Problem Solving

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!

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!!!