Question 1
(a) 5.4 Discrete random variables
(b) 5.4 Discrete random variables
Nicola throws an ordinary fair six-sided dice. The random variable X is the number of throws that she takes to obtain a 6.
(a) Find \( P(X < 8) \).
(b) Find the probability that Nicola obtains a 6 for the second time on her 8th throw.
▶️Answer/Explanation
Solution :-
Part (a): Finding \( P(X < 8) \)
Method 1:
\( P(X < 8) = 1 – \left( \frac{5}{6} \right)^7 \)
\( P(X < 8) = 1 – 0.2790816… \)
\( P(X < 8) = 0.721 \)
Alternative exact form: \( \frac{201811}{279936} \)
Method 2:
\( P(X < 8) = \frac{1}{6} + \left( \frac{5}{6} \right) \frac{1}{6} + \left( \frac{5}{6} \right)^2 \frac{1}{6} + \left( \frac{5}{6} \right)^3 \frac{1}{6} + \left( \frac{5}{6} \right)^4 \frac{1}{6} + \left( \frac{5}{6} \right)^5 \frac{1}{6} + \left( \frac{5}{6} \right)^6 \frac{1}{6} \)
\( P(X < 8) = 0.721 \)
Alternative exact form: \( \frac{201811}{279936} \)
Part (b): Finding the probability of the second 6 on the 8th throw.
Probability = \( \left( \frac{5}{6} \right)^6 \times \left( \frac{1}{6} \right)^2 \times 7 \)
Probability = 0.0651
Question 2
(a) Topic: 5.4 Discrete random variables
(b) Topic: 5.4 Discrete random variables
The random variable \(X\) takes the values \(-2, -1, 0, 2, 3\). It is given that \(P(X = x) = k(x^2 + 2)\), where \(k\) is a positive constant.
(a) Draw up the probability distribution table for \(X\), giving the probabilities as numerical fractions.
(b) Find the value of \(\text{Var}(X)\).
▶️Answer/Explanation
Solution :-
Part (a): Probability Distribution Table
Probabilities are \(6k, 3k, 2k, 6k, 11k\) so \(28k = 1\).
\(k = \frac{1}{28}\)
| \(x\) | -2 | -1 | 0 | 2 | 3 |
|---|---|---|---|---|---|
| \(P(X=x)\) | \(\frac{6}{28}\) | \(\frac{3}{28}\) | \(\frac{2}{28}\) | \(\frac{6}{28}\) | \(\frac{11}{28}\) |
| Decimal | 0.2143 | 0.1071 | 0.0714 | 0.2143 | 0.3929 |
Part (b): Variance of \(X\) (\(\text{Var}(X)\))
\[ E(X) = \left[ (-2) \times \frac{6}{28} + (-1) \times \frac{3}{28} + 0 \times \frac{2}{28} + 2 \times \frac{6}{28} + 3 \times \frac{11}{28} \right] \]
\[ E(X) = \frac{1}{28} (-12 – 3 + 12 + 33) = \frac{15}{14} \]
\[ \text{Var}(X) = \frac{6(-2)^2 + 3(-1)^2 + 2(0)^2 + 6(2)^2 + 11(3)^2}{28} – \left(\frac{15}{14}\right)^2 \]
\(\text{Var}(X) = 4.21\) or \(4 \frac{41}{196}\)
Question 3
(a) Topic: 5.1 Representation of data
(b) Topic: 5.1 Representation of data
(c) Topic: 5.1 Representation of data
The time taken, in minutes, to walk to school was recorded for 200 pupils at a certain school. These times are summarised in the following table.
Time taken (minutes): \(t \le 15\), \(t \le 25\), \(t \le 30\), \(t \le 40\), \(t \le 50\), \(t \le 70\)
Cumulative frequency: 18, 46, 88, 140, 176, 200
(a) Draw a cumulative frequency graph to illustrate the data.
(b) Use your graph to estimate the median and the interquartile range of the data.
(c) Calculate an estimate for the mean value of the times taken by the 200 pupils to walk to school.
▶️Answer/Explanation
Solution :-
Part (a): Cumulative Frequency Graph
Plot at least 4 points within tolerance at upper bounds. Use linear cf scale 0 to 200 and linear time scale 0 to 70, with at least 3 values identified on each. Minimum scale uses at least ½ the grid.
