Question

A bag contains 8 red and 5 white balls. Three balls are drawn at random. Find the probability

(i) All the three balls are while.

(ii) All the three balls are red.

(iii) One ball is red and two balls are white.

Answer 1

$\because$ Number of red balls = 8 and number of white balls = 5
Total number of possible outcomes=${}^{13}{C}_3$
i.e., selecting three balls out of 13 balls.
As we know that the probability of an event is $\frac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}}$

(i) P (all the three balls are white)

Number of favourable out come=${}^5{C}_3$
Thus, the probability of selecting all the three balls are white $=\frac{{}^5{C}_3}{{}^{13}{C}_3}=\frac{\frac{5!}{3!2!}}{\frac{13!}{3!10!}}=\frac{5!}{3!2!}\times \frac{3!10!}{13!}$

$=\frac{5\times 4\times 3\times 2!}{2!}\times \frac{10!}{13\times 12\times 11\times 10!}=\frac{5\times 4\times 3}{13\times 12\times 11}=\frac{5}{13\times 11}=\frac{5}{143}$

(ii) P (all the three balls are red)

Number of favourable out come=${}^8{C}_3$
Thus, the probability of selecting all the three balls are red$=\frac{{}^8{C}_3}{{}^{13}{C}_3}=\frac{\frac{8!}{3!5!}}{\frac{13!}{3!10!}}=\frac{8!}{3!5!}\times \frac{3!10!}{13!}$

$=\frac{8\times 7\times 6\times 5!}{5!}\times \frac{10!}{13\times 12\times 11\times 10!}=\frac{8\times 7\times 6}{13\times 12\times 11}=\frac{28}{143}$

(iii) P (one ball is red and two balls are white)

Number of favourable outcomes${=}^8{C}_1{\times }^5{C}_2$
Thus, required probability $=\frac{{}^8{C}_1{\times }^5{C}_2}{{}^{13}{C}_3}=\frac{8\times 10}{13\times 6\times 11}=\frac{40}{143}$