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Frank has a box containing 6 doughnuts: 3 jellies and 3 cinnamon twist

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Frank has a box containing 6 doughnuts: 3 jellies and 3 cinnamon twist  [#permalink]

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New post Updated on: 22 Jul 2018, 04:25
2
00:00
A
B
C
D
E

Difficulty:

  15% (low)

Question Stats:

77% (01:08) correct 23% (01:07) wrong based on 97 sessions

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Frank has a box containing 6 doughnuts: 3 jellies and 3 cinnamon twists. If Frank eats 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

Difficulty Level: 550

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Originally posted by SajjadAhmad on 02 May 2017, 10:38.
Last edited by SajjadAhmad on 22 Jul 2018, 04:25, edited 1 time in total.
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Re: Frank has a box containing 6 doughnuts: 3 jellies and 3 cinnamon twist  [#permalink]

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New post 02 May 2017, 10:49
1
The total possibilities of finding a doughnut is 6c3 or 20
Similarly if we need to find if all the doughnuts picked by frank are jelly doughnuts, the possibility is 3c3 or 1

Probability of such an event occuring is 1/20(Option D)
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Re: Frank has a box containing 6 doughnuts: 3 jellies and 3 cinnamon twist  [#permalink]

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New post 03 May 2017, 11:35
1
Hi SajjadAhmad,

This question can be solved with straight-forward 'probability math', without the need of the Combination Formula.

We're told that there are 6 doughnuts - 3 jellies and 3 cinnamon twists. Choosing 3 donuts at random - and having them all be jellies - can be broken down into small steps:

1st is a jelly = 3/6
2nd is a jelly = 2/5
3rd is a jelly = 1/4

Probability of getting 3 jellies in 3 tries = (3/6)(2/5)(1/4) = 6/120 = 1/20

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Re: Frank has a box containing 6 doughnuts: 3 jellies and 3 cinnamon twist  [#permalink]

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New post 27 Jun 2018, 17:13
1
SajjadAhmad wrote:
Frank has a box containing 6 doughnuts: 3 jellies and 3 cinnamon twists. If Frank eats 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
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Re: Frank has a box containing 6 doughnuts: 3 jellies and 3 cinnamon twist &nbs [#permalink] 27 Jun 2018, 17:13
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