Question

A bin contains 71 light bulbs, of which 9 are defective. If 6 light bulbs are randomly selected from the bin without replacement of the selected bulbs, what is the probability that all of the selected bulbs are good?

Answer 1

`(62c6)/(71c6)`

Explanation:

There are 71 light bulbs in the bin of which 9 are defective.
So 62 of them are good.
6 bulbs are selected and we want all of them to be good.
So, the good bulbs must come from the 62 bulbs and this is possible in `(62c6)` ways. The total possibilities are `(71c6)` since we select 6 bulbs from the 71 bulbs available for selection. Finally, the probability is obtained as the ratio mentioned.

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This Problem is to be solved using Binomial Distribution.

If a ball is taken randomly from the bin the ball being defective is `q=9/71`
The ball not being defective is `p =1 - 9/91=62/71`

The desired event is not being defective.

Let `x` be the number of desired event.

We expect `x=6`

`P_(x=6)="^6C_6(p^x)(q^(n-x))`
`P_(x=6)="^6C_6(62/71)^6(9/71)^(6-6)`
`P_(x=6)="^6C_6(62/71)^6(9/71)^0`
`P_(x=6)=(1)(62/71)^6(1)`
`P_(x=6)=(1)(62/71)^6(1)`
`P_(x=6)=(1)(0.44)(1)`

`P_(x=6)=0.44`