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

P(selecting first bowl) = 1/3

P(selecting second bowl) = 1/3

P(selecting third bowl) = 1/3

P(both balls are red | first bowl was selected) = 3/5 * 2/4 = 3/10

P(both balls are red | second bowl was selected) = 2/3 * 1/2 = 1/3

P(both balls are red | third bowl was selected) = 0

P(both balls are red) = P(both balls are red | first bowl was selected) * P(selecting first bowl) + P(both balls are red | second bowl was selected) * P(selecting second bowl) + P(both balls are red | third bowl was selected) * P(selecting third bowl)

                                  = 3/10 * 1/3 + 1/3 * 1/3 + 1/3 * 0

                                  = 19/20 or 0.2111