Question

a box contains 10 transistors, 3 of which are defective. if 3 are selected at random, find the probability of the statements below. a. all are defective b. none are defective

Explanation:

Step1: Calculate total number of ways to select 3 transistors

The total number of transistors is $n = 10$, and we want to select $r=3$. Using the combination formula $C(n,r)=\frac{n!}{r!(n - r)!}$, we have $C(10,3)=\frac{10!}{3!(10 - 3)!}=\frac{10\times9\times8}{3\times2\times1}=120$.

Step2: Calculate number of ways for all - defective case (a)

There are 3 defective transistors, and we want to select 3 of them. Using the combination formula, $C(3,3)=\frac{3!}{3!(3 - 3)!}=1$. So the probability that all are defective is $P(\text{all defective})=\frac{C(3,3)}{C(10,3)}=\frac{1}{120}$.

Step3: Calculate number of ways for none - defective case (b)

There are $10 - 3=7$ non - defective transistors, and we want to select 3 of them. Using the combination formula, $C(7,3)=\frac{7!}{3!(7 - 3)!}=\frac{7\times6\times5}{3\times2\times1}=35$. So the probability that none are defective is $P(\text{none defective})=\frac{C(7,3)}{C(10,3)}=\frac{35}{120}=\frac{7}{24}$.

Answer:

a. $\frac{1}{120}$
b. $\frac{7}{24}$