Unit 18 Exam Practice
Discrete Random Variables and the Binomial Distribution
These questions are written in the style of IB examination questions, and they are harder than the review exercises. The marks for each part are shown in square brackets on the right, for example [3]. Paper 1 means no calculator; Paper 2 means a graphic display calculator (GDC) is allowed. Try each question under exam conditions first, then open the markscheme.
| Mark | Meaning |
|---|---|
| M | Method mark: a correct method is used, even with a small slip |
| A | Accuracy mark: a correct answer or line of working (usually needs the M mark before it) |
| R | Reasoning mark: a clear reason or justification is given |
| N | Mark for a correct final answer with no working shown |
| AG | “Answer given” in the question, so no extra mark is awarded for writing it |
The number after the letter is the number of marks, so A2 is worth 2 marks. In the exam, give answers exactly or correct to 3 significant figures (s.f.) unless the question says otherwise.
SL questions
SL 1 Paper 1
The probability distribution of a discrete random variable \(X\) is shown in the table, where \(k\) is a constant.
| \(x\) | \(1\) | \(2\) | \(3\) |
|---|---|---|---|
| \(\mathrm{P}(X = x)\) | \(2k^{2}\) | \(k\) | \(4k^{2}\) |
(a) Find the value of \(k\). [3]
(b) Find \(\mathrm{E}(X)\). [2]
Two independent observations of \(X\) are made.
(c) Find the probability that the total of the two observations is \(4\). [3]
Total 8 marks
(a)
\(2k^{2} + k + 4k^{2} = 1 \;\Rightarrow\; 6k^{2} + k - 1 = 0\) M1
\((3k - 1)(2k + 1) = 0\), so \(k = \dfrac{1}{3}\) or \(k = -\dfrac{1}{2}\) A1
\(\mathrm{P}(X = 2) = k\) cannot be negative, so \(k = \dfrac{1}{3}\) R1
Do not award the R1 if \(k = -\dfrac{1}{2}\) is not rejected, or is rejected with no reason.
[3 marks]
(b)
\(\mathrm{E}(X) = 1 \times \dfrac{2}{9} + 2 \times \dfrac{1}{3} + 3 \times \dfrac{4}{9}\) M1
\(= \dfrac{2 + 6 + 12}{9} = \dfrac{20}{9}\) A1
[2 marks]
(c)
a total of \(4\) comes from \((1, 3)\), \((3, 1)\) or \((2, 2)\) M1
\(2 \times \dfrac{2}{9} \times \dfrac{4}{9} + \left(\dfrac{1}{3}\right)^{2} = \dfrac{16}{81} + \dfrac{9}{81}\) A1
\(= \dfrac{25}{81}\) A1
[3 marks]
See 18.1.
SL 2 Paper 1
A spinner has three coloured sectors. When it is spun, the probability that it lands on red is \(\dfrac{1}{2}\), on green is \(\dfrac{1}{4}\) and on blue is \(\dfrac{1}{4}\). In a game, the spinner is spun twice. If both spins land on the same colour, the organiser pays the player $\(k\). Otherwise, the player pays the organiser $3.
(a) Show that the probability that both spins land on the same colour is \(\dfrac{3}{8}\). [2]
(b) Find the value of \(k\) for which the game is fair. [2]
The rules are changed. If both spins land on red, the organiser pays the player $2. If both spins land on green, or both land on blue, the organiser pays the player $\(m\). Otherwise, the player pays the organiser $3.
(c) Find the value of \(m\) for which the new game is fair. [3]
Total 7 marks
(a)
\(\mathrm{P}(\text{same}) = \mathrm{P}(RR) + \mathrm{P}(GG) + \mathrm{P}(BB)\) M1
\(= \dfrac{1}{4} + \dfrac{1}{16} + \dfrac{1}{16} = \dfrac{6}{16} = \dfrac{3}{8}\) A1 AG
Award M1 A0 for \(\left(\dfrac{1}{2}\right)^{2} + \left(\dfrac{1}{4}\right)^{2} + \left(\dfrac{1}{4}\right)^{2}\) with no further working.