All points plotted correctly. Curve drawn and joined to (0, 0). Axes labelled cumulative frequency (cf), time (t) and minutes (min) or a suitable title.
Part (b): Median and Interquartile Range
Median = 33 (Must be identified. Evidence of use of graph must be seen. Strict FT ± ½ square on time axis.)
[IQR =] 42 – 26 (41 < LQ < 43 – 25 < LQ < 27. If outside of range FT ± ½ square on time axis.)
16
Part (c): Estimate for the Mean
Midpoint: 7.5, 20, 27.5, 35, 45, 60 (At least 5 correct midpoints)
Frequency: 18, 28, 42, 52, 36, 24 (5 correct frequencies seen.)
Mean = \(\frac{18 \times 7.5 + 28 \times 20 + 42 \times 27.5 + 52 \times 35 + 36 \times 45 + 24 \times 60}{200}\) (Correct mean formula using their 6 midpoints (must be within class, not upper bound, not lower bound) condone 1 error and their 6 frequencies (not cumulative frequencies).)
Mean = 33.65, 33 \(\frac{13}{20}\) (Accept 33.7, not \(\frac{673}{20}\))
Question 4
(a) Topic: 5.3 Probability
(b) Topic: 5.3 Probability
Rahul has two bags, X and Y. Bag X contains 4 red marbles and 2 blue marbles. Bag Y contains 3 red marbles and 4 blue marbles. Rahul also has a coin which is biased so that the probability of obtaining a head when it is thrown is 1/4.
Rahul throws the coin.
- If he obtains a head, he chooses at random a marble from bag X. He notes the colour and replaces the marble in bag X. He then chooses at random a second marble from bag X.
- If he obtains a tail, he chooses at random a marble from bag Y. He notes the colour and discards the marble. He then chooses at random a second marble from bag Y.
(a) Find the probability that the two marbles that Rahul chooses are the same colour.
(b) Find the probability that the two marbles that Rahul chooses are both from bag Y given that both marbles are blue.
▶️Answer/Explanation
Solution :-
Part (a): Probability of same colour marbles
P(HRR) = \(\frac{1}{4} \times \frac{4}{6} \times \frac{4}{6} = \frac{16}{144} = \frac{4}{36}\)
P(TRR) = \(\frac{3}{4} \times \frac{3}{7} \times \frac{2}{6} = \frac{18}{168} = \frac{3}{28}\)
P(HBB) = \(\frac{1}{4} \times \frac{2}{6} \times \frac{2}{6} = \frac{4}{144} = \frac{1}{36}\)
P(TBB) = \(\frac{3}{4} \times \frac{4}{7} \times \frac{3}{6} = \frac{36}{168} = \frac{6}{28} = \frac{3}{14}\)
\(\frac{4}{36} + \frac{3}{28} + \frac{1}{36} + \frac{6}{28} = \frac{29}{63}\) or 0.460
Part (b): Probability of two marbles from Y given both are blue
P(T|BB) = \(\frac{P(T \cap BB)}{P(BB)}\)
P(T|BB) = \(\frac{\frac{3}{4} \times \frac{4}{7} \times \frac{3}{6}}{\frac{1}{36} + \frac{6}{28}}\)
P(T|BB) = \(\frac{\frac{3}{14}}{\frac{61}{252}}\)
P(T|BB) = \(\frac{54}{61}\) or 0.885
Question 5
(a) Topic: 5.5 The normal distribution
(b) Topic: 5.5 The normal distribution
The weights of the green apples sold by a shop are normally distributed with mean 90 grams and standard deviation 8 grams.
(a) Find the probability that a randomly chosen green apple weighs between 83 grams and 95 grams.
(b) The shop also sells red apples. 60% of the red apples sold by the shop weigh more than 80 grams. 160 red apples are chosen at random from the shop. Use a suitable approximation to find the probability that fewer than 105 of the chosen red apples weigh more than 80 grams.
