Discussion

Explanation:

Let us calculate P (selecting 1^st urn and drawing a red ball from it).

Out of three urns probability of selecting 1^st urn = 1/3.

P (drawing a red ball from 1^st urn) = $\frac{6}{10}$ = $\frac{3}{5}$

P (selecting 1^st urn and drawing a red ball from it) = $\frac{1}{3}\times \frac{3}{5}$ = $\frac{3}{15}$

Similarly,

P (selecting 2^nd urn and drawing a red ball from it) = $\frac{1}{3}\times \frac{2}{5}$ = $\frac{2}{15}$

P (selecting 3^rd urn and drawing a red ball from it) = $\frac{1}{3}\times \frac{1}{2}$ = $\frac{1}{6}$

P (If the ball drawn is red, find the probability that it is drawn from the first urn) 

= $\frac{\mathrm{P}\left(\mathrm{red}\mathrm{ }\mathrm{ball}\mathrm{ }\mathrm{from}\mathrm{ }1\mathrm{st}\mathrm{ }\mathrm{urn}\right)}{\mathrm{P}\left(\mathrm{red}\mathrm{ }\mathrm{ball}\mathrm{ }\mathrm{from}\mathrm{ }1\mathrm{st}\mathrm{ }\mathrm{urn}\right)+\mathrm{P}\left(\mathrm{red}\mathrm{ }\mathrm{ball}\mathrm{ }\mathrm{from}\mathrm{ }2\mathrm{nd}\mathrm{ }\mathrm{urn}\right)+\mathrm{P}\left(\mathrm{red}\mathrm{ }\mathrm{ball}\mathrm{ }\mathrm{from}\mathrm{ }3\mathrm{rd}\mathrm{ }\mathrm{urn}\right)}$ 

= $\frac{\frac{3}{15}}{\frac{3}{15}+\frac{2}{15}+\frac{1}{6}}$ = $\frac{\frac{3}{15}}{\frac{6+4+5}{30}}$ = $\frac{3}{15}\times \frac{30}{15}$ = $\frac{2}{5}$

Hence, option (a).

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