[2 marks]
(b)
\(\mathrm{E}(\text{gain}) = \dfrac{3}{8}k - \dfrac{5}{8} \times 3 = 0\) M1
\(k = 5\) A1
[2 marks]
(c)
\(\mathrm{P}(RR) = \dfrac{1}{4}\), \(\mathrm{P}(GG \text{ or } BB) = \dfrac{1}{8}\), \(\mathrm{P}(\text{different}) = \dfrac{5}{8}\) A1
\(2 \times \dfrac{1}{4} + m \times \dfrac{1}{8} - 3 \times \dfrac{5}{8} = 0\) M1
\(\dfrac{m}{8} = \dfrac{15}{8} - \dfrac{4}{8}\), so \(m = 11\) A1
[3 marks]
See 18.1.
SL 3 Paper 2
On average, \(62\%\) of a tennis player’s first serves are “in”. In one set she hits \(20\) first serves. Let \(X\) be the number of these first serves that are “in”.
(a) State one assumption that is needed for \(X\) to be modelled by a binomial distribution. [1]
(b) Find the probability that fewer than \(12\) of her first serves are “in”. [2]
(c) Given that at least \(10\) of her first serves are “in”, find the probability that at most \(14\) of them are “in”. [4]
Total 7 marks
(a)
the probability that a first serve is “in” is \(0.62\) for every serve (or: the serves are independent of each other) R1
[1 mark]
(b)
\(X \sim \mathrm{B}(20,\, 0.62)\), \(\;\mathrm{P}(X < 12) = \mathrm{P}(X \le 11)\) M1
\(= 0.334\) \((0.334082\ldots)\) (using Binomial Cdf on a GDC) A1
[2 marks]
(c)
\(\mathrm{P}(X \le 14 \mid X \ge 10) = \dfrac{\mathrm{P}(10 \le X \le 14)}{\mathrm{P}(X \ge 10)}\) M1
\(\mathrm{P}(10 \le X \le 14) = 0.740601\ldots\) A1
\(\mathrm{P}(X \ge 10) = 0.907671\ldots\) A1
\(= 0.816\) \((0.815936\ldots)\) A1
[4 marks]
See 18.3.
SL 4 Paper 2
An aircraft has \(180\) seats. The airline knows that each person who has a ticket for a flight turns up for it with probability \(0.94\), independently of all other ticket holders. For one flight, the airline sells \(185\) tickets. Let \(X\) be the number of ticket holders who turn up.
(a) Find the mean and the standard deviation of \(X\). [3]
(b) Find the probability that more than \(180\) ticket holders turn up. [2]
(c) Given that there are enough seats for everyone who turns up, find the probability that at least \(175\) ticket holders turn up. [3]
The airline wants the probability that more than \(180\) ticket holders turn up to be less than \(0.05\).
(d) Find the greatest number of tickets that the airline can sell for a flight. [4]
The airline sells \(185\) tickets for each of \(30\) such flights. Assume that the flights are independent.
