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Chapter 23: Problem 14

The probability that a component is acceptable is \(0.91\). Ten components are picked at random. Calculate the probability that (a) eight are acceptable (b) more than eight are acceptable (c) three are not acceptable.

Short Answer

Expert verified
Question: In a manufacturing process, 91% of the components produced are acceptable. If 10 components are randomly selected, find the probability of the following scenarios: (a) Exactly 8 components are acceptable. (b) More than 8 components are acceptable. (c) Exactly 3 components are not acceptable. Answer: (a) The probability that 8 components are acceptable is approximately 0.1941. (b) The probability that more than 8 components are acceptable is approximately 0.7007. (c) The probability that exactly 3 components are not acceptable is approximately 0.0571.

Step by step solution

01

Identify the relevant parameters in the given problem

The problem involves a binomial distribution with n=10 trials, where the probability of success (an acceptable component) is p=0.91.
02

Recall the formula for the probability in a binomial distribution

The probability of getting exactly k successes in n trials is given by: P(X=k) = (n!/(k!(n-k)!)) * (p^k) * ((1-p)^(n-k)), where n is the total number of trials, k is the number of successes, and p is the probability of success.
03

Calculate the probability of 8 acceptable components

Using the formula, we substitute n=10, k=8, and p=0.91: P(X=8) = (10!/(8!(10-8)!)) * (0.91^8) * ((1-0.91)^(10-8)) P(X=8) ≈ 0.1941
04

Calculate the probability of more than eight acceptable components

To find the probability of 9 or 10 acceptable components, sum their individual probabilities: P(X>8) = P(X=9) + P(X=10) Substitute n=10, k=9 for P(X=9) and n=10, k=10 for P(X=10): P(X>8) ≈ (10!/(9!(10-9)!)) * (0.91^9) * (0.09^1) + (10!/(10!(10-10)!)) * (0.91^10) * (0.09^0) P(X>8) ≈ 0.7007
05

Calculate the probability of 3 unacceptable components

Since we want exactly 3 unacceptable components, that means only 7 components are acceptable: P(X=7) = (10!/(7!(10-7)!)) * (0.91^7) * ((1-0.91)^(10-7)) P(X=7) ≈ 0.0571 So, the probability of each scenario is: (a) Probability that 8 components are acceptable: 0.1941 (b) Probability that more than 8 components are acceptable: 0.7007 (c) Probability that 3 components are not acceptable: 0.0571

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