80.6k views
4 votes
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
by
7.6k points

No related questions found

Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.4m questions

12.2m answers

Categories