(e) Find the probability that on at least one of these flights, more than \(180\) ticket holders turn up. [2]
(f) Many people travel with their families. Comment on the validity of the model in this context. [1]
Total 15 marks
(a)
\(X \sim \mathrm{B}(185,\, 0.94)\), \(\;\mathrm{E}(X) = 185 \times 0.94 = 173.9\) A1
\(\mathrm{Var}(X) = 185 \times 0.94 \times 0.06 = 10.434\) M1
standard deviation \(= \sqrt{10.434} = 3.23\) \((3.23017\ldots)\) A1
[3 marks]
(b)
\(\mathrm{P}(X > 180) = \mathrm{P}(X \ge 181) = 1 - \mathrm{P}(X \le 180)\) M1
\(= 0.0121\) \((0.0121378\ldots)\) (using Binomial Cdf on a GDC) A1
[2 marks]
(c)
\(\mathrm{P}(X \ge 175 \mid X \le 180) = \dfrac{\mathrm{P}(175 \le X \le 180)}{\mathrm{P}(X \le 180)}\) M1
\(= \dfrac{0.431824\ldots}{0.987862\ldots}\) A1
\(= 0.437\) \((0.437130\ldots)\) A1
[3 marks]
(d)
let \(n\) tickets be sold; we need \(\mathrm{P}(X \ge 181) < 0.05\), that is \(\mathrm{P}(X \le 180) > 0.95\) M1
Inverse Binomial N on a GDC (Cumulative Prob \(0.95\), Prob Success \(0.94\), Successes \(180\)) gives \(187\), the least \(n\) with \(\mathrm{P}(X \le 180) < 0.95\) M1
\(n = 186\): \(\mathrm{P}(X \le 180) = 0.970\); \(\;n = 187\): \(\mathrm{P}(X \le 180) = 0.936\) A1
the greatest number of tickets is \(186\) A1
Accept a table of \(\mathrm{P}(X \le 180)\) against \(n\) for the second M1. The GDC gives the least \(n\) with \(\mathrm{P}(X \le 180) < 0.95\), so the answer is one less. Do not award the final A1 for \(187\).
[4 marks]
(e)
\(Y \sim \mathrm{B}(30,\, 0.0121378\ldots)\), \(\;\mathrm{P}(Y \ge 1) = 1 - (1 - 0.0121378\ldots)^{30}\) M1
\(= 0.307\) \((0.306749\ldots)\) A1
[2 marks]
(f)
members of a family usually all turn up or all fail to turn up, so the ticket holders are not independent and the binomial model may not be valid R1
[1 mark]
SL 5 Paper 1
A bag contains \(3\) white balls and \(2\) black balls. In a game, balls are taken from the bag at random, one at a time and without replacement, until a black ball is taken. Let \(X\) be the number of balls taken, including the black ball.
(a) Show that \(\mathrm{P}(X = 3) = \dfrac{1}{5}\). [2]
(b) Find the probability distribution of \(X\). [2]
(c) Find \(\mathrm{E}(X)\). [2]
A player pays $\(c\) to play the game and then receives $\(2^{X}\).
(d) Find the value of \(c\) for which the game is fair. [3]
The game is played \(5\) times.
(e) Find the probability that the first ball taken is black in exactly \(2\) of these games. [3]
The game is played \(50\) times. Let \(Y\) be the number of these games in which the first ball taken is black.
(f) Find \(\mathrm{E}(Y)\) and \(\mathrm{Var}(Y)\). [2]
Total 14 marks
(a)
\(X = 3\) means white, white, black M1
\(\mathrm{P}(X = 3) = \dfrac{3}{5} \times \dfrac{2}{4} \times \dfrac{2}{3} = \dfrac{12}{60} = \dfrac{1}{5}\) A1 AG
[2 marks]
(b)
\(\mathrm{P}(X = 1) = \dfrac{2}{5}\), \(\;\mathrm{P}(X = 2) = \dfrac{3}{5} \times \dfrac{2}{4} = \dfrac{3}{10}\) A1
\(\mathrm{P}(X = 4) = \dfrac{3}{5} \times \dfrac{2}{4} \times \dfrac{1}{3} = \dfrac{1}{10}\) A1
| \(x\) | \(1\) | \(2\) | \(3\) | \(4\) |
|---|---|---|---|---|
| \(\mathrm{P}(X = x)\) | \(\frac{2}{5}\) | \(\frac{3}{10}\) | \(\frac{1}{5}\) | \(\frac{1}{10}\) |
[2 marks]
(c)
\(\mathrm{E}(X) = 1 \times \dfrac{4}{10} + 2 \times \dfrac{3}{10} + 3 \times \dfrac{2}{10} + 4 \times \dfrac{1}{10}\) M1
\(= \dfrac{4 + 6 + 6 + 4}{10} = 2\) A1
[2 marks]
(d)
\(\mathrm{E}\left(2^{X}\right) = 2 \times \dfrac{4}{10} + 4 \times \dfrac{3}{10} + 8 \times \dfrac{2}{10} + 16 \times \dfrac{1}{10}\) M1
\(= \dfrac{8 + 12 + 16 + 16}{10} = 5.2\) A1
fair when \(\mathrm{E}\left(2^{X}\right) - c = 0\), so \(c = 5.20\) ($5.20) A1
Do not award the M1 for \(2^{\mathrm{E}(X)} = 4\).
