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

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by AAPL » Mon Mar 12, 2018 5:32 am
A box contains one dozen of 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.

We know that the probability for the first selection will be, 4/12 = 1/3. Then for the second selection will be, 3/11.

Finally, the combined probability will be, (1/3)*(3/11) = 3/33 = 1/11. Option A.

Is there another strategic approach to solve this PS question? Can any experts help, please? Thanks!
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by GMATinsight » Mon Mar 12, 2018 6:03 am
AAPL wrote:A box contains one dozen of 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.

We know that the probability for the first selection will be, 4/12 = 1/3. Then for the second selection will be, 3/11.

Finally, the combined probability will be, (1/3)*(3/11) = 3/33 = 1/11. Option A.

Is there another strategic approach to solve this PS question? Can any experts help, please? Thanks!
Method 1:

4C2 / 12C2 = 6 / 66 = 1/11

Method 2:

Probability of first being Jelly Donut = 4/12
Probability of Second being Jelly Donut = 3/11

therefore Favourable Probability = (4/12)*(3/11) = 1/11

Answer: option A

I hope this helps!!!
"GMATinsight"Bhoopendra Singh & Sushma Jha
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Whatsapp/Mobile: +91-9999687183 l [email protected]
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by GMATinsight » Mon Mar 12, 2018 6:04 am
GMATinsight wrote:
AAPL wrote:A box contains one dozen of 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.

We know that the probability for the first selection will be, 4/12 = 1/3. Then for the second selection will be, 3/11.

Finally, the combined probability will be, (1/3)*(3/11) = 3/33 = 1/11. Option A.

Is there another strategic approach to solve this PS question? Can any experts help, please? Thanks!
Method 1:

4C2 / 12C2 = 6 / 66 = 1/11
4C2 - Number of ways of picking 2 Jelly donuts out of 4 Jelly Donuts
12C2 - Number of ways of picking 2 donuts out of 12 Donuts

Method 2:

Probability of first being Jelly Donut = 4/12
Probability of Second being Jelly Donut = 3/11

therefore Favourable Probability = (4/12)*(3/11) = 1/11

Answer: option A

I hope this helps!!!
"GMATinsight"Bhoopendra Singh & Sushma Jha
Most Comprehensive and Affordable Video Course 2000+ CONCEPT Videos and Video Solutions
Whatsapp/Mobile: +91-9999687183 l [email protected]
Contact for One-on-One FREE ONLINE DEMO Class Call/e-mail
Most Efficient and affordable One-On-One Private tutoring fee - US$40-50 per hour
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by Brent@GMATPrepNow » Mon Mar 12, 2018 6:13 am
AAPL wrote:A box contains one dozen of 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
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

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
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by Jeff@TargetTestPrep » Tue Mar 13, 2018 4:10 pm
AAPL wrote:A box contains one dozen of 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 probability of two jelly donuts being selected is:

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

Answer: A

Jeffrey Miller
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