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tatuchka [14]
3 years ago
15

A stationary store has decided to accept a large shipment of ball-point pens if an inspection of 17 randomly selected pens yield

s no more than two defective pens. (a) Find the probability that this shipment is accepted if 10% of the total shipment is defective. (Use 3 decimal places.)
Mathematics
1 answer:
Leno4ka [110]3 years ago
8 0

Answer:

0.762 = 76.2% probability that this shipment is accepted

Step-by-step explanation:

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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