[3 marks]
(e)
the number of such games is \(W \sim \mathrm{B}\left(5, \dfrac{2}{5}\right)\) M1
\(\mathrm{P}(W = 2) = \dbinom{5}{2}\left(\dfrac{2}{5}\right)^{2}\left(\dfrac{3}{5}\right)^{3} = 10 \times \dfrac{4}{25} \times \dfrac{27}{125}\) A1
\(= \dfrac{1080}{3125} = \dfrac{216}{625}\) \((0.3456)\) A1
[3 marks]
(f)
\(Y \sim \mathrm{B}(50,\, 0.4)\), \(\;\mathrm{E}(Y) = 50 \times 0.4 = 20\) A1
\(\mathrm{Var}(Y) = 50 \times 0.4 \times 0.6 = 12\) A1
[2 marks]
HL questions
HL questions on this unit also use the AHL content: the variance of a discrete random variable, \(\mathrm{E}(aX + b) = a\,\mathrm{E}(X) + b\) and \(\mathrm{Var}(aX + b) = a^{2}\,\mathrm{Var}(X)\), and some combine the binomial distribution with conditional probability and Bayes’ theorem from Unit 16.
HL 1 Paper 1
The discrete random variable \(X\) has probability distribution \[ \mathrm{P}(X = x) = a + bx, \qquad x = -1,\ 0,\ 1,\ 2, \] where \(a\) and \(b\) are constants. It is given that \(\mathrm{E}(X) = 0.8\).
(a) Find the value of \(a\) and the value of \(b\). [3]
(b) Find \(\mathrm{Var}(X)\). [2]
Let \(Y = c - 3X\), where \(c\) is a positive constant. It is given that \(\mathrm{E}\left(Y^{2}\right) = 26.44\).
(c) Find the value of \(c\). [3]
Total 8 marks
(a)
sum: \((a - b) + a + (a + b) + (a + 2b) = 4a + 2b = 1\)
mean: \(-(a - b) + (a + b) + 2(a + 2b) = 2a + 6b = 0.8\) M1
\(b = 0.06\) A1
\(a = 0.22\) A1
Award M1 for both equations, even if not simplified.
[3 marks]
(b)
probabilities \(0.16,\ 0.22,\ 0.28,\ 0.34\); \(\;\mathrm{E}\left(X^{2}\right) = 0.16 + 0.28 + 4 \times 0.34 = 1.8\) M1
\(\mathrm{Var}(X) = 1.8 - 0.8^{2} = 1.16\) A1
[2 marks]
(c)
\(\mathrm{E}(Y) = c - 3 \times 0.8 = c - 2.4\) and \(\mathrm{Var}(Y) = (-3)^{2} \times 1.16 = 10.44\) A1
\(\mathrm{E}\left(Y^{2}\right) = \mathrm{Var}(Y) + [\mathrm{E}(Y)]^{2}\), so \(10.44 + (c - 2.4)^{2} = 26.44\) M1
\((c - 2.4)^{2} = 16\), so \(c = 6.4\) or \(c = -1.6\); \(\;c > 0\), so \(c = 6.4\) A1
[3 marks]
HL 2 Paper 1
The random variable \(X\) has the distribution \(\mathrm{B}(n, p)\). It is given that \(\mathrm{E}(2X + 3) = 11\) and \(\mathrm{Var}(2X + 3) = 12\).
