A bag contains 3 red and 3 green balls and a person draws out 3 at random. He then drops 3 blue balls into the bag and again draws out 3 at random. The chance that the 3 later balls being all of different colors is `15 %` b. `20 %` c. `27 %` d. `40 %`

Options

A. `15%`
B. `20%`
C. `27%`
D. `40%`

Answer

C

Explanation

The required probability is
`(""^(3)C_(1)""^(3)C_(2))/(""^(6)C_(3))(""^(2)C_(1)""^(1)C_(1)""^(3)C_(1))/(""^(6)C_(3))+(""^(3)C_(2)""^(3)C_(1))/(""^(6)C_(3))(""^(1)C_(1)""^(2)C_(1)""^(3)C_(1))/(""^(6)C_(3))`
`=2(9)/(20)xx(6)/(20)=(27)/(100)`