Question

at a manufacturing plant, it is known that 8% of the computer chips produced are defective. a random sample of 20 chips is taken. what is the probability that 3 of those chips are defective? 0.08 0.14 0.24 0.78

Explanation:

Step1: Identify the binomial probability formula

The binomial probability formula is \(P(X = k)=C(n,k)\times p^{k}\times(1 - p)^{n - k}\), where \(n\) is the number of trials, \(k\) is the number of successes, \(p\) is the probability of success in a single trial, and \(C(n,k)=\frac{n!}{k!(n - k)!}\)

Step2: Assign values to the formula

Here, \(n = 20\), \(k = 3\), \(p=0.08\), and \(1-p = 0.92\)

First, calculate \(C(20,3)=\frac{20!}{3!(20 - 3)!}=\frac{20\times19\times18}{3\times2\times1}=1140\)

Then, \(p^{k}=(0.08)^{3}=0.000512\)

And \((1 - p)^{n - k}=(0.92)^{17}\approx0.249\)

Step3: Calculate the probability

\(P(X = 3)=C(20,3)\times(0.08)^{3}\times(0.92)^{17}\)

\(P(X = 3)=1140\times0.000512\times0.249\approx0.14\)

Answer:

0.14