# Balls in a bag

• Mar 6th 2011, 07:03 AM
smallfry
[SOLVED] Balls in a bag
Q: There are 5 black balls and 3 white balls in a bag. Three balls are chosen at random from the bag without replacement. Given that of the three balls chosen, at least one black ball and at least one white ball was chosen, what is the probability that two black balls were chosen?

My answer: 2/3. Can anybody try to disprove this? Other possible answers: 11/56, 45/56, 4/9, 1/2 (multiple choice question).
• Mar 6th 2011, 04:44 PM
Soroban
[soze=3]Hello, smallfry![/size]

Quote:

There are 5 black balls and 3 white balls in a bag.
Three balls are chosen at random from the bag without replacement.

Given that of the three balls chosen, at least one black ball
and at least one white ball was chosen,
what is the probability that two black balls were chosen?

My answer: 2/3 . I agree!

Can anybody try to disprove this?
Other possible answers: 11/56, 45/56, 4/9, 1/2 (multiple choice question).

Here is my reasoning (probably different from yours) . . .

Three balls are chosen from eight: . ${8\choose3} \,=\, 56$ possible outcomes.

There are ${5\choose3} = 10$ ways to get 3 black balls.
There is ${3\choose3} = 1$ way to get 3 white balls.
. . So there are: $10 + 1 \,=\,11$ ways to get 3 balls of one color.

Hence, there are: $56-11 \,=\,\boxed{45}$ ways to get 3 balls of mixed colors.

Of these 45 outcomes, how many have 2 blacks and 1 white?

There are: ${5\choose2}{3\choose1} \,=\,10\cdot3 \,=\,\boxed{30}$ such outcomes.

$\displaystyle\text{Therefore: }\;\text{Probability} \:=\:\frac{30}{45} \:=\:\frac{2}{3}$