Question

a box contains 15 transistors, 5 of which are defective. if 5 are selected at random, find the probability of the statements below.
a. all are defective
b. none are defective
a. the probability is \frac{1}{3003}
(type a fraction. simplify your answer.)
b. the probability is \square
(type a fraction simplify your answer)

Explanation:

Step1: Calculate the number of non - defective transistors

The total number of transistors is \(n = 15\), and the number of defective transistors is \(d=5\). So the number of non - defective transistors is \(15 - 5=10\).

Step2: Use the combination formula \(C(n,r)=\frac{n!}{r!(n - r)!}\)

The total number of ways to select \(r = 5\) transistors out of \(n=15\) is \(C(15,5)=\frac{15!}{5!(15 - 5)!}=\frac{15\times14\times13\times12\times11}{5\times4\times3\times2\times1}=3003\).
The number of ways to select \(5\) non - defective transistors out of \(10\) non - defective transistors is \(C(10,5)=\frac{10!}{5!(10 - 5)!}=\frac{10\times9\times8\times7\times6}{5\times4\times3\times2\times1}=252\).

Step3: Calculate the probability

The probability \(P\) that none of the selected transistors is defective is \(P=\frac{C(10,5)}{C(15,5)}\).
Substitute \(C(10,5) = 252\) and \(C(15,5)=3003\) into the formula: \(P=\frac{252}{3003}=\frac{12}{143}\).

Answer:

\(\frac{12}{143}\)