Question 1
(a) Topic: 5.3 Probability
(b) Topic: 5.5 The Normal Distribution
30% of the residents of Wimfield own an electric car. Three residents are chosen at random.
(a) Find the probability that either all three own an electric car or none of them owns an electric car.
(b) A random sample of 125 of the residents of Wimfield is selected. Use a suitable approximation to find the probability that more than 45 of these residents own an electric car.
▶️Answer/Explanation
Solution :-
Part (a): Probability of all or none owning an electric car
Probability that all three own an electric car: \( (0.3)^3 \)
Probability that none of them own an electric car: \( (0.7)^3 \)
Total probability: \( (0.3)^3 + (0.7)^3 = 0.027 + 0.343 = 0.37 \)
Therefore, the probability is \( 0.37 \) or \( \frac{37}{100} \).
Part (b): Probability using Normal Approximation
Mean: \( 125 \times 0.3 = 37.5 \)
Variance: \( 125 \times 0.3 \times 0.7 = 26.25 \)
Standard Deviation: \( \sqrt{26.25} \approx 5.12 \) (or \( \frac{\sqrt{105}}{2} \))
Using the normal approximation with continuity correction:
\( P(X > 45) = P\left(Z > \frac{45.5 – 37.5}{\sqrt{26.25}}\right) \)
\( Z = \frac{8}{\sqrt{26.25}} \approx 1.5614 \)
\( P(Z > 1.5614) = 1 – \Phi(1.5614) \approx 1 – 0.9407 = 0.0593 \)
Therefore, the probability is approximately \( 0.0593 \).
Question 2
(a) 5.4 Discrete Random Variables
(b) 5.4 Discrete Random Variables
A red fair six-sided dice has faces labelled 1, 1, 1, 2, 2, 2. A blue fair six-sided dice has faces labelled 1, 1, 2, 2, 3, 3. Both dice are thrown. The random variable X is the product of the scores on the two dice.
(a) Draw up the probability distribution table for X.
(b) Find E(X).
▶️Answer/Explanation
Solution :-
Part (a): Probability Distribution Table for X
| x | 1 | 2 | 3 | 4 | 6 |
|---|---|---|---|---|---|
| P(X=x) | \( \frac{6}{36} \) | \( \frac{12}{36} \) | \( \frac{6}{36} \) | \( \frac{6}{36} \) | \( \frac{6}{36} \) |
| \( \frac{1}{6} \) | \( \frac{1}{3} \) | \( \frac{1}{6} \) | \( \frac{1}{6} \) | \( \frac{1}{6} \) | |
| 0.167 | 0.333 | 0.167 | 0.167 | 0.167 |
Part (b): Finding E(X)
\[ E(X) = \frac{1}{36} (6 + 24 + 18 + 24 + 36) = 3 \]
Question 3
(a) 5.5 The Normal Distribution
(b) 5.5 The Normal Distribution
In Molimba, the heights, in cm, of adult males are normally distributed with mean 176 cm and standard deviation 4.8 cm.
(a) Find the probability that a randomly chosen adult male in Molimba has a height greater than 170 cm.
60% of adult males in Molimba have a height between 170 cm and \( k \) cm, where \( k \) is greater than 170.
(b) Find the value of \( k \), giving your answer correct to 1 decimal place.
▶️Answer/Explanation
Solution :-
Part (a): Finding Probability
Using the standardization formula:
\[ P(X > 170) = P\left(Z > \frac{170 – 176}{4.8}\right) \]
\[ Z = \frac{170 – 176}{4.8} = -1.25 \]
From the normal distribution table:
\[ P(Z > -1.25) = 0.894 \]
Thus, the probability that a randomly chosen adult male in Molimba has a height greater than 170 cm is 0.894.
Part (b): Finding \( k \)
We are given that 60% of adult males have a height between 170 cm and \( k \) cm.
First, find \( P(h < 170) \):
\[ P(h < 170) = 1 – 0.894 = 0.1056 \]
Now, find \( k \) using the standardization formula:
\[ \frac{k – 176}{4.8} = \Phi^{-1}(0.1056 + 0.6) \]
From the normal distribution table:
\[ \Phi^{-1}(0.7056) = 0.541 \]
Solving for \( k \):
\[ k = 176 + (0.541 \times 4.8) \]
\[ k = 178.6 \]
Thus, the value of \( k \) is 178.6 cm (correct to 1 decimal place).
