Unit 17 Review Exercises

Discrete Random Variables and the Binomial Distribution

Review exercises on Discrete Random Variables and the Binomial Distribution for IB Math AA SL and HL, Paper 1 and Paper 2 style, with full worked solutions.

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 \(\mathrm{E}(X)\).

\(x\)	\(0\)	\(1\)	\(2\)	\(3\)
\(\mathrm{P}(X = x)\)	\(0.1\)	\(0.3\)	\(k\)	\(2k\)

Solution

\(0.1 + 0.3 + 3k = 1 \Rightarrow k = 0.2\).

\(\mathrm{E}(X) = 0 + 1(0.3) + 2(0.2) + 3(0.4) = 0.3 + 0.4 + 1.2 = 1.9\). See 17.1.

2 Paper 1 AHL For the random variable \(X\) in question 1, find \(\mathrm{Var}(X)\) and \(\mathrm{Var}(2X - 3)\).

Solution

\(\mathrm{E}(X^2) = 1(0.3) + 4(0.2) + 9(0.4) = 0.3 + 0.8 + 3.6 = 4.7\).

\(\mathrm{Var}(X) = 4.7 - 1.9^2 = 1.09\), and \(\mathrm{Var}(2X - 3) = 4 \times 1.09 = 4.36\). See 17.2.

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.

Solution

The primes are \(2, 3, 5, 7\), so \(\mathrm{P}(\text{prime}) = 0.4\).

Expected profit \(= 0.4k - 2 = 0\), so \(k = 5\). See 17.1.

4 Paper 1 \(X \sim \mathrm{B}(5,\, 0.2)\). Find \(\mathrm{P}(X = 1)\) and \(\mathrm{P}(X \le 1)\).

Solution

\(\mathrm{P}(X = 1) = 5 \times 0.2 \times 0.8^4 = 5 \times 0.2 \times 0.4096 = 0.4096\).

\(\mathrm{P}(X \le 1) = 0.8^5 + 0.4096 = 0.32768 + 0.4096 = 0.73728\). See 17.3.

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.

Solution

\(X \sim \mathrm{B}(20,\, 0.85)\).

(a) \(\mathrm{P}(X \ge 18) =\) binomCdf\((20, 0.85, 18, 20) = 0.405\).

(b) \(\mathrm{P}(X \le 14) =\) binomCdf\((20, 0.85, 0, 14) = 0.0673\). See 17.3.

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 \(0.5\).

Solution

\(1 - 0.98^n > 0.5 \Rightarrow 0.98^n < 0.5 \Rightarrow n > \dfrac{\ln 0.5}{\ln 0.98} = 34.3\ldots\), so \(n = 35\).

(Or invBinomN(0.5, 0.02, 0) \(= 35\).) See 17.3.

7 Paper 1 \(X \sim \mathrm{B}(n, p)\) has mean \(3.6\) and variance \(2.52\). Find \(n\) and \(p\).

Solution

\(1 - p = \dfrac{2.52}{3.6} = 0.7\), so \(p = 0.3\) and \(n = \dfrac{3.6}{0.3} = 12\). See 17.4.

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.

Solution

(a) \(X \sim \mathrm{B}(10,\, 0.35)\). \(\;\mathrm{P}(X \ge 5) =\) binomCdf\((10, 0.35, 5, 10) = 0.249\) (3 s.f.).

(b) The number of such days is \(Y \sim \mathrm{B}(30,\, 0.2485\ldots)\), so \(\mathrm{E}(Y) = 30 \times 0.2485\ldots = 7.46\) days. See 17.3 and 17.4.

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 \(\mathrm{Var}(10 - 3X)\).

Solution

(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 17.1 and 17.2.

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 \(\mathrm{E}(X)\).

Solution

Table of \(\mathrm{P}(X \ge 3) =\) binomCdf\((n, 0.4, 3, n)\):

\(n\)	\(10\)	\(11\)	\(12\)
\(\mathrm{P}(X \ge 3)\)	\(0.833\)	\(0.881\)	\(0.917\)

(Or \(\mathrm{P}(X \le 2) < 0.1\) with Inverse Binomial N: invBinomN(0.1, 0.4, 2) \(= 12\).)

The least value is \(n = 12\), and then \(\mathrm{E}(X) = 12 \times 0.4 = 4.8\). See 17.3 and 17.4.