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One bag contains $5$ white and $4$ black balls. Another bag contains $7$ white and $9$ black balls. A ball is transferred from the first bag to the second and then a ball is drawn from second. The probability that the ball is white, is
$\frac{8}{{17}}$
$\frac{{40}}{{153}}$
$\frac{5}{9}$
$\frac{4}{9}$
Solution
(d) Let $ a$ white ball be transferred from the first bag to the second.
The Probability of selecting a white ball from the first bag $ = \frac{5}{9}.$
Now the second bag has $8$ white and $9$ black.
The probability of selecting white ball from the second bag $ = \frac{8}{{17}}$.
Hence required probability $ = \frac{5}{9} \times \frac{8}{{17}} = \frac{{40}}{{153}}$
If a black ball be transferred from the first bag to the second, then the probability $ = \frac{4}{9} \times \frac{7}{{17}} = \frac{{28}}{{153}}$
Therefore required probability $ = \frac{{40}}{{153}} + \frac{{28}}{{153}} = \frac{4}{9}.$
