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1. 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 ?

A. $\frac{12}{21}$

B. $\frac{13}{17}$

C. $\frac{11}{30}$

D. $\frac{13}{18}$

Solution:

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

\begin{aligned} = \frac{4}{18} \times \frac{11}{20} + \frac{6}{18} \times \frac{11}{20} \\ = \frac{11}{20} [\frac{4}{18} + \frac{6}{18}] \\ = \frac{11}{20} \times \frac{10}{18} \\ = \frac{11}{36} \end{aligned}

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