A box contains one dozen donuts. Four of the donuts are

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Source: Manhattan Prep

A box contains one dozen donuts. Four of the donuts are chocolate, four are glazed, and four are jelly. If two donuts are randomly selected from the box, one after the other, what is the probability that both will be jelly donuts?

A.1/11
B. 1/9
C. 1/3
D. 2/3
E. 8/9

The OA is A
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by Brent@GMATPrepNow » Thu Aug 22, 2019 12:43 pm
BTGmoderatorLU wrote:Source: Manhattan Prep

A box contains one dozen donuts. Four of the donuts are chocolate, four are glazed, and four are jelly. If two donuts are randomly selected from the box, one after the other, what is the probability that both will be jelly donuts?

A.1/11
B. 1/9
C. 1/3
D. 2/3
E. 8/9

The OA is A
P(both are jelly) = P(1st selection is jelly AND 2nd selection is jelly)
= P(1st selection is jelly) x P(2nd selection is jelly)
= 4/12 x 3/11
= 1/11

Answer: A

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by swerve » Thu Aug 22, 2019 2:41 pm
BTGmoderatorLU wrote:Source: Manhattan Prep

A box contains one dozen donuts. Four of the donuts are chocolate, four are glazed, and four are jelly. If two donuts are randomly selected from the box, one after the other, what is the probability that both will be jelly donuts?

A.1/11
B. 1/9
C. 1/3
D. 2/3
E. 8/9

The OA is A
Probability of getting the first Jelly doughnut - \(\frac{4}{12}\)

Probability of getting the second Jelly doughnut - \(\frac{3}{11}\)

Total probability \(= \frac{4}{12} \cdot \frac{3}{11} = \frac{1}{11}\)

Therefore, the correct option is __A__

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by Scott@TargetTestPrep » Sat Aug 24, 2019 4:19 am
BTGmoderatorLU wrote:Source: Manhattan Prep

A box contains one dozen donuts. Four of the donuts are chocolate, four are glazed, and four are jelly. If two donuts are randomly selected from the box, one after the other, what is the probability that both will be jelly donuts?

A.1/11
B. 1/9
C. 1/3
D. 2/3
E. 8/9

The OA is A
The probability of selecting two jelly donuts is:

4/12 x 3/11 = 1/3 x 3/11 = 1/11

Answer: A

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