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 E_{1} be the event that bag I is selected, E_{2} be the event that bag II is selected and E_{3} 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 E_{4} 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(E_{4}) = P(E_{1})P(E_{4}|E_{1}) + P(E_{2})P(E_{4}|E_{2}) + P(E_{3})P(E_{4}|E_{3})

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

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