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by [email protected] » Tue Aug 12, 2014 3:38 pm
Hi guys,

Really stuck on these two questions. I paused my mock paper to spend 15 mins working each of them out and still didn't get anywhere! Would appreciate any guidance...

Thanks
Shakira


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by GMATinsight » Tue Aug 12, 2014 6:19 pm
[email protected] wrote:Hi guys,

Really stuck on these two questions. I paused my mock paper to spend 15 mins working each of them out and still didn't get anywhere! Would appreciate any guidance...

Thanks
Shakira


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SOLUTION TO QUESTION 2

Favorable Out comes = 3 Pies and 2 cakes = 5C3 x 6C2 = 10 x 15 = 150

Total Outcomes = any 5 out of 11 pieces = 11C5 = 462

Probability = 150/462 = 75/231 = 25/77

Answer: Option B
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by GMATGuruNY » Tue Aug 12, 2014 8:15 pm
Each of 4 bags contains 25 blue disks, 25 green disks, 25 orange disks, 25 yellow disks, and nothing else. If one disk is chosen at random from each of the four bags, what is the probability that the number of blue disks chosen will be no less than 1 and no greater than 3?
a. 1/16
b. 41/128
c. 87/128
d. 225/256
e.255/256

OA is C
Here:
A GOOD outcome is selecting 1, 2, or 3 blue disks.
A BAD outcome is selecting no blue disks or 4 blue disks.

P(good outcome) = 1 - P(bad outcome).

In each bag, 1/4 of the disks are blue, while 3/4 are not blue.
P(4 non-blue disks) = 3/4 * 3/4 * 3/4 * 3/4 = 81/256.
P(4 blue disks) = 1/4 * 1/4 * 1/4 * 1/4 = 1/256.
Thus:
P(good outcome) = 1 - (81/256 + 1/256) = 174/256 = 87/128.

The correct answer is C.
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by [email protected] » Wed Aug 13, 2014 3:03 pm
GMATinsight wrote:
[email protected] wrote:Hi guys,

Really stuck on these two questions. I paused my mock paper to spend 15 mins working each of them out and still didn't get anywhere! Would appreciate any guidance...

Thanks
Shakira


Image
SOLUTION TO QUESTION 2

Favorable Out comes = 3 Pies and 2 cakes = 5C3 x 6C2 = 10 x 15 = 150

Total Outcomes = any 5 out of 11 pieces = 11C5 = 462

Probability = 150/462 = 75/231 = 25/77

Answer: Option B
Thank you! I'm not sure why I didn't use combinatorics to begin with.
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by [email protected] » Wed Aug 13, 2014 3:07 pm
GMATGuruNY wrote:
Each of 4 bags contains 25 blue disks, 25 green disks, 25 orange disks, 25 yellow disks, and nothing else. If one disk is chosen at random from each of the four bags, what is the probability that the number of blue disks chosen will be no less than 1 and no greater than 3?
a. 1/16
b. 41/128
c. 87/128
d. 225/256
e.255/256

OA is C
Here:
A GOOD outcome is selecting 1, 2, or 3 blue disks.
A BAD outcome is selecting no blue disks or 4 blue disks.

P(good outcome) = 1 - P(bad outcome).

In each bag, 1/4 of the disks are blue, while 3/4 are not blue.
P(4 non-blue disks) = 3/4 * 3/4 * 3/4 * 3/4 = 81/256.
P(4 blue disks) = 1/4 * 1/4 * 1/4 * 1/4 = 1/256.
Thus:
P(good outcome) = 1 - (81/256 + 1/256) = 174/256 = 87/128.

The correct answer is C.
Thank you! Much appreciated
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by shrivats » Wed Aug 13, 2014 7:32 pm
For the first question:

no less than 1 blue disk means, it cannot be zero
no more than 3 means, it canot be all four.

P( zero blue) = (3/4)* (3/4)*(3/4)*(3/4) because 75/100 is 3/4
P(all 4 bue) = (25/100)^4

required probability is = 1 - [ (3/4)^4 * (1/4)^4

= 1 - [ (81/256 + 1/256) ]
= 1 - [ 82 /256 ]
= 174/256 = 87/128
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by shrivats » Wed Aug 13, 2014 7:37 pm
The second is straight forward.
there are 5 pies and 6 cakes

total of 5 items selected out of 11, 3 of which are pies, implies are 2 cakes.

so the probability is = 5c3 * 6c2
----------- = 150/462 = 25/77
11c5
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