Question

We have 3 bowls,

1) The first bowl contains 3 red and 2 green balls

2) the second bowl contains 2 red and 1 white balls

3) the third one contains 1 red and 3 green balls

One bowl by the random is selected and then 2 balls will be drawn. what is the probability that both of these balls will be red?

Answer 1

There are 3 bowls. 1st bowl contains 3 red and 2 green balls, i.e., 5 balls in total. 2nd bowl contains 2 red and 1 white balls, i.e., 3 balls in total. 3rd bowl contains 1 red and 3 green balls, i.e., 4 balls in total.

Now any bowl can be selected equally likely, i.e., any bowl can be selected with probability 1/3.

Now, given 1st bowl is selected, 2 red balls from 3 red balls can be selected in ³C₂ possible ways. Now, 2 balls from total 5 balls can be selected in ⁵C₂ possible ways.

Thus, P(selecting 2 red balls | 1st bowl is selected) =

³C₂/⁵C₂=3/10=0.3.

Given the 2nd bowl is selected, 2 red balls from 2 red balls can be selected in ²C₂=1 possible way. And 2 balls from total of 3 balls can be selected in ³C₂=3 possible ways.

Thus, P(selecting 2 red balls | 2nd bowl is selected) =1/3.

Again, given the 3rd bowl is selected, 2 red balls from 1 red ball can be selected in 0 way.

So, P(selecting 2 red balls | 3rd bowl is selected) =0.

Thus, P(selecting 2 red balls)

=P[(selecting 2 red balls and the 1st bowl) (selecting 2 red balls and the 2nd bowl) (selecting 2 red balls and the 3rd bowl) ]

=P(selecting 2 red balls and the 1st bowl) +P(selecting 2 red balls and the 2nd bowl) +P(selecting 2 red balls and the 3rd bowl) [since the events are mutually exclusive events]

=P(selecting 2 red balls | 1st bowl is selected)*P(selecting the qst bowl) +P(selecting 2 red balls | 2nd bowl is selected)*P(selecting the 2nd bowl) +P(selecting 2 red balls | 3rd bowl is selected)*P(selecting the 3rd bowl)

={(3/10)*(1/3) }+{(1/3) *(1/3) }+{0*(1/3) }

=19/90

=0.211.

Thus the required probability is 0.211.