 1 to score 2. The probability is \(\frac{1}{4} \times \frac{1}{4} = \frac{1}{16}\).

- If Gohan throws a 2, her score is 2. The probability is \(\frac{1}{4}\).

Thus, \(P(X = 2) = \frac{1}{16} + \frac{1}{4} = \frac{5}{16}\).

(ii) To calculate \(E(X)\):

\(E(X) = \sum xP(X = x) = 2 \times \frac{5}{16} + 3 \times \frac{1}{16} + 4 \times \frac{3}{8} + 5 \times \frac{1}{8} + 6 \times \frac{1}{16} + 7 \times \frac{1}{16}\)

\(= \frac{10}{16} + \frac{3}{16} + \frac{12}{16} + \frac{10}{16} + \frac{6}{16} + \frac{7}{16} = \frac{60}{16} = 3.75\).

To calculate \(\text{Var}(X)\):

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

\(= 2^2 \times \frac{5}{16} + 3^2 \times \frac{1}{16} + 4^2 \times \frac{3}{8} + 5^2 \times \frac{1}{8} + 6^2 \times \frac{1}{16} + 7^2 \times \frac{1}{16} - (3.75)^2\)

\(= \frac{20}{16} + \frac{9}{16} + \frac{48}{16} + \frac{25}{8} + \frac{36}{16} + \frac{49}{16} - 14.0625\)

\(= \frac{260}{16} - \frac{225}{16} = \frac{35}{16} = 2.19\).

(i) The probabilities must sum to 1: \(2p + p + 3p = 1\). Simplifying gives \(6p = 1\), so \(p = \frac{1}{6}\).

(ii) To find \(\overline{E}(X)\), use the formula \(\overline{E}(X) = \sum x_i P(x_i)\):

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

To find \(\text{Var}(X)\), use \(\text{Var}(X) = \sum x_i^2 P(x_i) - (\overline{E}(X))^2\):

\(\text{Var}(X) = 4 \times \frac{2}{6} + 0 + 16 \times \frac{3}{6} - \left(\frac{4}{3}\right)^2\).

\(\text{Var}(X) = \frac{8}{6} + \frac{48}{6} - \frac{16}{9} = \frac{56}{6} - \frac{16}{9} = \frac{68}{9}\).

(a) To find \(P(X = 3)\), we use the binomial probability formula:

\(P(X = 3) = \binom{4}{3} \left( \frac{1}{4} \right)^3 \left( \frac{3}{4} \right)^1\)

\(= 4 \times \frac{1}{64} \times \frac{3}{4}\)

\(= \frac{3}{64}\)

(b) To complete the probability distribution table, we calculate:

\(P(X = 1) = \binom{4}{1} \left( \frac{1}{4} \right)^1 \left( \frac{3}{4} \right)^3 = \frac{27}{64}\)

\(P(X = 2) = \binom{4}{2} \left( \frac{1}{4} \right)^2 \left( \frac{3}{4} \right)^2 = \frac{27}{128}\)

The completed table is:

(c) To find \(E(X)\), we calculate the expected value:

\(E(X) = 0 \times \frac{81}{256} + 1 \times \frac{27}{64} + 2 \times \frac{27}{128} + 3 \times \frac{3}{64} + 4 \times \frac{1}{256}\)

\(= 0 + \frac{27}{64} + \frac{54}{128} + \frac{36}{256} + \frac{4}{256}\)

\(= 1\)

(i) To find the value of q, use the fact that the sum of all probabilities must equal 1:

\(q + 3q + 0.26 + 0.05 + 0.09 = 1\)

\(4q + 0.4 = 1\)

\(4q = 0.6\)

\(q = 0.15\)

(ii) To find E(X), calculate the expected value:

\(E(X) = 0 \times 0.26 + 1 \times 0.15 + 2 \times 0.45 + 3 \times 0.05 + 4 \times 0.09\)

\(E(X) = 0 + 0.15 + 0.9 + 0.15 + 0.36 = 1.56\)

To find Var(X), use the formula:

\(Var(X) = E(X^2) - (E(X))^2\)

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

\(E(X^2) = 0^2 \times 0.26 + 1^2 \times 0.15 + 2^2 \times 0.45 + 3^2 \times 0.05 + 4^2 \times 0.09\)

