Question

A bag contains 6 red, 4 white and 8 blue balls. If three balls are drawn at random, find the probability that :

(i) one is red and two arc white

(ii) two are blue and one is red

(iii) one is red.

Answer 1

Bag, 6-red ball, 4-white ball, 8-blue ball
Since three ball are drawn

$\therefore n\left(S\right){=}^{18}{C}_3$

(i) Let E be the event that one red and two white ball are drawn.

$\therefore n\left(E\right){=}^6{C}_1{\times }^4{C}_2$

$\therefore P\left(E\right)=\frac{{}^6{C}_1{\times }^4{C}_2}{{}^{18}{C}_3}=\frac{\frac{6!}{1!\times 5!}\times \frac{4!}{2!\times 2!}}{\frac{18!}{3!\times 5!}}=\frac{6\times 4\times 3}{2}\times \frac{3\times 2}{18\times 17\times 16}$

$PE=\frac{3}{68}$

(ii) Let E be the event that two blue and one red ball was drawn.

$\therefore n\left(E\right){=}^8{C}_2{\times }^6{C}_1$

$\therefore P\left(E\right)=\frac{{}^8{C}_2{\times }^6{C}_1}{{}^{18}{C}_3}=\frac{\frac{8!}{2!\times 6!}\times \frac{6!}{1!\times 5!}}{\frac{18!}{3!\times 5!}}=\frac{8\times 7}{2}\times 6\times \frac{3\times 2\times 1}{18\times 17\times 16}=\frac{7}{34}$

$P\left(E\right)=\frac{7}{34}$

(iii) Let E be the event that one of the ball must be red.

$\therefore$ E = {(R, W, B) or (R, W, W) or (R, B, B)}

$\therefore n\left(E\right){=}^6{C}_1{\times }^4{C}_1{\times }^8{C}_1{+}^6{C}_1{\times }^4{C}_2{+}^6{C}_1{\times }^8{C}_2$

$\therefore P\left(E\right)=\frac{{}^6{C}_1{\times }^4{C}_1{\times }^8{C}_1{+}^6{C}_1{\times }^4{C}_2{+}^6{C}_1{\times }^8{C}_2}{{}^{18}{C}_3}=\frac{396}{816}=\frac{33}{68}$