Example 5.9

EXAMPLE 5.9 Inspecting Switches

CASE 5.1 Consider the scenario of Example 5.6 (pages 248–249) in which the number $X$ of switches that fail inspection closely follows the binomial distribution with $n = 15$ and $p = 0.08$ .

The probability that no more than one switch fails is

$\begin{array}{l} P (X \leq 1) & = P (X = 0) + P (X = 1) \\ = (\begin{matrix} 15 \\ 0 \end{matrix}) {(0.08)}^{0} {(0.92)}^{15} + (\begin{matrix} 15 \\ 1 \end{matrix}) {(0.08)}^{1} {(0.92)}^{14} \\ = \frac{15!}{0! 15!} (1) (0.2863) + \frac{15!}{1! 14!} (0.08) (0.3112) \\ = (1) (1) (0.2863) + (15) (0.08) (0.3112) \\ = 0.2863 + 0.3734 = 0.6597 \end{array}$

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The calculation used the facts that $0! = 1$ and that $a^{0} = 1$ for any number $a \neq 0$ . The result agrees with that obtained from Table C in Example 5.6.