▶️Answer/Explanation
Solution :-
Part (a): Probability for Green Apples
P(83 < X < 95) = P(\(\frac{83 – 90}{8}\) < Z < \(\frac{95 – 90}{8}\))
P(83 < X < 95) = P(-0.875 < Z < 0.625)
P(83 < X < 95) = [Φ(0.625) + Φ(0.875) – 1]
P(83 < X < 95) = 0.7340 + 0.8092 – 1
P(83 < X < 95) = 0.543
Part (b): Probability for Red Apples
Mean = 160 × 0.6 = 96
Var = 160 × 0.6 × 0.4 = 38.4
P(X < 105) = P(Z < \(\frac{104.5 – 96}{\sqrt{38.4}}\))
P(X < 105) = P(Z < 1.372) = Φ(1.372)
P(X < 105) = 0.915[0]
Question 6
(a) Topic: 5.4 Discrete random variables and
(b) Topic: 5.5 The normal distribution
The heights of the female students at Breven college are normally distributed.
- 90% of the female students have heights less than 182.7 cm.
- 40% of the female students have heights less than 162.5 cm.
(a) Find the mean and the standard deviation of the heights of the female students at Breven college.
(b) Ten female students are chosen at random from those at Breven college. Find the probability that fewer than 8 of these 10 students have heights more than 162.5 cm.
▶️Answer/Explanation
Solution :-
Part (a): Mean and Standard Deviation
\(\frac{182.7 – \mu}{\sigma}\) = 1.282
\(\frac{162.5 – \mu}{\sigma}\) = -0.253
Solve, obtaining values for μ and σ
μ = 165.8
σ = 13.2
Part (b): Probability for Binomial Distribution
Method 1:
[P(X < 8) =] 1 – [P(8, 9, 10)] = 1 – [\({}^{10}C_8\) (0.6)⁸ (0.4)² + \({}^{10}C_9\) (0.6)⁹ (0.4)¹ + (0.6)¹⁰]
[= 1 – (0.12093 + 0.040311 + 0.0060466)]
= 0.833
Method 2:
(0.4)¹⁰ + \({}^{10}C_1\) (0.6)¹ (0.4)⁹ + \({}^{10}C_2\) (0.6)² (0.4)⁸ + \({}^{10}C_3\) (0.6)³ (0.4)⁷ + \({}^{10}C_4\) (0.6)⁴ (0.4)⁶ + \({}^{10}C_5\) (0.6)⁵ (0.4)⁵ + \({}^{10}C_6\) (0.6)⁶ (0.4)⁴ + \({}^{10}C_7\) (0.6)⁷ (0.4)³
[1.0486 × 10⁻⁴ + 1.5729 × 10⁻³ + … + 0.21499]
= 0.833
Question 7
(a) Topic: 5.2 Permutations and combinations
(b) Topic: 5.2 Permutations and combinations
(c) Topic: 5.2 Permutations and combinations
The word INTELLECT has 9 letters.
(a) How many different arrangements are there of the 9 letters in the word INTELLECT in which the two Ts are together?
(b) How many different arrangements are there of the 9 letters in the word INTELLECT in which there is a T at each end and the two Es are not next to each other?
Four letters are selected at random from the 9 letters in the word INTELLECT.
(c) Find the percentage of the possible selections which contain at least one E and exactly one T.
▶️Answer/Explanation
Solution :-
Part (a): Arrangements with two Ts together
\[\frac{8!}{2! \times 2!} = 10080\]
Part (b): Arrangements with Ts at ends and Es not together
Method 1:
Number of ways with no restriction on Es – ways with Es together
\[\frac{7!}{2! \times 2!} – \frac{6!}{2!} = 1260 – 360 = 900\]
Method 2:
\(T \_ \_ \_ \_ \_ \_ T\) with Es inserted in gaps
\[\frac{5!}{2!} \times \frac{6 \times 5}{2} = 60 \times 15 = 900\]
Part (c): Selections with at least one E and exactly one T
Method 1: addition
\[T E \_ \_ = {}^2C_1 \times {}^2C_1 \times {}^7C_2 = 40\]
\[T E E \_ = {}^2C_1 \times {}^2C_2 \times {}^7C_1 = 10\]
Probability = \(\frac{40 + 10}{ {}^9C_4 }\) = \(\frac{50}{126}\)
Percentage = \(\frac{50}{126} \times 100 = 39.7\%\)
Method 2: subtraction
\[T \_ \_ \_ = {}^2C_1 \times {}^7C_3 = 70\]
\[T * * * = {}^2C_1 \times {}^7C_4 = 20\]
Probability = \(\frac{70 – 20}{ {}^9C_4 }\) = \(\frac{50}{126}\)
Percentage = \(\frac{50}{126} \times 100 = 39.7\%\)