Question

a factory tests a random sample of 27 transistors for defects. the probability that a particular transistor will be defective has been established by past experience as 0.06. what is the probability that there are no defective transistors in the sample? the probability that there are no defective transistors in the sample is (round to four decimal places as needed.)

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: Determine the values of \(n\), \(k\), and \(p\)

Here, \(n = 27\) (number of transistors), \(k = 0\) (no defective transistors), \(p=0.06\) (probability of a transistor being defective). Then \(1-p = 0.94\)

Step3: Calculate \(C(n,k)\)

When \(k = 0\), \(C(27,0)=\frac{27!}{0!(27-0)!}=1\)

Step4: Calculate the probability

Substitute into the formula: \(P(X = 0)=C(27,0)\times(0.06)^{0}\times(0.94)^{27}\)
Since \((0.06)^{0}=1\), then \(P(X = 0)=1\times1\times(0.94)^{27}\)
\((0.94)^{27}\approx0.1804\)

Answer:

\(0.1804\)