Question

A box contains 10 red marbles, 20 blue marbles and 30 green marbles. 5 marbles are drawn at random. From the box, what is the probability that :

(i) all are blue?

(ii) at least one is green?

Answer 1

Out of 60 marbles, five marbles can be draw in ${}^{60}{C}_5$ ways.

$\therefore$ Total number of elementary events ${=}^{60}{C}_5$

(i) Out of 20 blue marbles, five blue marbles can be chosen in ${}^{20}{C}_5$ ways.

$\therefore$ Favourable number of events ${=}^{20}{C}_5$ ways

Hence, the required probability is given by
$\frac{{}^{20}{C}_5}{{}^{60}{C}_5}=\frac{20\times 19\times 18\times 17\times 16}{60\times 59\times 58\times 57\times 56}$

$=\frac{19\times 6\times 17}{59\times 29\times 57\times 7}=\frac{2\times 17}{59\times 29\times 7}=\frac{37}{11977}$

(ii) P (no green) =$\frac{\text{Favorable outcomes}}{\text{Total outcomes}}=\frac{{}^{30}{C}_5}{{}^{60}{C}_5}$

Thus, P (at least one green)=1-P (no green)

$=1-\frac{{}^{30}{C}_5}{{}^{60}{C}_5}$

$=1-\frac{117}{4484}$

$=\frac{4484-117}{4484}=\frac{4367}{4484}$