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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[email protected]
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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
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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GMATinsight
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SOLUTION TO QUESTION 2[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
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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Here: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
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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[email protected]
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Thank you! I'm not sure why I didn't use combinatorics to begin with.GMATinsight wrote:SOLUTION TO QUESTION 2[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
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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Thank you! Much appreciatedGMATGuruNY wrote:Here: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
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.
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
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
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
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
