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by amyhussein » Tue Mar 26, 2013 8:26 am
If we have 8 numbers from 1 to 8, we r choosing 4 numbers . What is the probability that the chosen numbers are 1,2,4,5?
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by Anju@Gurome » Tue Mar 26, 2013 8:36 am
amyhussein wrote:If we have 8 numbers from 1 to 8, we r choosing 4 numbers . What is the probability that the chosen numbers are 1,2,4,5?
Number of ways to select 4 numbers from 8 numbers = 8C4 = 8!/[(4!)*(4!)] = (8*7*6*5)/(4*3*2) = 70

Only one of these selections will be the combination {1, 2, 4, 5}
Hence, required probability = 1/70
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by amyhussein » Tue Mar 26, 2013 8:44 am
Should we take into consideration the different arrangement for 1 2, 4,5?
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by Anju@Gurome » Tue Mar 26, 2013 9:06 am
amyhussein wrote:Should we take into consideration the different arrangement for 1 2, 4,5?
8C4 is the number of selections of 4 numbers from 8.
It doesn't include different arrangements of a particular selection.
So, if you use this formula then there is no need to consider different arrangements.
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by GMATGuruNY » Tue Mar 26, 2013 9:41 am
amyhussein wrote:If we have 8 numbers from 1 to 8, we r choosing 4 numbers . What is the probability that the chosen numbers are 1,2,4,5?
Good numbers are 1, 2, 4, and 5.

P(1st number is good) = 4/8. (Of the 8 numbers, any of the 4 good numbers may be selected.)
P(2nd number is good) = 3/7. (Of the 7 remaining numbers, any of the 3 remaining good numbers may be selected.)
P(3rd number is good) = 2/6. (Of the 6 remaining numbers, either of the 2 remaining good numbers may be selected.)
P(4th number is good) = 1/5. (Of the 5 remaining numbers, the 1 remaining good number must be selected.)
Since we want all of these events to happen together, we multiply the probabilities:
4/8 * 3/7 * 2/6 * 1/5 = 1/70.
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