A bag contains 3 white, 3 black and 2 red balls. One by one, three balls are drawn without replacing them. Find the probability that the third ball is red.

Answer

`1//4`

Explanation

If third ball is red, then in first two draws, there will be either no red ball or one red ball.
Let events
`R_(0)=` in first two draws, red ball is drawn
`R_(1)` = in first two draws, one red ball is drawn.
`R=` in third draw , red ball is drawn.
So, from total probability theorem,
`P(R)=P(R_(0))P(R//R_(0))+P(R_(1))P(R//R_(1))`
`=(""^(6)C_(2))/(""^(8)C_(2)).(""^(2)C_(1))/(""^(6)C_(1))+(""^(6)C_(1)""^(2)C_(1))/(""^(8)C_(2)).(""^(1)C_(1))/(""^(6)C_(1))`
`=15/28.(2)/(6)+(6xx2)/(28).1/6=1/4`