Question

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.70/323
c.91/323
d.938/969

Answer 1

Explanation:

938/969 is right just took test got a 5/5

Answer 2

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.