Question

A box contains 3 white and 2 red balls. A ball is drawn and another ball is drawn without replacing first ball, then the probability of second ball to be red is

• A. $\frac { 8 } { 25 }$
• B. $\frac { 2 } { 5 }$
• C. $\frac { 3 } { 5 }$
• D. $\frac { 21 } { 25 }$

Answer 1

Answer: (B)

$\frac { 2 } { 5 }$

Answer 2

The second ball can be red in two different ways

(i) First is white and second red $P ( A ) = \frac { 3 } { 5 } \times \frac { 2 } { 4 } = \frac { 6 } { 20 }$

(ii) First is red and second is also red $P ( B ) = \frac { 2 } { 5 } \times \frac { 1 } { 4 } = \frac { 2 } { 20 }$

Both are mutually exclusive events, hence required probability is $\frac { 6 } { 20 } + \frac { 2 } { 20 } = \frac { 8 } { 20 } = \frac { 2 } { 5 }$