Question

A box contains 5 red balls, 6 white balls and 9 black balls. Two balls are drawn at

random. (The first ball is not replaced). Find the probability that both balls are of
the same colour

Answer 1

`P(Same)=(61)/(190)`

Explanation:

Given

`Red = 5`

`White = 6`

`Black = 9`

Required

The probability of selecting 2 same colors when the first is not replaced

The total number of ball is:

`Total = 5 + 6 + 9`

`Total = 20`

This is calculated as:

`P(Same)=P(Red\ and\ Red) + P(White\ and\ White) + P(Black\ and\ Black)`

So, we have:

`P(Same)=(n(Red))/(Total) * (n(Red) - 1)/(Total - 1) + (n(White))/(Total) * (n(White) - 1)/(Total - 1) + (n(Black))/(Total) * (n(Black) - 1)/(Total - 1)`

Note that: 1 is subtracted because it is a probability without replacement

`P(Same)=(5)/(20) * (5 - 1)/(20- 1) + (6)/(20) * (6 - 1)/(20- 1) + (9)/(20) * (9- 1)/(20- 1)`

`P(Same)=(5)/(20) * (4)/(19) + (6)/(20) * (5)/(19) + (9)/(20) * (8)/(19)`

`P(Same)=(20)/(380) + (30)/(380) + (72)/(380)`

`P(Same)=(20+30+72)/(380)`

`P(Same)=(122)/(380)`

`P(Same)=(61)/(190)`