A bag contains 4 red balls, 6 blue balls and 8 pink balls. One ball is drawn at random and replaced with 3 pink balls. A probability that the first ball drawn was either red or blue in colour and the second drawn was pink in colour ?
Correct Answer: None of these
Number of Red balls = 4
Number of Blue balls = 6
Number of Pink balls = 8
Total number of balls = 4 + 6 + 8 = 18
Required probability
$$\eqalign{ & = \frac{4}{{18}} \times \frac{{11}}{{20}} + \frac{6}{{18}} \times \frac{{11}}{{20}} \cr & = \frac{{11}}{{20}}\left \cr & = \frac{{11}}{{20}} \times \frac{{10}}{{18}} \cr & = \frac{{11}}{{36}} \cr} $$
Number of Blue balls = 6
Number of Pink balls = 8
Total number of balls = 4 + 6 + 8 = 18
Required probability
$$\eqalign{ & = \frac{4}{{18}} \times \frac{{11}}{{20}} + \frac{6}{{18}} \times \frac{{11}}{{20}} \cr & = \frac{{11}}{{20}}\left \cr & = \frac{{11}}{{20}} \times \frac{{10}}{{18}} \cr & = \frac{{11}}{{36}} \cr} $$