Question 4
(a) 5.1 Representation of Data
(b) 5.1 Representation of Data
(c) 5.1 Representation of Data
On a certain day, the heights of 150 sunflower plants grown by children at a local school are measured, correct to the nearest cm. These heights are summarised in the following table.
| Height (cm) | 10-19 | 20-29 | 30-39 | 40-44 | 45-49 | 50-54 | 55-59 |
|---|---|---|---|---|---|---|---|
| Frequency | 10 | 18 | 32 | 42 | 28 | 14 | 6 |
(a) Draw a cumulative frequency graph to illustrate the data.
(b) Use your graph to estimate the 30th percentile of the heights of the sunflower plants.
(c) Calculate estimates for the mean and the standard deviation of the heights of the 150 sunflower plants.
▶️Answer/Explanation
Solution :-
(a) Cumulative Frequency Graph:
Graph Explanation:
- B1: Correct cumulative frequency values seen (10, 28, 60, 102, 130, 144, 150).
- B1: Linearly scaled axes correctly labeled (cumulative frequency and height).
- M1: At least 4 points plotted at upper boundary +0.5.
- A1: All points plotted correctly, curve drawn.
(b) 30th Percentile:
- \( 150 \times 0.3 = 45 \)
- Line drawn from 45 on cf axis to meet graph at \( h = 36 \)
Graph Explanation:
- M1: Use of graph must be seen.
- A1 FT: Must be an increasing cf graph. Expect an answer in range 35 ≤ h ≤ 37 for a correct graph.
(c) Mean and Standard Deviation:
- Midpoints: 14.5, 24.5, 34.5, 42, 47, 52, 57
- Mean: \( \frac{10 \times 14.5 + 18 \times 24.5 + 32 \times 34.5 + 42 \times 42 + 28 \times 47 + 14 \times 52 + 6 \times 57}{150} = \frac{5840}{150} = 38.9 \)
- Standard Deviation: \( \sqrt{\frac{\sum fx^2}{n} – \bar{x}^2} = \sqrt{\frac{244285}{150} – (\frac{5840}{150})^2} = \sqrt{112.76 – 15.13} = \sqrt{112.76 – 38.9^2} = \sqrt{112.76 – 1513.21} = \sqrt{112.76-1513.21} = 10.6 \)
Calculations and Explanations:
- B1: At least 6 correct midpoints seen.
- M1: Correct unsimplified mean formula with their midpoints.
- A1: Accept answers wrt 38.9.
- M1: Correct unsimplified variance formula with their midpoint.
- A1: AWRT 10.6.
Question 5
(a) 5.3 Probability
(b) 5.3 Probability
(c) 5.3 Probability
A factory produces chocolates. 30% of the chocolates are wrapped in gold foil, 25% are wrapped in red foil and the remainder are unwrapped.
Indigo chooses 8 chocolates at random from the production line.
(a) Find the probability that she obtains no more than 2 chocolates that are wrapped in red foil.
Jake chooses chocolates one at a time at random from the production line.
(b) Find the probability that the first time he obtains a chocolate that is wrapped in red foil is before the 7th choice.
Keifu chooses chocolates one at a time at random from the production line.
(c) Find the probability that the second chocolate chosen is the first one wrapped in gold foil given that the fifth chocolate chosen is the first unwrapped chocolate.
