One bag A contains 4 white and 5 black balls another bag B contains 5 white and 6 black balls. One bag is selected at random and a ball is drawn from it. The probability that it is white is

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$\frac{89}{198}$

$\frac{47}{198}$

$\frac{17}{198}$

$\frac{97}{198}$

answer is B.

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Detailed Solution

$\begin{array}{l} Bag A : 4 W 5 B \\ Bag B : 5 W 6 B \end{array}$

Probability of selecting bag $= \frac{1}{2}$

Required probability $= (\frac{1}{2} \times \frac{4}{9}) + (\frac{1}{2} \times \frac{5}{11})$

$= \frac{1}{2} (\frac{44 + 45}{99})$

$= \frac{89}{198}$

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