In a box containing $10$ bulbs, $2$ are defective. What is the probability that among $5$ bulbs chosen at random, none is defective.
$5$ bulbs can be chosen from $10$ bulbs in
${^{10}}{C}_{5} = \dfrac{10!}{5! \times 5!} = \dfrac{10 \times 9 \times 8 \times 7 \times 6}{5 \times 4 \times 3 \times 2 \times 1} = 252$ ways
$\therefore \;$ Number of elements in sample space $S = n\left(S\right) = 252$
$\because \;$ $2$ bulbs are defective, number of bulbs which are not defective $= 8$
Let $A$ be the event of selecting $5$ bulbs which are not defective.
Now, $5$ bulbs can be selected from $8$ non-defective bulbs in
${^{8}}{C}_{5} = \dfrac{8!}{3! \times 5!} = \dfrac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56$ ways
$\therefore \;$ Number of elements in $A = n\left(A\right) = 56$
$\therefore \;$ Probability of event $A = P\left(A\right) = \dfrac{n\left(A\right)}{n \left(S\right)} = \dfrac{56}{252} = \dfrac{2}{9}$