Question

When circuit boards used in the manufacture of compact disc players are tested, the long-run percentage of defectives is 5%. If 25 such discs are tested, what is the probability that:

20 are not defective?

A) 0.061
B) 0.042
C) 0.072
D) 0.081

Answer 1

The probability that 20 out of 25 circuit boards tested are not defective is found using the binomial distribution formula. It turns out to be 0.081, which corresponds to answer choice D.

Step-by-step explanation:

The question deals with finding the probability that 20 out of 25 circuit boards tested are not defective, given that the long-run percentage of defectives is 5%. This is a problem that can be addressed using the binomial distribution, where the number of successes in a sequence of n independent experiments is counted, with each experiment having a success probability p and failure probability (1-p).

We define X as the random variable representing the number of non-defective circuit boards. We are looking to find P(X = 20), with n = 25 and p = 0.95 (since 100% - 5% = 95%, the probability of a board not being defective).

Using the binomial probability formula:

P(X = k) = nCk * p^k * (1-p)^(n-k)

Where nCk is the binomial coefficient, k is the number of successes (non-defective boards), p is the probability of success, and n is the total number of trials.

Plugging in the values:

P(X = 20) = 25C20 * (0.95)^20 * (0.05)^(25-20)

Calculating this, we get P(X = 20) = 0.081, which corresponds to answer choice D.