Question

An urn contains 4 white and 4 black balls. We randomly choose 4 balls. If 2 of them are white and 2 are black, we stop. If not, we replace the balls in the urn and again randomly select 4 balls. This continues until exactly 2 of the 4 chosen are white. What is the probability that we shall make exactly n selections?

Answer 1

The probability that we shall make exactly n selections is
`P(X = n)=((17)/(35))^(n-1)(18)/(35)`.

Explanation:

It is given that an urn contains 4 white and 4 black balls and we randomly choose 4 balls. If 2 of them are white and 2 are black, we stop.

The total number of ways to select exactly 2 white and 2 black balls.

`^4C_2* ^4C_2=(4!)/(2!(4-2)!)* (4!)/(2!(4-2)!)=6* 6=36`

The total number of ways to select 4 balls from 8 balls is

`^8C_4=(8!)/(4!(8-4)!)=(8* 7* 6* 5* 4!)/(4* 3* 2* 1* !4!)=70`

The probability of selecting exactly 2 white and 2 black balls is

`p=(36)/(70)=(18)/(35)`

The probability of not selecting exactly 2 white and 2 black balls is

`q=1-p=1-(18)/(35)=(17)/(35)`

If we not get exactly 2 white and 2 black balls, then we replace the balls in the urn and again randomly select 4 balls.

The probability that we shall make exactly n selections is

`P(X = n)=(q)^(n-1)p`

`P(X = n)=((17)/(35))^(n-1)(18)/(35)`

Therefore the probability that we shall make exactly n selections is
`P(X = n)=((17)/(35))^(n-1)(18)/(35)`.