Question

In a bag, there are six white balls and four black balls. I take out two balls at random, one at a time, without replacement. What is the probability that the second ball I take is black

Answer 1

`Probability = 0.4`

Explanation:

Given

`White = 6`

`Black = 4`

Required

Probability that second is black

This selection can be represented as:

(Black and Black) or (White and Black)

The probability of Black and Black is:

`P(Black\ Only) = (Black)/(Total) * (Black - 1)/(Total - 1)`

1 is subtracted from the second fraction because it is probability without replacement

`P(Black\ Only) = (4)/(10) * (4- 1)/(10 - 1)`

`P(Black\ Only) = (4)/(10) * (3)/(9)`

`P(Black\ Only) = (2)/(5) * (1)/(3)`

`P(Black\ Only) = (2)/(15)`

The probability of White and Black is:

`P(Black\ and\ White) = (Black)/(Total) * (White)/(Total - 1)`

1 is subtracted from the second fraction because it is probability without replacement

`P(Black\ and\ White) = (4)/(10) * (6)/(10- 1)`

`P(Black\ and\ White) = (4)/(10) * (6)/(9)`

`P(Black\ and\ White) = (2)/(5) * (2)/(3)`

`P(Black\ and\ White) = (4)/(15)`

So, the required probability is:

`Probability = P(Black\ Only) + P(Black\ and\ White)`

`Probability = (2)/(15) + (4)/(15)`

`Probability = (2+4)/(15)`

`Probability = (6)/(15)`

`Probability = 0.4`