Question

A box contains 20 lightbulbs, of which 5 are defective.

Answer 1

Answer:

The correct option is d.

Step-by-step explanation:

It is given that the box contains 20 light bulbs, of which 5 are defective.

The possible ways of selecting r items from total n.

$^nC_r=\frac{n!}{r!(n-r)!}$

The total possible ways to select 4 light bulbs form 20 bulbs is

$^{20}C_4=4845$

We have to find the probability that at most 2 of them are defective.

Total possibility of selecting at most 2 defective bulbs.

Total = 0 bulb is defective + 1 bulb is defective +2 bulb is defective

$Possibility=^{15}C_4\times ^{5}C_0+^{15}C_3\times ^{5}C_1+^{15}C_2\times ^{5}C_2$

$Possibility=1365\times 1+455\times 5+105\times 10=4690$

The probability of selecting at most 2 defective bulbs is

$P=\frac{\text{Required possibility}}{\text{Total possibility}}$

$P=\frac{4690}{4845}$

Divide the numerator and denominator by 5.

$P=\frac{938}{969}$

The correct option is d.