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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

1 Answer

4 votes

Answer:


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)

User Bill Reiss
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