▶️Answer/Explanation
Solution :-
(a) Probability of no more than 2 red foil chocolates
Method 1
[P(0, 1, 2)] = 8C2(0.75)6(0.25)2 + 8C1(0.75)7(0.25)1 + (0.75)8
[~0.31146 + 0.26697 + 0.10011] = 0.679
Method 2
[P(0, 1, 2)] = 1 – [P(3, 4, 5, 6, 7, 8)] = 1 – [8C3(0.75)5(0.25)3 + 8C4(0.75)4(0.25)4 + 8C5(0.75)3(0.25)5 + 8C6(0.75)2(0.25)6 + 8C7(0.75)(0.25)7 + (0.25)8]
= 0.679
(b) Probability of first red foil before 7th choice
Method 1
1 – 0.756 = 0.822, 3367/4096
Method 2
0.25 + 0.25 × 0.75 + 0.25 × 0.752 + 0.25 × 0.753 + 0.25 × 0.754 + 0.25 × 0.755 = 0.822
Method 3
1 – 0.756 – 0.25 × 0.756 = 0.822
(c): Conditional probability
Method 1
P(2nd gold | 5th is first unwrapped). R G R or G R or G U
0.25 × 0.3 × 0.55 × 0.55 × 0.45 [= 0.01021]
Method 2
P(2nd gold | 5th is first unwrapped) 4 possible scenarios
\[ 0.25 \times 0.3 \times 0.25 \times 0.3 \times 0.45 = 0.00253125 \]
\[ 0.25 \times 0.3 \times 0.25 \times 0.25 \times 0.45 = 0.002109375 \]
\[ 0.25 \times 0.3 \times 0.3 \times 0.25 \times 0.45 = 0.00253125 \]
\[ 0.25 \times 0.3 \times 0.3 \times 0.3 \times 0.45 = 0.0030375 \]
[Total 0.010209375]
For either approach
\[ P(5th \text{ is first unwrapped}) = (0.55)^4 (0.45) = 0.041178 \]
\[ P(2nd \text{ is first gold} | 5th \text{ is first unwrapped}) = \frac{0.010209375}{0.0411778125} \]
\[ = \frac{30}{121} = 0.248 \]
Method 3
First chocolate is red and second gold
\[ P(\text{red given that it is wrapped}) \times P(\text{gold given that it is wrapped}) = \frac{0.25}{0.55} \times \frac{0.3}{0.55} \]
\[ = \frac{30}{121} = 0.248 \]
Question 6
(a) 5.2 Permutations and Combinations
(b) 5.2 Permutations and Combinations
(c) 52. Permutations and Combinations
(d) 5.3 Probability
The word is HAPPINESS.
(a) Find the number of different arrangements of the 9 letters in the word HAPPINESS.
(b) Find the number of different arrangements of the 9 letters in the word HAPPINESS in which the first and last letters are not the same as each other.
(c) Find the number of different arrangements of the 9 letters in the word HAPPINESS in which the two Ps are together and there are exactly two letters between the two Ss.
The 9 letters in the word HAPPINESS are divided at random into a group of 5 and a group of 4.
(d) Find the probability that both Ps are in one group and both Ss are in the other group.
▶️Answer/Explanation
Solution:
(a)
The word HAPPINESS has 9 letters with 2 Ps and 2 Ss.
Number of arrangements = \( \frac{9!}{2! \times 2!} = 90720 \)
(b)
Method 1: Total arrangements – arrangements with repeated letters at ends.
Total arrangements = \( \frac{9!}{2! \times 2!} = 90720 \)
Arrangements with repeated letters at ends:
If both ends are P: \( \frac{7!}{2!} = 2520 \)
If both ends are S: \( \frac{7!}{2!} = 2520 \)
Total with repeated letters at ends = \( 2520 \times 2 = 5040 \)
Arrangements with different letters at ends = \( 90720 – 5040 = 85680 \)
(c)
Method 1: Arrangements with PP between Ss \([S P P S \wedge \wedge \wedge \wedge]\) add arrangements with PP not between Ss \([S \wedge \wedge S \wedge P P \wedge]\)
\(6! + 5 \times 5 \times 4\)
Total = 3120
(d)
Method 1:
Either PP in the group of 5 or PP in the group of 4
\( \frac{\binom{2}{2} \times \binom{7}{3}}{\binom{9}{5}} + \frac{\binom{2}{2} \times \binom{7}{4}}{\binom{9}{4}} \)
Probability = \( \frac{20}{126} = \frac{10}{63} = 0.159 \)
Method 2:
Considering the positions of P and then S
\( \left( \frac{5}{9} \times \frac{4}{8} \times \frac{4}{7} \times \frac{3}{6} \right) + \left( \frac{5}{7} \times \frac{4}{6} \times \frac{4}{9} \times \frac{3}{8} \right) \)
= \( \frac{10}{63} \)