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Question

A bag X contains 2 white and 3 black balls and another bag Y contains 4 white and 2 black balls. One bag is selected at random and a ball is drawn from it. Then, the probability chosen to be white is
(a) 2/15
(b) 7/15
(c) 8/15
(d) 14/15

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Solution

(c) 8/15

A white ball can be drawn in two mutually exclusive ways:
(I) Selecting bag X and then drawing a white ball from it.
(II) Selecting bag Y and then drawing a white ball from it.

Let E1, E2 and A be three events as defined below:
E1 = Selecting bag X
E2 = Selecting bag Y
A = Drawing a white ball

We know that one bag is selected randomly.

PE1=12 PE2=12 PA/E1=25PA/E2=46=23Using the law of total probability, we getRequired probability = PA=PE1PA/E1+ PE2PA/E2 =12×25+12×23 =15+13 =3+515=815

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