another confusing probability problem

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another confusing probability problem

by knight247 » Thu Jun 23, 2011 5:59 am
Of all the five-digit integers formed by permutation of the digits 1, 2, 3, 4 and 5 without repition. what is the probability that a five-digit integer drawn randomly will be divisible by 4
(A) 1/10
(B )4/15
(C) 13/30
(D) 1/5
(E) 3/20

Hope i can get a detailed explanation on this one. OA is D
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by Frankenstein » Thu Jun 23, 2011 6:08 am
Hi,
For the number to be divisible by 4, last two digits should be divisible by 4.
So, possible combinations are 12,32,52,24 --> 4 pairs
Last 2 digits can be arranged from 5 digits in 5P2 = 20 ways
So, probability is 4/20 = 1/5

Hence, D
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by Anurag@Gurome » Thu Jun 23, 2011 11:35 pm
knight247 wrote:Of all the five-digit integers formed by permutation of the digits 1, 2, 3, 4 and 5 without repition. what is the probability that a five-digit integer drawn randomly will be divisible by 4
(A) 1/10
(B )4/15
(C) 13/30
(D) 1/5
(E) 3/20

Hope i can get a detailed explanation on this one. OA is D
Total no. of ways of arranging 5-digit integers without repetition = 5! = 5 * 4 * 3 * 2 * 1 = 120
An integer is divisible by 4 if the last 2 digits are divisible by 4.
In this case, the possibilities of a 5-digit integer divisible by 4 can be the following:

_ _ _ 12, where 3 blank spaces can be filled by any of the numbers 3, 4, and 5. So, this can be done in 3! = 6 ways
_ _ _ 32, where 3 blank spaces can be filled by any of the numbers 1, 4, and 5. So, this can be done in 3! = 6 ways
_ _ _ 42, where 3 blank spaces can be filled by any of the numbers 1, 3, and 5. So, this can be done in 3! = 6 ways
_ _ _ 52, where 3 blank spaces can be filled by any of the numbers 1, 3, and 4. So, this can be done in 3! = 6 ways

So, the possible 5-digit integers that are divisible by 4 and none of the digits repeat = 6 * 4 = 24

Hence, the required probability = 24/120 = [spoiler]1/5[/spoiler]

The correct answer is D.
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by smackmartine » Fri Jun 24, 2011 12:06 am
IMO D

5 digit # without repetition can be arranged in 5!=120

For a # to be divisible by 4 ,last 2 digits should be divisible by 4. This leave 4 #s 12,24,32,52

First three digits can be arranged in 3! ways and last two digits in 4 ways. Total ways
3.2.1. 4 = 24
_ _ _ __

The probability that a five-digit integer drawn randomly will be divisible by 4 = 24/120 =1/5
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