13.2 Calculating binomial probabilities — Practise with help

Work each question closed book first. Say the inequality in words before you calculate. Keep every term to four decimal places or more, and round once.

Warm-up: remember

The random variable $X \sim B\left(5, \tfrac12\right)$ . Find $P(X = 2)$ .
The probability that a bus is late is $0.35$ . Find the probability that it is not late.
Write $P(X \ge 4)$ in terms of an at-most probability.

Answers

$P(X = 2) = \dbinom{5}{2}\left(\tfrac12\right)^5 = 10 \times \tfrac{1}{32} = \tfrac{10}{32} = \tfrac{5}{16}$ .
Use the complement: $1 - 0.35 = 0.65$ .
$P(X \ge 4) = 1 - P(X \le 3)$ .

If question 3 was hard, revise the complement from Section 12.1.

Practise with help

1. The random variable $X \sim B\left(8, \tfrac12\right)$ . Find $P(X \le 2)$ .

Hint

"At most 2" means 0, 1 or 2. Work out three terms and add them.

Next step

Use $p = \tfrac12$ , so both $p^r$ and $(1-p)^{n-r}$ are powers of $\tfrac12$ , and every term has denominator $2^8 = 256$ .

Full working

P(X = 0) = \binom{8}{0}\left(\tfrac12\right)^8 = \tfrac{1}{256}, \qquad P(X = 1) = \binom{8}{1}\left(\tfrac12\right)^8 = \tfrac{8}{256},

P(X = 2) = \binom{8}{2}\left(\tfrac12\right)^8 = \tfrac{28}{256}.

Add the three terms:

P(X \le 2) = \tfrac{1 + 8 + 28}{256} = \tfrac{37}{256} \approx 0.1445.

2. The random variable $X \sim B(10, 0.2)$ . Find $P(X \ge 3)$ .

Hint

Adding the terms from 3 to 10 uses eight terms. The complement uses three.

Next step

Use $P(X \ge 3) = 1 - P(X \le 2)$ , and work out the three terms for 0, 1 and 2.

Full working

P(X = 0) = (0.8)^{10} = 0.1073742 \ (4 \text{ d.p.}),

P(X = 1) = 10(0.2)(0.8)^9 = 0.2684355 \ (4 \text{ d.p.}),

P(X = 2) = 45(0.2)^2(0.8)^8 = 0.3019899 \ (4 \text{ d.p.}).

Add and subtract from 1:

P(X \ge 3) = 1 - (0.1073742 + 0.2684355 + 0.3019899) = 1 - 0.6777996 = 0.3222 \ (4 \text{ d.p.}).

3. The random variable $X \sim B\left(6, \tfrac12\right)$ . Find $P(2 \le X \le 4)$ .

Hint

The values are 2, 3 and 4. Include both ends.

Next step

Every term has denominator $2^6 = 64$ . The binomial coefficients for $r = 2, 3 \text{ and } 4$ are 15, 20 and 15.

Full working

P(2 \le X \le 4) = \frac{\binom{6}{2} + \binom{6}{3} + \binom{6}{4}}{64} = \frac{15 + 20 + 15}{64} = \frac{50}{64} = \frac{25}{32} = 0.78125.

4. The random variable $X \sim B(15, 0.4)$ . Find $P(X < 5)$ .

Hint

$X$ takes whole-number values, so $X < 5$ is the same as $X \le 4$ .

Next step

Work out the five terms for 0, 1, 2, 3 and 4, then add them.

Full working

Keep four decimal places in each term.

P(X = 0) = (0.6)^{15} = 0.0004702, \qquad P(X = 1) = 15(0.4)(0.6)^{14} = 0.0047018,

P(X = 2) = 105(0.4)^2(0.6)^{13} = 0.0219418, \qquad P(X = 3) = 455(0.4)^3(0.6)^{12} = 0.0633874,

P(X = 4) = 1365(0.4)^4(0.6)^{11} = 0.1267748.

Add:

P(X < 5) = 0.0004702 + 0.0047018 + 0.0219418 + 0.0633874 + 0.1267748 = 0.2173 \ (4 \text{ d.p.}).

5. The random variable $X \sim B(20, 0.3)$ . Find $P(X > 12)$ .

Hint

$X > 12$ is the same as $X \ge 13$ . That is a long tail, so use the complement.

Next step

$P(X > 12) = 1 - P(X \le 12)$ . Work out the sum from 0 to 12 with a calculator, then subtract.

Full working

P(X > 12) = 1 - P(X \le 12).

The cumulative probability is $P(X \le 12) = 0.9987 \ (4 \text{ d.p.})$ , so

P(X > 12) = 1 - 0.9987 = 0.0013 \ (4 \text{ d.p.}).

A value as high as 13 is very unlikely when $p = 0.3$ and $n = 20$ .

6. A machine makes parts, and 10% of them are faulty. A quality inspector picks 12 parts at random. Let $X$ be the number of faulty parts. Find the probability that at least one part is faulty.

Hint

"At least one" is the complement of "none". It is shorter to work out the probability that no part is faulty.

Next step

Model $X \sim B(12, 0.1)$ . Then $P(X = 0) = (0.9)^{12}$ , and $P(X \ge 1) = 1 - P(X = 0)$ .

Full working

The parts are chosen at random, so the trials are independent and $p = 0.1$ is constant. Model $X \sim B(12, 0.1)$ .

P(X = 0) = (0.9)^{12} = 0.2824 \ (4 \text{ d.p.}).

P(X \ge 1) = 1 - 0.2824 = 0.7176 \ (4 \text{ d.p.}).

So at least one faulty part appears in about 71.8% of samples of 12.