A bag contains 7 white, 5 black and 4 red balls. If two balls are drawn at random, find the probability that: 1. Both the balls are white 2. One ball is black and the other red 3. Both the balls are of the same colour.

BAG 7-white ball 5-black ball 4-blue ball Two balls are drawn (S)=^ 16 C_2 1. Let E be the event that both the balls are white (E)=^ 7 C_2 (E)= ^ 7 C_2 ^ 16 C_2 =7 6 16 15 =7 40 (E)=7 40 2. Let E be the event that, one ball and one red ball is drawn (E)=^5C_1 ^4C_1 (E)= ^ 5 C_1 ^ 4 C_1 ^ 16 C_1 =5 4 2 16 15 =1 6 (E)=1 6 3. Let E be the event that both the balls are of the same colour. (E)=^7C_2 or ^ 5 C_2 or ^4C_2 (E)= ^ 7 C_2+^ 5 C_2+^4C_2 ^ 16 C_2 =7 6+5 4+4 2 16 15 =70 240 =7 24

A bag contains 7 white, 5 black and 4 red balls. If two balls are…

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Question

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

Both the balls are white
One ball is black and the other red
Both the balls are of the same colour.

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Answer

BAG 7-white ball 5-black ball 4-blue ball $\because$ Two balls are drawn $\therefore\text{n}(\text{S})=^{16}\text{C}_2$

Let E be the event that both the balls are white

$\therefore\text{n}(\text{E})=^{7}\text{C}_2$
$\therefore\text{p}(\text{E})=\frac{^{7}\text{C}_2}{^{16}\text{C}_2}=\frac{7\times6}{16\times15}=\frac{7}{40}$
$\therefore\text{p}(\text{E})=\frac{7}{40}$

Let E be the event that, one ball and one red ball is drawn

$\therefore\text{n}(\text{E})=^5\text{C}_1\times^4\text{C}_1$
$\therefore\text{p}(\text{E})=\frac{^{5}\text{C}_1\times^{4}\text{C}_1}{^{16}\text{C}_1}=\frac{5\times4\times2}{16\times15}=\frac{1}{6}$
$\therefore\text{p}(\text{E})=\frac{1}{6}$

Let E be the event that both the balls are of the same colour.

$\therefore\text{n}(\text{E})=^7\text{C}_2\text{ or }^{5}\text{C}_2\text{ or }^4\text{C}_2$
$\therefore\text{p}(\text{E})=\frac{^{7}\text{C}_2+^{5}\text{C}_2+^4\text{C}_2}{^{16}\text{C}_2}$
$=\frac{7\times6+5\times4+4\times2}{16\times15}=\frac{70}{240}=\frac{7}{24}$

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