Question

The contents of 3 bags w.r.t green and red marbles is as given in the table shown. A child randomly selects one of the bags, and draws a marble from it and retains it. If the marble is green, the child draws the second marbles randomly from one of the two remaining bags. If the first marble drawn is red the child draws one more marble from the same bag. The probability that the second drawn marble is green is expressed as $\frac{m}{n}$ (where $m$ and $n$ are coprime). Find the value of $(m+n)$.

Answer 1

Answer: 217

Let $E$ : second drawn marble is green.
$P(E)=\frac{1}{3}\left[\left\{\frac{1}{4}\cdot 1+\frac{3}{4}\cdot \frac{1}{2}\left(\frac{1}{2}+\frac{1}{4}\right)\right\}+\frac{1}{2}\left\{\frac{2}{3}+\frac{1}{2}\left(\frac{3}{4}+\frac{1}{4}\right)\right\}+\left\{\frac{3}{4}\cdot \frac{1}{3}+\frac{1}{4}\cdot \frac{1}{2}\left(\frac{3}{4}+\frac{1}{2}\right)\right\}\right]$
$=\frac{1}{3}\left[\frac{1}{4}+\frac{3}{8}\cdot \frac{3}{4}+\frac{1}{2}\left(\frac{2}{3}+\frac{1}{2}\right)+\left(\frac{3}{4}\cdot \frac{1}{3}+\frac{1}{8}\cdot \frac{5}{4}\right)\right]=\frac{1}{3}\left[\frac{1}{4}+\frac{9}{32}+\frac{7}{12}+\frac{13}{32}\right]$
$=\frac{1}{3}\left[\frac{24+27+56+39}{96}\right]=\frac{1}{3}\left[\frac{146}{96}\right]=\frac{1}{3}\left[\frac{73}{48}\right]=\frac{73}{144}$