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  • A bag contains 4 white and 5 black balls. Another bag contains 9 white and 7 black balls. A ball is transferred from the first bag to the second and then a ball is drawn at random from the second bag. Find the probability that the ball drawn is white.

A bag contains 4 white and 5 black balls. Another bag contains 9 white and 7 black balls. A ball is transferred from the first bag to the second and then a ball is drawn at random from the second bag. Find the probability that the ball drawn is white.

Answers (1)

Given-

W_1= [4 white balls] and B_1= [5 black balls]

 W_2= [9 white balls] and B_2= [7 black balls]

Let E_1 be the event that the ball transferred from the first bag is white and

E_2 be the event that the ball transferred from the bag is black.

E is the event that the ball drawn from the second bag is white.

\begin{aligned} &\therefore \mathrm{P}\left(\mathrm{E} \mid \mathrm{E}_{1}\right)=\frac{0}{17}, \mathrm{P}\left(\mathrm{E} \mid \mathrm{E}_{2}\right)=\frac{9}{17}\\ &\text { And }\\ &\mathrm{P}\left(\mathrm{E}_{1}\right)=\frac{4}{9} \text { and } \mathrm{P}\left(\mathrm{E}_{2}\right)=\frac{5}{9}\\ &\therefore P(E)=P\left(E_{1}\right) \cdot P\left(E \mid E_{1}\right)+P\left(E_{2}\right) \cdot P\left(E \mid E_{2}\right)\\ &=\frac{4}{9} \times \frac{10}{17}+\frac{5}{9} \times \frac{9}{17}\\ &=\frac{40+45}{153}\\ &=\frac{85}{153}\\ &=\frac{5}{9} \end{aligned}

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