Question

suppose that a computer chip company has just shipped 10,000 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 } { 10,000 } = 0.001 = 0.1 \\% \\). compute the probability that two randomly selected chips are defective under the assumption of independent events. (a) the probability is 0.00000090 (round to eight decimal places as needed ) (b) the probability is (round to eight decimal places as needed.)

Explanation:

Step1: Recall the formula for independent events

For two independent events \(A\) and \(B\), \(P(A\cap B)=P(A)\times P(B)\)

Step2: Identify the probabilities

Let \(A\) be the event that the first chip is defective and \(B\) be the event that the second chip is defective. Given \(P(A)=\frac{10}{10000} = 0.001\). Since we assume independence, \(P(B)=P(A)=0.001\)

Step3: Calculate the probability

\(P(A\cap B)=P(A)\times P(B)=0.001\times0.001\)
\(P(A\cap B)= 0.00000100\)

Answer:

\(0.00000100\)