Question

In a box containing $15$ bulbs, $5$ are defective. If $5$ bulbs are selected at random from the box, find the probability of the event, that
None of them is defective

Answer 1

Out of $15$ balls $5$ are defective. Probability of selecting a defective balls $=\frac{5}{15}=\frac{1}{3}$

Now we are selecting a $5$ balls at random

$i.e{;}^{15}{C}_5$

$1$ none of them is defective:-

Now we must select from $10$ good balls $i.e{;}^{10}{C}_5$

$P\left(A\right)=\frac{{}^{10}{C}_5}{{}^{15}{C}_5}=\frac{\frac{10!}{(10-5)!5!}}{\frac{15!}{(15-5)!5!}}=\frac{10!}{5!5!}\times \frac{10!\times 5!}{15!}$

$=\frac{5!\times 6\times 7\times 8\times 9\times 10}{5!}\times \frac{10!}{10!\times 11\times 12\times 13\times 14\times 15}$

$P\left(A\right)=\frac{3\times 4}{11\times 13}=\frac{12}{143}$

$2$ only one is defective:-

So we should have $4$ good and $1$ defective

$i.e;\frac{\left({}^{10}{C}_4\right)\left({}^5{C}_1\right)}{{}^{15}{C}_5}=\frac{\frac{10!}{(10-4)!4!}\frac{5!}{(5-1)!1!}}{\frac{15!}{(15-5)!5!}}=\frac{10!5!}{6!4!4!}\times \frac{10!5!}{15!}$

$=\frac{10!5!}{5!\times 6!4!4!}\times \frac{10!5!\times 5}{10!\times 15\times 14\times 13\times 12\times 11}=\frac{5\times 6\times 7\times 8\times 9\times 10\times 5}{6\times 11\times 12\times 13\times 14\times 15}$

$=\frac{50}{143}$

$3$ Atleast one is defective:-

Probability atleast one of them is defective

$=P\left(A\right)\Rightarrow 1-P\left(A\right)$

$1-\frac{12}{143}=\frac{143-12}{143}=\frac{131}{143}$