# A bin contains 71 light bulbs, of which 9 are

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?
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Kristin Ferrell
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.
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 $x=6$
${P}_{x=6}=6{C}_{6}\left({p}^{x}\right)\left({q}^{n-x}\right)$
${P}_{x=6}=6{C}_{6}{\left(\frac{62}{71}\right)}^{6}{\left(\frac{9}{71}\right)}^{6-6}$
${P}_{x=6}=6{C}_{6}{\left(\frac{62}{71}\right)}^{6}{\left(\frac{9}{71}\right)}^{0}$
${P}_{x=6}=\left(1\right){\left(\frac{62}{71}\right)}^{6}\left(1\right)$
${P}_{x=6}=\left(1\right){\left(\frac{62}{71}\right)}^{6}\left(1\right)$
${P}_{x=6}=\left(1\right)\left(0.44\right)\left(1\right)$
${P}_{x=6}=0.44$