For each integrated circuit, there are only two possible outcomes. Either they are defective, or they are not. The integrated circuits are independent. This means that 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.

An electronic product contains 48 integrated circuits.

This means that $n = 48$

The probability that any integrated circuit is defective is 0.01.

This means that $p = 0.01$

The product operates only if there are no defective integrated circuits. What is the probability that the product operates?

This is P(X = 0). So

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

$P(X = 0) = C_{48,0}.(0.01)^{0}.(0.99)^{48} = 0.6173$

0.6173 = 61.73% probability that the product operates.

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3 years ago