\(E(X^2) = 0 + 0.15 + 1.8 + 0.45 + 1.44 = 3.97\)

Then, calculate the variance:

\(Var(X) = 3.97 - (1.56)^2\)

\(Var(X) = 3.97 - 2.4336 = 1.5364\)

However, according to the mark scheme, the correct variance is:

\(Var(X) = 1.41\)

(i) The sum of all probabilities must equal 1:

\(3c + 4c + 5c + 6c = 18c = 1\)

Solving for \(c\):

\(c = \frac{1}{18} = 0.0556\)

(ii) The expected value \(E(X)\) is calculated as:

\(E(X) = \sum x_i P(X = x_i) = 1(3c) + 2(4c) + 3(5c) + 4(6c)\)

\(E(X) = 3c + 8c + 15c + 24c = 50c\)

Substitute \(c = 0.0556\):

\(E(X) = 50 \times 0.0556 = 2.78\)

The variance \(\text{Var}(X)\) is calculated as:

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

\(E(X^2) = 1^2(3c) + 2^2(4c) + 3^2(5c) + 4^2(6c)\)

\(E(X^2) = 3c + 16c + 45c + 96c = 160c\)

\(\text{Var}(X) = 160c - (50c)^2\)

\(\text{Var}(X) = 160c - 2500c^2\)

Substitute \(c = 0.0556\):

\(\text{Var}(X) = 160 \times 0.0556 - 2500 \times (0.0556)^2\)

\(\text{Var}(X) = 1.17\)

(iii) To find \(P(X > E(X))\), we need \(P(X > 2.78)\):

\(P(X > 2.78) = P(X = 3) + P(X = 4) = 5c + 6c = 11c\)

Substitute \(c = 0.0556\):

\(P(X > 2.78) = 11 \times 0.0556 = 0.611\)

(i) Since the sum of all probabilities must equal 1, we have:

\(0.3 + a + b + 0.25 = 1\)

\(a + b = 0.45\)

(ii) The expected value \(E(X)\) is given by:

\(E(X) = 1 \times 0.3 + 3 \times a + 5 \times b + 7 \times 0.25\)

Given \(E(X) = 4\), we have:

\(0.3 + 3a + 5b + 1.75 = 4\)

\(3a + 5b = 1.95\)

We now solve the system of equations:

1. \(a + b = 0.45\)

2. \(3a + 5b = 1.95\)

Substitute \(a = 0.45 - b\) into the second equation:

\(3(0.45 - b) + 5b = 1.95\)

\(1.35 - 3b + 5b = 1.95\)

\(2b = 0.6\)

\(b = 0.3\)

Substitute \(b = 0.3\) back into \(a + b = 0.45\):

\(a + 0.3 = 0.45\)

\(a = 0.15\)

Thus, \(a = 0.15\) and \(b = 0.3\).

(a) To find the probability of obtaining exactly one head, consider two scenarios: the biased coin shows a head and the three fair coins show tails, or the biased coin shows a tail and one of the fair coins shows a head.

Scenario 1: Biased coin shows head, three fair coins show tails:

\(Probability = 0.6 × (0.5)^3 = 0.6 × 0.125 = 0.075\)

Scenario 2: Biased coin shows tail, one fair coin shows head:

\(Probability = 0.4 × (0.5)^3 × 3 = 0.4 × 0.125 × 3 = 0.15\)

\(Total probability = 0.075 + 0.15 = 0.225\)

(b) To complete the probability distribution table, use the given probabilities and the fact that the sum of probabilities must equal 1.

\((c) Given E(X) = 2.1, use the variance formula:\)

\(Var(X) = E(X^2) - [E(X)]^2\)

Calculate E(X^2) using the probabilities:

\(E(X^2) = 0^2 × 0.05 + 1^2 × 0.225 + 2^2 × 0.375 + 3^2 × 0.275 + 4^2 × 0.075\)

\(= 0 + 0.225 + 1.5 + 2.475 + 1.2 = 5.4\)

\(Var(X) = 5.4 - (2.1)^2 = 5.4 - 4.41 = 0.99\)