A box contains 20 lightbulbs, of which 5 are defective. If 4 lightbulbs are picked from the box randomly, the probability that at most 2 of them are defective is a.31/969 b.70/3…

Question

asked Aug 12, 2016 89.1k views

2 Answers

Liauchuk Ivan · Answer 1 · 2016-08-17T12:09:13+0000

Answer:

The correct option is d.

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=(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* ^(5)C_0+^(15)C_3* ^(5)C_1+^(15)C_2* ^(5)C_2

Possibility=1365* 1+455* 5+105* 10=4690

The probability of selecting at most 2 defective bulbs is

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

P=(4690)/(4845)

Divide the numerator and denominator by 5.

P=(938)/(969)

The correct option is d.