Probability

[swerve]

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. \(\dfrac{1}{2}\)

B. \(\dfrac{1}{3}\)

C. \(\dfrac{1}{8}\)

D. \(\dfrac{1}{20}\)

E. \(\dfrac{1}{36}\)

The OA is D

Source: Princeton

[Jay@ManhattanReview]

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. \(\dfrac{1}{2}\)

B. \(\dfrac{1}{3}\)

C. \(\dfrac{1}{8}\)

D. \(\dfrac{1}{20}\)

E. \(\dfrac{1}{36}\)

The OA is D

Source: Princeton

No. of ways any 3 doughnuts can be chosen out of the 6 = 6C3 = (6.5.4)/(1.2.3) = 20

No. of ways of choosing 3 jelly doughnuts out of 3 jelly doughnuts = 1

The probability that he eats 3 jelly doughnuts = 1/20

The correct answer: D

Hope this helps!

-Jay
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[Scott@TargetTestPrep]

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. \(\dfrac{1}{2}\)

B. \(\dfrac{1}{3}\)

C. \(\dfrac{1}{8}\)

D. \(\dfrac{1}{20}\)

E. \(\dfrac{1}{36}\)

The OA is D

Source: Princeton

Solution:

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 to choose 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