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Probability Disks

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by yourshail123 » Sat Oct 20, 2012 1:24 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
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by TheGmatTutor » Sat Oct 20, 2012 2:51 pm
I'd rephrase the question as "What is the probability you will pick 1, 2, or 3 blue disks?" In order to answer this question, I would start with 1 and subtract off the probability of zero blue disks and the probability of 4 blue disks.

1 - P(0 blue disks) - P(4 blue disks)

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

1 - (81/256) - (1/256)

1 - (82/256)

174/256

87/128

Answer C
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by yourshail123 » Sun Oct 21, 2012 9:39 am
I am expecting some one to try with the conventional way - Calculating out the particular probability for picking 1, 2, or 3 blue disks. Anyone?
Agree with TheGmatTutor - your way is best way possible.
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by beingNeha » Sat Jul 06, 2013 10:23 pm
Probability of getting 0 blue disks p(0) = 3/4 * 3/4 * 3/4 * 3/4 = 81/256
Probability of getting 4 blue disks p(4) = 1/4 * 1/4 * 1/4 * 1/4 = 1/256

Probability of getting 1,2, or 3 blue disks p(1, 2, 3) = 1 - 81/256 - 1/256 = 174/256 = 87/128 (C). B-)
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by Uva@90 » Mon Jul 08, 2013 8:19 pm
Hi All,
Could any one please explain me how probability of zero blue disk is (3/4)^4 and probability of 4 blue disk is (1/4)^4.

Thanks in advance.
Regards,
Uva.
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by Jeff@TargetTestPrep » Wed Nov 08, 2017 4:40 pm
yourshail123 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

We should first note that the condition "the number of blue disks chosen is no less than 1 and no greater than 3" is equivalent to the condition "the number of blue disks is neither 0 nor 4."

We can use the following equation:

P(the number of blue disks chosen will be no less than 1 and no greater than 3) = 1 - P(selecting 0 blue disks) - P(selecting 4 blue disks)

Let's first determine P(selecting 0 blue disks):

75/100 x 75/100 x 75/100 x 75/100 = (3/4)^4 = 81/256

Next let's determine P(selecting 4 blue disks):

25/100 x 25/100 x 25/100 x 25/100 = (1/4)^4 = 1/256

Thus:

P(the number of blue disks chosen will be no less than 1 and no greater than 3) is:

1 - 81/256 - 1/256 = 1 - 82/256 = 174/256 = 87/128

Answer: C

Jeffrey Miller
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by Matt@VeritasPrep » Thu Nov 09, 2017 7:06 pm
Uva@90 wrote:Hi All,
Could any one please explain me how probability of zero blue disk is (3/4)^4 and probability of 4 blue disk is (1/4)^4.

Thanks in advance.
Regards,
Uva.
p(zero blue disks) = p(first disk isn't blue) * p(second disk isn't blue) * p(third disk isn't blue) * p(fourth disk isn't blue)

In any given bag, the ratio of blues to nonblues is 25 : 75, so the probability of getting a nonblue disk is 75/(75 + 25), or 3/4. With that in mind, p(zero blue) = 3/4 * 3/4 * 3/4 * 3/4.
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