# A bag contains 8 red and 5 white balls.

Question:

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

(a) All the three balls are white

(b) All the three balls are red

(c) One ball is red and two balls are white

Solution:

Given that, number of red balls = 8

Number of white balls = 5

∴ Total balls, n = 13

It is given that 3 balls are drawn at random

⇒ r = 3

∴ n(S) = nCr = 13C3

(a) All the three balls are white

We know that,

$P(A)=\frac{n(A)}{n(S)}=\frac{\text { Number of favourable outcomes }}{\text { Sample Space }}$

Total white balls are $=5$

$P$ (all the three balls are white) $=\frac{{ }^{5} C_{3}}{{ }^{13} C_{3}}$

$=\frac{\frac{5 !}{3 !(5-3) !}}{\frac{13 !}{3 !(13-3) !}}\left[\because{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}}=\frac{\mathrm{n} !}{(\mathrm{n}-\mathrm{r}) ! \mathrm{r} !}\right]$

$=\frac{\frac{5 \times 4 \times 3 !}{3 ! 2 !}}{\frac{13 \times 12 \times 11 \times 10 !}{3 \times 2 \times 1 \times 10 !}}$

$=\frac{\frac{5 \times 4}{2 \times 1}}{13 \times 2 \times 11}$

$=\frac{5}{143}$

(b) All the three balls are red We know that,

$P(A)=\frac{n(A)}{n(S)}=\frac{\text { Number of favourable outcomes }}{\text { Sample Space }}$

Total red balls are $=8$

$\mathrm{P}$ (all the three balls are red) $=\frac{{ }^{8} \mathrm{C}_{3}}{{ }^{13} \mathrm{C}_{3}}$

$=\frac{\frac{8 !}{3 !(8-3) !}}{\frac{13 !}{3 !(13-3) !}}\left[\because{ }^{\mathrm{n}} C_{\mathrm{r}}=\frac{\mathrm{n} !}{(\mathrm{n}-\mathrm{r}) ! \mathrm{r} !}\right]$

$=\frac{\frac{8 \times 7 \times 6 \times 5 !}{3 ! 5 !}}{\frac{13 \times 12 \times 11 \times 10 !}{3 \times 2 \times 1 \times 10 !}}$

$=\frac{\frac{8 \times 7 \times 6}{3 \times 2}}{13 \times 2 \times 11}$

$=\frac{28}{143}$

$=\frac{\frac{8 \times 7 \times 6}{3 \times 2}}{13 \times 2 \times 11}$

$=\frac{28}{143}$

(c) One ball is red and two balls are white We know that,

$P(A)=\frac{n(A)}{n(S)}=\frac{\text { Number of favourable outcomes }}{\text { Sample Space }}$

$P$ (one ball is red and two balls are white) $=\frac{{ }^{8} C_{1} \times{ }^{5} C_{2}}{{ }^{13} C_{3}}$

$=\frac{\frac{8 !}{1 !(8-1) !} \times \frac{5 !}{2 !(5-2) !}}{\frac{13 !}{3 !(13-3) !}}$

$\left[\because{ }^{\mathrm{n}} C_{\mathrm{r}}=\frac{\mathrm{n} !}{(\mathrm{n}-\mathrm{r}) ! \mathrm{r} !}\right]$

$=\frac{\frac{8 \times 7 !}{7 !} \times \frac{5 \times 4 \times 3 !}{2 \times 1 \times 3 !}}{\frac{13 \times 12 \times 11 \times 10 !}{3 \times 2 \times 1 \times 10 !}}$

$=\frac{8 \times 10}{13 \times 2 \times 11}$

$=\frac{40}{143}$