Question

An urn contains 9 red, 7 white and 4 black balls. If two balls are drawn at random, find the
probability that
(i) both the balls are red.
(ii) one is white and other is red.
(iii) the balls are of
same colour.​

Answer 1

Concept

• Number of ways in which we can choose n materials out of m materials is $^mC_n$.

Solution

Total number of balls = 20 balls.

Total number of ways in which we can choose 2 balls out of 20 balls is $^{20}C_2$.

(i) Number of ways in which we can choose two red balls is $^9C_2$. Hence, probability =

$\dfrac{ ^9C_2}{^{20}C_2} \\ \\ \\ = \: \dfrac{18}{95}$

(ii) Number of ways in which we can choose one white ball and one red ball is $^9C_1 \times ^7C_1$

Hence, Probability =

$\dfrac{^9C_1 \times ^7C_1}{^{20}C_2} \\ \\ \\ = \: \dfrac{63}{20}$

(iii) The balls can be of either red colour , or of white colour or of black colour. Hence, Total ways of choosing them is $^9C_2\:+\:^7C_2+\:^4C_2$

Probability =

$\dfrac{ ^9C_2\:+\:^7C_2+\:^4C_2} {^{20}C_2} \\ \\ \\ = \: \dfrac{63}{190}$

Answer 2

Answer:

Given :

No. of red balls = 9

No. of white balls = 7

No. of black balls = 4

Have to find :

Probability of

(i) both the balls are red.

(ii) one is white and other is red.

(iii) the balls are of

Solution :

At first we know the formula of number of ways

$\bf{ }^{m} c _{n}$

Number of balls = 9 + 7 + 4 = 20

9C2/20C2 = 18/95

For second

For third

9C2 + 7C2 + 4C2/20C2

63/190

$\sf \dfrac{ {}^{9}C_1\times{}^{7} C_1}{20 C_2}$

=> 63/20