Question

A bag contains 4 red and 5 black balls. Another bag contains 5 red, 3 black balls. If one ball is drawn at random from each bag. Then the probability that one red and one black ball drawn is _______.
(a) 12/72 (b) 25/72 (c) 37/72 (d) 13/72

Answer 1

Let the two bags be A and B

A contains 4red 5 black balls

B contains 5 red and 3 black balls.

One ball is drawn from each bag. One must be red and one is black.
Red ball from A and black ball from B or black from A and red from B.

Probability =4/9*3/8+5/9*5/8=37/72.

Answer 2

Given: A bag contains 4 red and 5 black balls. Another bag contains 5 red, 3 black balls. If one ball is drawn at random from each bag.

To find: The probability that one red and one black ball drawn.

Solution:

In order to find the stated probability, first, the probability of getting a red ball from the first bag and getting a black ball from the second bag is to be found. Or, the probability of getting a black ball from the first bag and getting a red ball from the second bag is to be found. This can be written in the form of a formula as follows.

$P(E) = \frac{1}{2}*\frac{4}{9}*\frac{1}{2} *\frac{3}{8} + \frac{1}{2}*\frac{5}{9}*\frac{1}{2} *\frac{5}{8}$

         $= \frac{12+25}{288}$

         $= \frac{37}{288}$

Therefore, the probability that one red and one black ball is drawn is 37/288.