Question 84

# A bag contains 5 white and 3 black balls; another bag contains 4 white and 5 black balls. From any one of these bags a single draw of two balls is made. Find the probability that one of them would be white and another black ball.

Solution

Let P(B1) and P(B2) denote the probabilities of the events of drawing balls from the bag I or bag II.

Since both the bags are equally likely to be selected, P(B1) = P(B2) = $$\ \frac{\ 1}{2}$$

Let E denote the event of drawing two balls of different colours from a bag.

The required probability

= P(B1)P($$\ \frac{\ E}{B_1}$$)+ P(B2)P($$\ \frac{\ E}{B_2}$$)

P($$\ \frac{\ E}{B_1}$$) denote the probability of drawing the two balls from bag I and as that from bag II.

= $$\ \frac{\ 1}{2}\left[\ \frac{\ ^5C_1\ \times\ \ ^3C_1}{^8C_2}\ +\ \ \frac{\ ^5C_1\ \times\ \ ^4C_1}{^9C_2}\right]$$

= $$\ \frac{\ 1}{2}\left[\ \frac{\ 15}{28}+\ \frac{\ 5}{9}\right]$$

= $$\ \frac{\ 275}{504}$$

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