(a) Find the value of \(n\) and the value of \(p\). [4]
(b) Show that, for \(k = 0, 1, 2, \ldots, 15\), \[ \frac{\mathrm{P}(X = k + 1)}{\mathrm{P}(X = k)} = \frac{16 - k}{3(k + 1)} . \] [2]
(c) Hence find the most likely value of \(X\). [2]
Total 8 marks
(a)
\(2\,\mathrm{E}(X) + 3 = 11\) M1
\(\mathrm{E}(X) = np = 4\) A1
\(4\,\mathrm{Var}(X) = 12\), so \(\mathrm{Var}(X) = np(1 - p) = 3\) A1
\(1 - p = \dfrac{3}{4}\), so \(p = \dfrac{1}{4}\) and \(n = 16\) A1
Do not award the second A1 for \(\mathrm{Var}(X) = \dfrac{12}{2} = 6\) or \(\mathrm{Var}(X) = 12 - 3 = 9\).
[4 marks]
(b)
\(\dfrac{\mathrm{P}(X = k + 1)}{\mathrm{P}(X = k)} = \dfrac{\binom{16}{k + 1}\left(\frac{1}{4}\right)^{k + 1}\left(\frac{3}{4}\right)^{15 - k}}{\binom{16}{k}\left(\frac{1}{4}\right)^{k}\left(\frac{3}{4}\right)^{16 - k}}\) M1
\(\dfrac{\binom{16}{k + 1}}{\binom{16}{k}} = \dfrac{16!\;k!\,(16 - k)!}{(k + 1)!\,(15 - k)!\;16!} = \dfrac{16 - k}{k + 1}\) A1
\(\dfrac{\frac{1}{4}}{\frac{3}{4}} = \dfrac{1}{3}\), so the ratio is \(\dfrac{16 - k}{3(k + 1)}\) AG
[2 marks]
(c)
ratio \(> 1 \iff 16 - k > 3k + 3 \iff k < 3.25\) M1
so \(\mathrm{P}(X = k)\) increases up to \(k = 4\) and then decreases; the most likely value is \(4\) A1
Do not award any marks for \(4\) found only from \(\mathrm{E}(X) = 4\): the question says “hence”.
[2 marks]
HL 3 Paper 2
A café buys all its eggs from two farms. \(65\%\) of the eggs come from Farm \(P\) and the rest come from Farm \(Q\). Each egg from Farm \(P\) is cracked with probability \(0.04\), and each egg from Farm \(Q\) is cracked with probability \(0.10\), independently of all other eggs. The eggs arrive in trays of \(30\), and all the eggs in a tray come from the same farm. A tray is chosen at random.
(a) Find the probability that the tray contains at most one cracked egg. [5]
(b) Given that the tray contains at most one cracked egg, find the probability that it came from Farm \(Q\). [2]
A student says that the number of cracked eggs in a tray chosen at random can be modelled by \(\mathrm{B}(30,\, 0.061)\), because \(0.65 \times 0.04 + 0.35 \times 0.10 = 0.061\).
(c) Explain why this model is not valid. [1]
Total 8 marks
(a)
let \(C\) be the number of cracked eggs: \(C \sim \mathrm{B}(30,\, 0.04)\) for Farm \(P\) and \(C \sim \mathrm{B}(30,\, 0.10)\) for Farm \(Q\) M1
\(\mathrm{P}(C \le 1 \mid P) = 0.661179\ldots\) (using Binomial Cdf on a GDC) A1
\(\mathrm{P}(C \le 1 \mid Q) = 0.183695\ldots\) A1
\(\mathrm{P}(C \le 1) = 0.65 \times 0.661179\ldots + 0.35 \times 0.183695\ldots\) M1
\(= 0.494\) \((0.494060\ldots)\) A1
[5 marks]
(b)
\(\mathrm{P}(Q \mid C \le 1) = \dfrac{0.35 \times 0.183695\ldots}{0.494060\ldots}\) M1
\(= 0.130\) \((0.130132\ldots)\) A1
[2 marks]
(c)
all the eggs in a tray come from the same farm, so they are not independent: if one egg is cracked, the tray is more likely to be from \(Q\), so the other eggs are more likely to be cracked R1
Accept “the probability that an egg is cracked is not the same for every tray (it is \(0.04\) or \(0.10\))”.
