An urn contains 7 white, 5 black and 3 red balls. Two balls are d… - Vidyadip

Question

An urn contains 7 white, 5 black and 3 red balls. Two balls are drawn at random. Find the probability that

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

✓

Answer

Urn7-White balls
5-Black balls
3-Red balls
Since two balls are drawn at random
$\therefore\text{n}(\text{S})=\frac{15}{2}$

E be the event that both the balls are red

$\therefore\text{n}(\text{S})=^3\text{C}_2$
$\therefore\text{p}(\text{E})=\frac{^3\text{C}_2}{^{15}\text{C}_2}=\frac{3\times2}{15\times14}=\frac{1}{35}$

E be the event that one ball is red and other is black

$\therefore\text{n}(\text{E})=^3\text{C}_1\times^5\text{C}_1$
$\therefore\text{p}(\text{E})=\frac{^3\text{C}_1\times^5\text{C}_1}{^{15}\text{C}_2}$
$=\frac{3\times5\times2}{15\times16}=\frac{1}{7}$

E be the event that one ball is white

$\therefore\text{n}(\text{E})=^7\text{C}_1\times^8\text{C}_1$
$\therefore\text{p}(\text{E})=\frac{^7\text{C}_1\times^8\text{C}_1}{^{15}\text{C}_2}$
$=\frac{7\times6\times2}{14\times15}=\frac{8}{15}$

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