Answer to Question #119623 in Statistics and Probability for Kyei William Frimpong

Question

Suppose that 3% of computer chips produced by a certain machine are defective. The
chips are put into packages of 20 chips for distribution to retailers. What is the
probability that a randomly selected package of chips will contain at least 2 defective
chips?

Answer 1

Let X = the random variable denoting the number of defective computer chips in the randomly selected package, X = 0, 1, ..., 20

Since, 3% of the chips produced are defective, therefore, the probability that a chip is defective is 3% i.e. 0.03.

Therefore, X ~ bin(n = 20, p = 0.03)

The p.m.f. of X is given by,

P(X = x) = $\begin{cases} \dbinom{20}{x}(0.03)^x(1-0.03)^{20-x} &\text{for } x=0,1,...,20 \\ 0 &\text{otherwise } \end{cases}$

The probability that the randomly selected package of chips will contain at least 2 defective chips

= P(X $\geq$ 2)

= 1 - P(X < 2) [since the total probability is 1]

= 1 - [P(X = 0) + P(X = 1)]

= 1 -[$\dbinom{20}{0}(0.03)^0(1-0.03)^{20-0}+\dbinom{20}{1}(0.03)^1(1-0.03)^{20-1}$]

1 - [0.5437 + 0.3364]

= 0.1199

Answer: The probability that the randomly selected package of chips will contain at least 2 defective chips is 0.1199.