Q. 53.7( 3 Votes )

The contents of <

Answer :

Given:


Bag I contains 1 white, 2 black and 3 red balls


Bag II contains 2 white, 1 black and 1 red ball


Bag III contains 4 white, 5 black and 3 red balls


A bag is chosen and two balls are drawn from it.


There are three mutually exclusive ways to draw a white and a red ball from one of the three bags –


a. Bag I is selected, and then, a white and a red ball are drawn from bag I


b. Bag II is selected, and then, a white and a red ball are drawn from bag II


c. Bag III is selected, and then, a white and a red ball are drawn from bag III


Let E1 be the event that bag I is selected, E2 be the event that bag II is selected and E3 be the event that bag III is selected.


Since there are only three bags and each bag has an equal probability of being selected, we have



Let E4 denote the event that a white and a red ball are drawn.


Hence, we have







We also have







Similarly, we also have







Using the theorem of total probability, we get


P(E4) = P(E1)P(E4|E1) + P(E2)P(E4|E2) + P(E3)P(E4|E3)








Thus, the probability of the drawn balls being white and red is.


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