Question

A box contains 13 ​transistors, 4 of which are defective. If 4 are selected at​ random, find the probability that a. All are defective. b. None are defective.

Answer 1

Answer:

(a) The probability that all are defective is $\frac{1}{725}$.

(b) The probability that all are defective is $\frac{126}{725}$.

Step-by-step explanation:

It is given that a box contains 13 ​transistors, 4 of which are defective.

Total number of ​transistors = 13

Total number of defective ​transistors = 4

Total number of non-defective ​transistors = 13 - 4 = 9

Number of selected transistors = 4

Total possible ways to select 4 ​transistors from 13 ​transistors is

$\text{Possible outcomes}=^{13}C_4=\frac{13!}{4!(13-4)!}=725$

(a)

We need to find the probability that none are defective.

It means all 4 transistors are from defective ​transistors.

$\text{Required outcomes}=^4C_4=1$

The probability that all are defective is

$P=\frac{\text{Required outcomes}}{\text{Possible outcomes}}$

$P=\frac{1}{725}$

Therefore the probability that all are defective is $\frac{1}{725}$.

(b)

We need to find the probability that none are defective.

It means all 4 transistors are from non-defective ​transistors.

$\text{Required outcomes}=^9C_4=\frac{9!}{4!(9-4)!}=126$

The probability that none are defective is

$P=\frac{\text{Required outcomes}}{\text{Possible outcomes}}$

$P=\frac{126}{725}$

Therefore the probability that none are defective is $\frac{126}{725}$.