Question

suppose that a computer chip company has just shipped 5,000 computer chips to a computer company. unfortunately, 10 of the chips are defective
(a) compute the probability that two randomly selected chips are defective using conditional probability.
(b) the probability that the first randomly selected chip is defective is \\( \frac { 10 } { 5,000 } = 0.002 = 0.2 \\% \\). compute the probability that two randomly selected chips are defective under the
assumption of independent events.
(a) the probability is \\( \square \\)
(round to eight decimal places as needed.)

Explanation:

Step1: Calculate probability for part (a)

The probability that the first chip is defective is \(P(A)=\frac{10}{5000} = 0.002\). After one defective chip is taken out, there are \(9\) defective chips left and \(4999\) chips in total. So the probability that the second chip is defective given the first is defective is \(P(B|A)=\frac{9}{4999}\). By the formula for conditional probability \(P(A\cap B)=P(A)\times P(B|A)\), we have \(P(A\cap B)=\frac{10}{5000}\times\frac{9}{4999}=\frac{90}{24995000}\approx0.00000360\)

Step2: Calculate probability for part (b)

If we assume independent events (which is not actually the case here as the sampling is without replacement, but if we force the independence formula \(P(A\cap B)=P(A)\times P(B)\)). Since \(P(A) = P(B)=\frac{10}{5000}=0.002\), then \(P(A\cap B)=0.002\times0.002 = 0.000004\)

Answer:

(a) \(0.00000360\)
(b) \(0.000004\)