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

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A box contains one dozen donuts. Four of the donuts are  [#permalink]

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New post 01 Mar 2014, 12:04
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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.\(\frac{1}{11}\)

B. \(\frac{1}{9}\)

C. \(\frac{1}{3}\)

D. \(\frac{2}{3}\)

E. \(\frac{8}{9}\)

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Re: A box contains one dozen donuts. Four of the donuts are  [#permalink]

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New post 01 Mar 2014, 12:05
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To determine the probability that jelly donuts will be chosen on the first and second selections, we must find the probability of both events and multiply them together.

The probability of picking a jelly donut on the first pick is 4/12. However, the probability of picking a jelly donut on the second pick is NOT 4/12. If a jelly donut is selected on the first pick, the number of donuts in the box has decreased from 12 to 11, and the number of jelly donuts has decreased from 4 to 3. Thus, the probability of picking a jelly donut on the second pick is 3/11.

Since overall probability is calculated by multiplying the probabilities of both events, the probability of picking two jelly donuts is 4/12 × 3/11 = 12/132 = 1/11.

The correct answer is A.
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Re: A box contains one dozen donuts. Four of the donuts are  [#permalink]

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New post 01 Dec 2016, 11:43
12 Doughnuts

4/12 - Chance of first jelly
3/11 - Chance of second jelly

4/12 * 3/11

The numerator 3 divides into the denominator 12. The equation gets simplified to the below.

4/4 * 1/11 = 1/11
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Re: A box contains one dozen donuts. Four of the donuts are  [#permalink]

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New post 12 Mar 2018, 07:16
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Anshulmodi wrote:
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.\(\frac{1}{11}\)

B. \(\frac{1}{9}\)

C. \(\frac{1}{3}\)

D. \(\frac{2}{3}\)

E. \(\frac{8}{9}\)


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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Re: A box contains one dozen donuts. Four of the donuts are  [#permalink]

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New post 13 Mar 2018, 17:04
Anshulmodi wrote:
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.\(\frac{1}{11}\)

B. \(\frac{1}{9}\)

C. \(\frac{1}{3}\)

D. \(\frac{2}{3}\)

E. \(\frac{8}{9}\)


The probability of two jelly donuts being selected is:

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

Answer: A
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Re: A box contains one dozen donuts. Four of the donuts are   [#permalink] 13 Mar 2018, 17:04
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