# Probabilities

• January 27th 2007, 10:21 PM
gretchen
Probabilities
If two balls are drawn (without replacement) from an urn that contains four red balls, five green balls and seven white balls. Find the probability that both balls are white.

Possible solutions:
2/3
1/9
7/40
49/240
• January 27th 2007, 11:21 PM
CaptainBlack
Quote:

Originally Posted by gretchen
If two balls are drawn (without replacement) from an urn that contains four red balls, five green balls and seven white balls. Find the probability that both balls are white.

Possible solutions:
2/3
1/9
7/40
49/240

Probability that first ball is white=#of white balls/total # of balls=7/16

Probability that the second ball is white given that the first is white=
#of white balls remaining/total # of balls remaining=6/15.

Then the required prob. is the product of these=(7/16)(6/15)=7/40

RonL
• January 28th 2007, 12:55 AM
gretchen
Thanks for all your help Ron. What do you do for a living?

Gretch
• January 28th 2007, 01:18 AM
CaptainBlack
Quote:

Originally Posted by gretchen
Thanks for all your help Ron. What do you do for a living?

Algorithm design consultant

RonL
• January 28th 2007, 05:10 AM
Soroban
Hello, Gretchen!

If you familiar with "combinations", here's another approach . . .

Quote:

If two balls are drawn (without replacement) from an urn that contains:
4 red balls, 5 green balls and 7 white balls, find the probability that both balls are white.

Possible solutions: . $\frac{2}{3}\quad\frac{1}{9}\quad\frac{7}{40}\quad\ frac{49}{240}$

There are $16$ balls.

There are $\binom{16}{2}\,=\,120$ possible pairs of balls.

There are $\binom{7}{2}\,=\,21$ possible pairs of white balls.

Therefore: . $P(\text{both white})\:=\:\frac{21}{120} \:=\:\frac{7}{40}$