Problem Solving

Hi All,

We're told that Frank has a box containing 6 doughnuts: 3 jellies and 3 cinnamon twists - and that Frank eats 3 doughnuts (chosen randomly from the box). We're asked for the probability that he eats the 3 jelly doughnuts. This question can be solved as a Combination Formula question or as a Permutation.

Since there are 3 jelly donuts and 6 total donuts, we can keep track of the probability of choosing a donut with each 'draw':
1st donut = 3/6.... assuming he draws a jelly donut...
2nd donut = 2/5.... assuming he draws a jelly donut...
3rd donut = 1/4....

Thus, the probability of pulling the 3 jelly donuts is (3/6)(2/5)(1/4) = 6/120 = 1/20

Final Answer: D

GMAT assassins aren't born, they're made,
Rich

AAPL wrote:Frank has a box containing 6 doughnuts: 3 jellies and 3 cinnamon twists. If Frank eat 3 doughnuts, chosen randomly from the box, what is the probability that he eats 3 jelly doughnuts?

A. 1/2
B. 1/3
C. 1/8
D. 1/20
E. 1/36

The number of ways to choose 3 doughnuts out of 6 is 6C3 = (6 x 5 x 4)/(3 x 2) = 20.

The number of ways of chose 3 jelly doughnuts out of 3 is 3C3 = 1.

Thus, the probability is 1/20.

Alternate Solution:

The probability of choosing the first jelly doughnut is 3/6; the probability that the next doughnut chosen is jelly is 2/5, and the probability that the third doughnut is jelly is 1/4. We multiply these probabilities to get 3/6 x 2/5 x 1/4 = 6/120 = 1/20.

Answer: D