Question

A bag contains 3 red, 4 white and 5 blue balls. All balls are different. Two balls are drawn at random. The probability that they are of different colour is

• A. $\frac { 47 } { 66 }$
• B. $\frac { 10 } { 33 }$
• C. $\frac { 5 } { 22 }$
• D. None of these

Answer 1

Answer: (A)

$\frac { 47 } { 66 }$

Answer 2

We have the following three pattern :

(i) Red, white $P ( A ) = \frac { 3 \times 4 } { { } ^ { 12 } C _ { 2 } }$

(ii) Red, blue $P ( B ) = \frac { 3 \times 5 } { { } ^ { 12 } C _ { 2 } }$

(iii) Blue, white $P ( C ) = \frac { 4 \times 5 } { { } ^ { 12 } C _ { 2 } }$

Since all these cases are exclusive, so the required probability $= \frac { ( 12 + 15 + 20 ) } { { } ^ { 12 } C _ { 2 } } = \frac { ( 47 \times 2 ) } { ( 12 \times 11 ) } = \frac { 47 } { 66 }$.