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Permutation and Combination

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Permutation and Combination

by pareekbharat86 » Tue Dec 03, 2013 6:41 am
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

This is an SOS. I fare very poorly with probability and permutation & combinations. I just can't seem to get them right. Can someone please suggest some good material (book or online) so that I can stop fearing these questions. I really want to conquer this before the D-day.
Thanks,
Bharat.
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by GMATGuruNY » Tue Dec 03, 2013 7:06 am
pareekbharat86 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.
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by Mathsbuddy » Tue Dec 03, 2013 9:23 am
A = P(no blue) = (3/4)^4
B = P(4 blue) = (1/4)^4

P(A or B) = A + B

Also P(1, 2 or 3 blue) = P(Not A or B) = 1 - (A + B)

= 1 - (3/4)^4 - (1/4)^4

= (4^4 - 3^4 - 1^4)/(4^4)

= (256 - 81 - 1) /256

= 174/256 = 87/128

ANSWER C.
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