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The Bag a Contains 8 White and 7 Black Balls While the Bag B Contains 5 White and 4 Black Balls.Then a Ball is Randomly Drawn Out from It. Find the Probability that Ball Drawn is White. - Mathematics

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Question

The bag A contains 8 white and 7 black balls while the bag B contains 5 white and 4 black balls. One ball is randomly picked up from the bag A and mixed up with the balls in bag B. Then a ball is randomly drawn out from it. Find the probability that ball drawn is white.

Solution

A white ball can be drawn in two mutually exclusive ways:

(I) By transferring a black ball from bag A to bag B, then drawing a white ball
(II) By transferring a white ball from bag A to bag B, then drawing a white ball

Let E1E2 and A be events as defined below:
E1 = A black ball is transferred from bag A to bag B
E2 = A white ball is transferred from bag A to bag B
A = A white ball is drawn

\[\therefore P\left( E_1 \right) = \frac{7}{15}\]

\[ P\left( E_2 \right) = \frac{8}{15}\]

\[\text{ Now } , \]

\[P\left( A/ E_1 \right) = \frac{5}{10} = \frac{1}{2}\]

\[P\left( A/ E_2 \right) = \frac{6}{10} = \frac{3}{5}\]

\[\text{ Using the law of total probability, we get} \]

\[\text{ Required probability } = P\left( A \right) = P\left( E_1 \right)P\left( A/ E_1 \right) + P\left( E_2 \right)P\left( A/ E_2 \right)\]

\[ = \frac{7}{15} \times \frac{1}{2} + \frac{8}{15} \times \frac{3}{5}\]

\[ = \frac{7}{30} + \frac{8}{25}\]

\[ = \frac{35 + 48}{150} = \frac{83}{150}\]

 

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Chapter 31: Probability - Exercise 31.6 [Page 81]

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RD Sharma Mathematics [English] Class 12
Chapter 31 Probability
Exercise 31.6 | Q 8 | Page 81

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