# 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?

Nov 1, 2015

$\frac{62 c 6}{71 c 6}$

#### 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 $\left(62 c 6\right)$ ways. The total possibilities are $\left(71 c 6\right)$ 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 $q = \frac{9}{71}$
The ball not being defective is $p = 1 - \frac{9}{91} = \frac{62}{71}$

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
${P}_{x = 6} = \left(1\right) {\left(\frac{62}{71}\right)}^{6} \left(1\right)$
${P}_{x = 6} = \left(1\right) {\left(\frac{62}{71}\right)}^{6} \left(1\right)$
${P}_{x = 6} = \left(1\right) \left(0.44\right) \left(1\right)$

${P}_{x = 6} = 0.44$