Unit 18 Review Exercises
Discrete Random Variables and the Binomial Distribution
These questions mix the ideas from the whole unit. Paper 1 means no calculator; Paper 2 means a GDC is allowed.
1 Paper 1 The probability distribution of \(X\) is shown. Find \(k\) and \(\mathrm{E}(X)\).
| \(x\) | \(0\) | \(1\) | \(2\) | \(3\) |
|---|---|---|---|---|
| \(\mathrm{P}(X = x)\) | \(0.1\) | \(0.3\) | \(k\) | \(2k\) |
\(0.1 + 0.3 + 3k = 1 \Rightarrow k = 0.2\).
\(\mathrm{E}(X) = 0 + 1(0.3) + 2(0.2) + 3(0.4) = 0.3 + 0.4 + 1.2 = 1.9\). See 18.1.
2 Paper 1 AHL For the random variable \(X\) in question 1, find \(\mathrm{Var}(X)\) and \(\mathrm{Var}(2X - 3)\).
\(\mathrm{E}(X^2) = 1(0.3) + 4(0.2) + 9(0.4) = 0.3 + 0.8 + 3.6 = 4.7\).
\(\mathrm{Var}(X) = 4.7 - 1.9^2 = 1.09\), and \(\mathrm{Var}(2X - 3) = 4 \times 1.09 = 4.36\). See 18.2.
3 Paper 1 A player pays $2 and then takes one card at random from ten cards numbered \(1\) to \(10\). If the number is prime, the player receives $\(k\); otherwise the player receives nothing. Find \(k\) so that the game is fair.
The primes are \(2, 3, 5, 7\), so \(\mathrm{P}(\text{prime}) = 0.4\).
Expected profit \(= 0.4k - 2 = 0\), so \(k = 5\). See 18.1.
4 Paper 1 \(X \sim \mathrm{B}(5,\, 0.2)\). Find \(\mathrm{P}(X = 1)\) and \(\mathrm{P}(X \le 1)\).
\(\mathrm{P}(X = 1) = 5 \times 0.2 \times 0.8^4 = 5 \times 0.2 \times 0.4096 = 0.4096\).
\(\mathrm{P}(X \le 1) = 0.8^5 + 0.4096 = 0.32768 + 0.4096 = 0.73728\). See 18.3.
5 Paper 2 85% of the seeds in a large batch grow, independently. Seeds are sold in packets of 20. Find the probability that in one packet
(a) at least 18 seeds grow,
(b) fewer than 15 seeds grow.
\(X \sim \mathrm{B}(20,\, 0.85)\).
(a) $(X ) = $ binomCdf\((20, 0.85, 18, 20) = 0.405\).
(b) $(X ) = $ binomCdf\((20, 0.85, 0, 14) = 0.0673\). See 18.3.
6 Paper 2 2% of the parcels sent by a company are damaged, independently. Find the least number of parcels for which the probability that at least one is damaged is greater than \(0.5\).
\(1 - 0.98^n > 0.5 \Rightarrow 0.98^n < 0.5 \Rightarrow n > \dfrac{\ln 0.5}{\ln 0.98} = 34.3\ldots\), so \(n = 35\).
(Or invBinomN(0.5, 0.02, 0) \(= 35\).) See 18.3.
7 Paper 1 \(X \sim \mathrm{B}(n, p)\) has mean \(3.6\) and variance \(2.52\). Find \(n\) and \(p\).
\(1 - p = \dfrac{2.52}{3.6} = 0.7\), so \(p = 0.3\) and \(n = \dfrac{3.6}{0.3} = 12\). See 18.4.
8 Paper 2 Sara wins each game of an online puzzle with probability \(0.35\), independently. She plays 10 games every day.
(a) Find the probability that she wins at least 5 games on one day.
(b) Find the expected number of days, in a period of 30 days, on which she wins at least 5 games.
9 Paper 1 AHL \(\mathrm{P}(X = x) = \dfrac{x^2}{30}\) for \(x = 1, 2, 3, 4\).
(a) Show that this is a probability distribution. \(\quad\)
(b) Find \(\mathrm{E}(X)\) and \(\mathrm{Var}(X)\). \(\quad\)
(c) Find \(\mathrm{Var}(10 - 3X)\).
(a) Each probability is between \(0\) and \(1\), and \(\dfrac{1 + 4 + 9 + 16}{30} = 1\).
(b) \(\mathrm{E}(X) = \dfrac{1 + 8 + 27 + 64}{30} = \dfrac{10}{3}\). \(\quad \mathrm{E}(X^2) = \dfrac{1 + 16 + 81 + 256}{30} = \dfrac{59}{5}\).
\(\mathrm{Var}(X) = \dfrac{59}{5} - \dfrac{100}{9} = \dfrac{531 - 500}{45} = \dfrac{31}{45}\).
(c) \(\mathrm{Var}(10 - 3X) = 9 \times \dfrac{31}{45} = \dfrac{31}{5}\). See 18.1 and 18.2.
10 Paper 2 \(X \sim \mathrm{B}(n,\, 0.4)\). Find the least value of \(n\) for which \(\mathrm{P}(X \ge 3) > 0.9\). For this value of \(n\), write down \(\mathrm{E}(X)\).