1. [Maximum mark: 6]
Becky sometimes works in an office and sometimes works at home. The random variable \( X \) denotes the number of days that she works at home in any given week. It is given that:
\[ P(X = x) = kx(x + 1), \]
where \( k \) is a constant and \( x = 1, 2, 3 \) or \( 4 \) only.
(a) Draw up the probability distribution table for \( X \), giving the probabilities as numerical fractions. [3]
(b) Find \( E(X) \) and \( \text{Var}(X) \). [3]
▶️Answer/Explanation
(a) Using the sum of probabilities equal to 1:
\[ 2k + 6k + 12k + 20k = 1 \Rightarrow k = \frac{1}{40} \]
The probability distribution table is:
| \( X \) | 1 | 2 | 3 | 4 |
|---|---|---|---|---|
| \( P(X) \) | \(\frac{2}{40}\) | \(\frac{6}{40}\) | \(\frac{12}{40}\) | \(\frac{20}{40}\) |
(b)
\[ E(X) = \frac{1 \times 2 + 2 \times 6 + 3 \times 12 + 4 \times 20}{40} = \frac{130}{40} = 3.25 \]
\[ \text{Var}(X) = \frac{1^2 \times 2 + 2^2 \times 6 + 3^2 \times 12 + 4^2 \times 20}{40} – (3.25)^2 = \frac{454}{40} – 10.5625 = 0.7875 \]
2. [Maximum mark: 6]
The weights of large bags of pasta produced by a company are normally distributed with mean \( 1.5 \, \text{kg} \) and standard deviation \( 0.05 \, \text{kg} \).
(a) Find the probability that a randomly chosen large bag of pasta weighs between \( 1.42 \, \text{kg} \) and \( 1.52 \, \text{kg} \). [3]
The weights of small bags of pasta produced by the company are normally distributed with mean \( 0.75 \, \text{kg} \) and standard deviation \( \sigma \, \text{kg} \). It is found that 68% of these small bags have weight less than \( 0.9 \, \text{kg} \).
(b) Find the value of \( \sigma \). [3]
▶️Answer/Explanation
(a)
Standardize the values:
\[ Z_1 = \frac{1.42 – 1.5}{0.05} = -1.6 \]
\[ Z_2 = \frac{1.52 – 1.5}{0.05} = 0.4 \]
Using standard normal tables:
\[ P(-1.6 < Z < 0.4) = \Phi(0.4) + \Phi(1.6) – 1 = 0.6554 + 0.9452 – 1 = 0.6006 \]
(b)
Given \( P(X < 0.9) = 0.68 \), find \( z \) such that \( \Phi(z) = 0.68 \):
\[ z \approx 0.468 \]
\[ \frac{0.9 – 0.75}{\sigma} = 0.468 \Rightarrow \sigma = \frac{0.15}{0.468} \approx 0.321 \, \text{kg} \]
3. [Maximum mark: 7]
Tim has two bags of marbles, A and B.
- Bag A contains 8 white, 4 red, and 3 yellow marbles.
- Bag B contains 6 white, 7 red, and 2 yellow marbles.
Tim rolls a fair 6-sided die. If he rolls a 1 or 2, he chooses two marbles from bag A without replacement. If he rolls a 3, 4, 5, or 6, he chooses two marbles from bag B without replacement.
(a) Find the probability that both marbles are white. [3]
(b) Find the probability that the two marbles come from bag B given that one is white and one is red. [4]
▶️Answer/Explanation
(a)
Probability of choosing bag A: \( \frac{2}{6} \). Probability of two white marbles from A:
\[ \frac{8}{15} \times \frac{7}{14} = \frac{56}{210} \]
Probability of choosing bag B: \( \frac{4}{6} \). Probability of two white marbles from B:
\[ \frac{6}{15} \times \frac{5}{14} = \frac{30}{210} \]
Total probability:
\[ \frac{2}{6} \times \frac{56}{210} + \frac{4}{6} \times \frac{30}{210} = \frac{58}{315} \approx 0.184 \]
(b)
Probability of one white and one red from bag B:
\[ 2 \times \frac{6}{15} \times \frac{7}{14} = \frac{84}{210} \]
Probability of one white and one red from bag A:
\[ 2 \times \frac{8}{15} \times \frac{4}{14} = \frac{64}{210} \]
Using conditional probability:
\[ P(\text{Bag B} \mid \text{one white and one red}) = \frac{\frac{4}{6} \times \frac{84}{210}}{\frac{4}{6} \times \frac{84}{210} + \frac{2}{6} \times \frac{64}{210}} = \frac{168}{232} \approx 0.724 \]
4. [Maximum mark: 10]
The weights, \( x \, \text{kg} \), of 120 students in a sports college are recorded. The results are summarized in the following table:
| Weight (\( x \, \text{kg} \)) | \( x \leq 40 \) | \( x \leq 60 \) | \( x \leq 65 \) | \( x \leq 70 \) | \( x \leq 85 \) | \( x \leq 100 \) |
|---|---|---|---|---|---|---|
| Cumulative frequency | 0 | 14 | 38 | 60 | 106 | 120 |
(a) Draw a cumulative frequency graph to represent this information. [2]
(b) It is found that 35% of the students weigh more than \( W \, \text{kg} \). Use your graph to estimate the value of \( W \). [2]
(c) Calculate estimates for the mean and standard deviation of the weights of the 120 students. [6]
▶️Answer/Explanation
(a) Plot the cumulative frequency points and join them with a smooth curve.
