A box contains 5 red balls, 4 white balls and 3 blue balls . A ball is selected at random from the box . the colour of the ball is noted and then replaced back . Find the probability that out of 6 balls selected in this manner , 3 are red, 2 are white , 1 is blue.

Solution :

Let P(R) be the probability that red ball is selected i.e. P(R) = 5/12

Let P(W) be the probability that white ball is selected i.e. P(W) = 4/12

Let P(B) be the probability that blue ball is selected i.e. P(B) = 3/12

Now the probability that out of 6 balls selected 3 are red, 2 are white , 1 is blue

= P(R).P(R). P(R) + P(W).P(W) + P(B)

= 5/12 . 5/12. 5/12 + 4/12 . 4/12 + 3/12

= 144/12

= 12

_{(Is this correct ?)}