For each pen, there are only two possible outcomes. Either it is defective, or it is not. The probability of a pen being defective is independent from other pens. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

17 randomly selected pens

This means that $n = 17$

(a) Find the probability that this shipment is accepted if 10% of the total shipment is defective. (Use 3 decimal places.)

This is $P(X \leq 2)$ when $p = 0.1$ . So

$P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2)$

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{17,0}.(0.1)^{0}.(0.9)^{17} = 0.167$

$P(X = 1) = C_{17,1}.(0.1)^{1}.(0.9)^{16} = 0.315$

$P(X = 2) = C_{17,2}.(0.1)^{2}.(0.9)^{15} = 0.280$

$P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0.167 + 0.315 + 0.280 = 0.762$

0.762 = 76.2% probability that this shipment is accepted

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3 years ago