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  • Bag I contains 3 red, 4 black and 3 white balls and Bag II contains 2 red, 5 black and 2 white balls. One ball is transferred from Bag I to Bag II and then a ball is drawn from Bag II. The ball so

Bag I contains 3 red, 4 black and 3 white balls and Bag II contains 2 red, 5 black and 2 white balls. One ball is transferred from Bag I to Bag II and then a ball is drawn from Bag II. The ball so drawn is found to be black in colour. Then the probability, that the transferred ball is red, is :

Option: 1

\frac{4}{9}


Option: 2

\frac{5}{18}


Option: 3

\frac{1}{6}


Option: 4

\frac{3}{10}


Answers (1)

best_answer

\begin{aligned} &\text{Using Baye's Thearem, required probability}\mathrm{ =\frac{3}{10} \times \frac{5}{10}}\\ &\mathrm{ \frac{3}{10} \times \frac{5}{10}+\frac{4}{10} \times \frac{6}{10}+\frac{3}{10} \times \frac{5}{10} }\\ =&\mathrm{ \frac{15}{15+24+15}=\frac{15}{54}=\frac{5}{18}}\\ &\therefore \text{ option (B)} \end{aligned}

Posted by

Divya Prakash Singh

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