Question

The contents of three bags I, II and III are as follows:

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

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

Bag III: 4 white, 5 black and 3 red balls A bag is chosen at random and two balls are drawn. What is the probability that the balls drawn are white and red?

Answer 1

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

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 \(\cfrac{118}{495}\).