A) \[\frac{27}{49}\]
B) \[\frac{26}{49}\]
C) \[\frac{32}{49}\]
D) \[\frac{21}{49}\]
Correct Answer: C
Solution :
5 Red and 2 green balls P(one red ball) \[=\,\,\frac{5}{7}\] P(one green ball) \[=\,\,\frac{2}{7}\] Case I: If drawn ball is green than a red ball is added \[\left( \begin{align} & 6\,\,\operatorname{Re}d \\ & 1\,\,Green \\ \end{align} \right)P\,\,(red\,\,ball)\,\,=\,\,\frac{6}{7}\] Case II: If drawn ball is red then a green ball is added \[\left( \begin{align} & 4\,\,\operatorname{Re}d \\ & 3\,\,Green \\ \end{align} \right)P\,\,(red\,\,ball)\,\,=\,\,\frac{4}{7}\] \[P({{2}^{nd}}\,red\,\,ball)\,\,=\,\,\frac{5}{7}\,\,\times \,\,\frac{4}{7}\,\,+\frac{2}{7}\,\,\times \frac{6}{7}=\frac{32}{49}\]
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