[1 mark]
HL 4 Paper 1
A box contains \(3\) gold tokens and \(n\) silver tokens, where \(n \ge 2\). In a game, a player takes two tokens from the box at random, without replacement. Let \(X\) be the number of gold tokens taken.
(a) Find \(\mathrm{P}(X = 0)\) and \(\mathrm{P}(X = 2)\) in terms of \(n\), and show that \(\mathrm{P}(X = 1) = \dfrac{6n}{(n + 3)(n + 2)}\). [3]
(b) Show that \(\mathrm{E}(X) = \dfrac{6}{n + 3}\). [2]
The player pays $2 to play and receives $5 for each gold token taken. Let \(G\) be the player’s gain, in dollars, so that \(G = 5X - 2\).
(c) Find the value of \(n\) for which the game is fair. [3]
For the rest of this question, use the value of \(n\) found in part (c).
(d) Show that \(\mathrm{Var}(X) = \dfrac{52}{175}\). [3]
(e) Hence find the exact standard deviation of \(G\). [2]
(f) Explain why \(X\) does not have a binomial distribution. [1]
Total 14 marks
(a)
\(\mathrm{P}(X = 0) = \dfrac{n}{n + 3} \times \dfrac{n - 1}{n + 2} = \dfrac{n(n - 1)}{(n + 3)(n + 2)}\) A1
\(\mathrm{P}(X = 2) = \dfrac{3}{n + 3} \times \dfrac{2}{n + 2} = \dfrac{6}{(n + 3)(n + 2)}\) A1
\(\mathrm{P}(X = 1) = \dfrac{3}{n + 3} \times \dfrac{n}{n + 2} + \dfrac{n}{n + 3} \times \dfrac{3}{n + 2}\) M1
\(= \dfrac{6n}{(n + 3)(n + 2)}\) AG
Accept \(1 - \mathrm{P}(X = 0) - \mathrm{P}(X = 2) = \dfrac{n^{2} + 5n + 6 - n^{2} + n - 6}{(n + 3)(n + 2)}\) for the M1.
[3 marks]
(b)
\(\mathrm{E}(X) = 1 \times \dfrac{6n}{(n + 3)(n + 2)} + 2 \times \dfrac{6}{(n + 3)(n + 2)}\) M1
\(= \dfrac{6(n + 2)}{(n + 3)(n + 2)} = \dfrac{6}{n + 3}\) A1 AG
[2 marks]
(c)
\(\mathrm{E}(G) = 5\,\mathrm{E}(X) - 2 = \dfrac{30}{n + 3} - 2\) M1 A1
\(\dfrac{30}{n + 3} = 2 \;\Rightarrow\; n + 3 = 15\), so \(n = 12\) A1
[3 marks]
(d)
\(\mathrm{P}(X = 1) = \dfrac{72}{210} = \dfrac{12}{35}\), \(\;\mathrm{P}(X = 2) = \dfrac{6}{210} = \dfrac{1}{35}\), \(\;\mathrm{E}(X) = \dfrac{6}{15} = \dfrac{2}{5}\) A1
\(\mathrm{E}\left(X^{2}\right) = 1 \times \dfrac{12}{35} + 4 \times \dfrac{1}{35} = \dfrac{16}{35}\) A1
\(\mathrm{Var}(X) = \dfrac{16}{35} - \left(\dfrac{2}{5}\right)^{2}\) M1
\(= \dfrac{80 - 28}{175} = \dfrac{52}{175}\) AG
[3 marks]
(e)
\(\mathrm{Var}(G) = 5^{2} \times \dfrac{52}{175} = \dfrac{52}{7}\) M1
standard deviation \(= \sqrt{\dfrac{52}{7}}\) \(\left(= \dfrac{2\sqrt{91}}{7}\right)\) A1
Do not award the M1 for \(\mathrm{Var}(G) = 5^{2} \times \dfrac{52}{175} - 2\).