(b) 35% above \( W \) implies 65% below \( W \). From the graph, \( W \approx 76 \, \text{kg} \).
(c)
Frequencies: 14 (40-60), 24 (60-65), 22 (65-70), 46 (70-85), 14 (85-100).
Midpoints: 50, 62.5, 67.5, 77.5, 92.5.
Mean:
\[ \frac{14 \times 50 + 24 \times 62.5 + 22 \times 67.5 + 46 \times 77.5 + 14 \times 92.5}{120} = 71.2 \, \text{kg} \]
Variance:
\[ \frac{14 \times 50^2 + 24 \times 62.5^2 + 22 \times 67.5^2 + 46 \times 77.5^2 + 14 \times 92.5^2}{120} – (71.2)^2 = 138.23 \]
Standard deviation:
\[ \sqrt{138.23} \approx 11.8 \, \text{kg} \]
5. [Maximum mark: 6]
The probability that a driver passes an advanced driving test is 0.3 on any given attempt.
(a) Dipak keeps taking the test until he passes. The random variable \( X \) denotes the number of attempts required for Dipak to pass the test.
(i) Find \( P(2 \leq X \leq 6) \). [2]
(ii) Find \( E(X) \). [1]
(b) Five friends will each take their advanced driving test tomorrow. Find the probability that at least three of them will pass. [3]
▶️Answer/Explanation
(a)(i)
Geometric distribution:
\[ P(2 \leq X \leq 6) = (0.7 \times 0.3) + (0.7^2 \times 0.3) + \dots + (0.7^5 \times 0.3) = 0.582 \]
(a)(ii)
\[ E(X) = \frac{1}{0.3} \approx 3.33 \]
(b)
Binomial distribution with \( n = 5 \), \( p = 0.3 \):
\[ P(\text{at least 3 pass}) = \binom{5}{3}(0.3)^3(0.7)^2 + \binom{5}{4}(0.3)^4(0.7)^1 + \binom{5}{5}(0.3)^5(0.7)^0 = 0.163 \]
6. [Maximum mark: 10]
Jai and his wife Kaz are having a party. Jai has invited five friends, and each friend will bring his wife.
(a) At the beginning of the party, the 12 people will stand in a line for a photograph.
(i) How many different arrangements are there of the 12 people if Jai stands next to Kaz and each friend stands next to his own wife? [3]
(ii) How many different arrangements are there of the 12 people if Jai and Kaz occupy the two middle positions in the line, with Jai’s five friends on one side and the five wives of the friends on the other side? [2]
(b) For a competition during the party, the 12 people are divided at random into a group of 5, a group of 4, and a group of 3. Find the probability that Jai and Kaz are in the same group as each other. [5]
▶️Answer/Explanation
(a)(i)
Treat Jai-Kaz and each friend-wife pair as single units. There are 6 units, which can be arranged in \( 6! \) ways. Each pair can be arranged in 2 ways. Total arrangements:
\[ 6! \times 2^6 = 46080 \]
(a)(ii)
Jai and Kaz in the middle, friends on one side (5! arrangements), wives on the other side (5! arrangements). Total:
\[ 5! \times 5! \times 2 = 57600 \]
(b)
Total ways to divide 12 people into groups of 5, 4, and 3:
\[ \frac{12!}{5!4!3!} = 27720 \]
Favorable cases where Jai and Kaz are together in one of the groups:
– Both in group of 5: \( \binom{10}{3} \times \binom{7}{4} = 4200 \)
– Both in group of 4: \( \binom{10}{2} \times \binom{8}{5} = 2520 \)
– Both in group of 3: \( \binom{10}{1} \times \binom{9}{5} = 1260 \)
Total favorable: \( 4200 + 2520 + 1260 = 7980 \)
Probability:
\[ \frac{7980}{27720} = \frac{19}{66} \approx 0.288 \]