Question

A hag contains 7 white, 5 black and 4 red balls. If two balls are drawn at random, find the probability that :

(i) both the balls are white

(ii)one ball is black and the other red

(iii) both the balls are of the same colour.

Answer 1

Bag, 7-white ball, 5-black ball, 4-blue ball,
$\because$ Two balls are drawn

$\therefore n\left(S\right){=}^{16}{C}_2$

(i) Let E be the event that both the balls are white

$\therefore n(E{)}^7{C}_2$

$\therefore P\left(E\right)=\frac{{}^7{C}_2}{{}^{16}{C}_2}=\frac{\frac{7!}{2!\times 5!}}{\frac{16!}{2!\times 14!}}$

$\therefore P\left(E\right)=\frac{7}{40}$

(ii) Let E be the even that, one black ball and one red ball is drawn

$\therefore n\left(E\right){=}^{5}C\times 1{\times }^4{C}_1$

$\therefore P\left(E\right)=\frac{{}^{5}C\times 1{\times }^4{C}_1}{{}^{16}{C}_2}=\frac{\frac{5!}{1!\times 4!}\times \frac{4!}{1!\times 3!}}{\frac{16!}{2!\times 14!}}$

$=\frac{5\times 4}{\frac{16\times 15}{1}}=\frac{5\times 4\times 2}{16\times 15}=\frac{1}{6}$

$\therefore P\left(E\right)=\frac{1}{6}$

(iii) Let E be the even that both the balls are of the same colour

$\therefore n\left(E\right){=}^7{C}_2$ or ${}^5{C}_2$ or ${}^4{C}_2$

$\therefore P\left(E\right)=\frac{{}^7{C}_2{+}^5{C}_2{+}^4{C}_2}{{}^{16}{C}_2}$

$=\frac{\frac{7!}{2!\times 5!}+\frac{5!}{2!\times 3!}+\frac{4!}{2!\times 2!}}{\frac{16!}{2!\times 14!}}$

$=\frac{7\times 6+5\times 4+4\times 2}{16\times 15}=\frac{74}{240}=\frac{37}{120}$