[2 marks]
(f)
the tokens are taken without replacement, so the probability of gold on the second draw depends on the first draw (the trials are not independent) R1
[1 mark]
HL 5 Paper 2
In each round of a biathlon race, an athlete shoots at \(5\) targets. She hits each target with probability \(p\), where \(0 < p < 1\), independently of all other shots. Let \(X\) be the number of targets that she hits in one round.
(a) Show that \(\mathrm{P}(X \ge 4) = 5p^{4} - 4p^{5}\). [2]
It is given that \(\mathrm{P}(X \ge 4) = 0.8\).
(b) Find the value of \(p\). [2]
A race has \(4\) rounds, so the athlete shoots at \(20\) targets. Let \(M\) be the number of targets that she misses in a race.
(c) Write down the distribution of \(M\), and find \(\mathrm{E}(M)\) and \(\mathrm{Var}(M)\). [3]
Each miss adds a penalty, and the athlete’s total race time, \(T\) minutes, is modelled by \(T = 25 + 0.4M\).
(d) Find the mean and the standard deviation of \(T\). [4]
(e) Given that her race time is more than \(26\) minutes, find the probability that it is less than \(27\) minutes. [4]
(f) Suggest one reason why the binomial model for \(M\) might not be realistic. [1]
Total 16 marks
(a)
\(\mathrm{P}(X \ge 4) = \mathrm{P}(X = 4) + \mathrm{P}(X = 5) = \dbinom{5}{4}p^{4}(1 - p) + p^{5}\) M1
\(= 5p^{4} - 5p^{5} + p^{5} = 5p^{4} - 4p^{5}\) A1 AG
[2 marks]
(b)
\(5p^{4} - 4p^{5} = 0.8\) (for example, the intersection of the two graphs on a GDC, for \(0 < p < 1\)) M1
\(p = 0.831\) \((0.831391\ldots)\) A1
[2 marks]
(c)
\(M \sim \mathrm{B}(20,\, 1 - p)\), where \(1 - p = 0.168608\ldots\) A1
\(\mathrm{E}(M) = 20(1 - p) = 3.37\) \((3.37217\ldots)\) A1
\(\mathrm{Var}(M) = 20p(1 - p) = 2.80\) \((2.80359\ldots)\) A1
[3 marks]
(d)
\(\mathrm{E}(T) = 25 + 0.4\,\mathrm{E}(M)\) M1
\(= 26.3\) minutes \((26.3488\ldots)\) A1
\(\mathrm{Var}(T) = 0.4^{2}\,\mathrm{Var}(M) = 0.448575\ldots\) M1
standard deviation \(= 0.670\) minutes \((0.669757\ldots)\) A1
Do not award the second M1 for \(0.4\,\mathrm{Var}(M)\) or for \(25 + 0.4^{2}\,\mathrm{Var}(M)\).
[4 marks]
(e)
\(T > 26 \iff 0.4M > 1 \iff M \ge 3\) A1
\(T < 27 \iff 0.4M < 2 \iff M \le 4\) A1
\(\mathrm{P}(T < 27 \mid T > 26) = \dfrac{\mathrm{P}(3 \le M \le 4)}{\mathrm{P}(M \ge 3)} = \dfrac{0.440755\ldots}{0.679589\ldots}\) M1
\(= 0.649\) \((0.648560\ldots)\) A1
[4 marks]
(f)
for example: wind or tiredness changes the probability of a hit from round to round, so \(p\) may not be constant (or a miss may affect her next shot, so the shots may not be independent) R1
[1 mark]