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
Hello, Gretchen!
If you familiar with "combinations", here's another approach . . .
There are $\displaystyle 16$ balls.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: . $\displaystyle \frac{2}{3}\quad\frac{1}{9}\quad\frac{7}{40}\quad\ frac{49}{240}$
There are $\displaystyle \binom{16}{2}\,=\,120$ possible pairs of balls.
There are $\displaystyle \binom{7}{2}\,=\,21$ possible pairs of white balls.
Therefore: .$\displaystyle P(\text{both white})\:=\:\frac{21}{120} \:=\:\frac{7}{40}$