A bag contains 8 red and 5 white balls.

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) $=\frac{{ }^{8} C_{1} \times{ }^{5} C_{2}}{{ }^{13} C_{3}}$

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

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

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

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

$=\frac{40}{143}$