Posted by **SuckatProb** on Tuesday, November 20, 2012 at 8:02am.

The digits 2, 3, 4, 7, and 8 are each used once in a random order to form a five-digit number. What is the probability that the resulting number is divisible by 4? Express your answer as a common fraction.

- Probability please help! -
**joanna**, Tuesday, November 20, 2012 at 8:14am
It means that numbers 2,3,7 have to be first so combination choose 3 out of 5 and then either 4 or 8 can be in the end which can be done in 2! ways. So 20ways out of possible 5!=120 so 20/120= 2/12=1/6

- Probability please help! -
**Reiny**, Tuesday, November 20, 2012 at 8:27am
if any number is divisible by 4 , then its last 2 digits must be divisible by 4

e.g 43728 is divisible by 4 because 28 is divisible by 4

whereas 72438 is not , even though its last digit is divisible by 4

obviously it must be also be even,

so the last two digits could be

xxx28

xxx48

xxx24

xxx84

the number of cases for each of these

= 3x2x1

= 6

e.g. for xxx28

34728 , 37428 , 43728 , 47328 , 73428 , 74328

so the number of cases which are divisible by 4 are 3(6) = 24

but the total number of arrangements with no restrictions are 5! = 120

prob of divisible by 4 = 24/120 = 1/5

- Probability please help! -
**SuckatProb**, Tuesday, November 20, 2012 at 10:22am
thanks

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**arg**, Saturday, January 19, 2013 at 9:59am
none of these answers are correct lol

- Probability please help! -
**hahaha**, Tuesday, November 5, 2013 at 3:23pm
You are all wrong!!!! Its 3/10

- Probability please help! -
**bella**, Saturday, November 16, 2013 at 4:42pm
hahaha is RIGHT!!!!!!! ur all worng!!

:) :(

- Probability please help! -
**person**, Saturday, November 29, 2014 at 11:41pm
For last 2 digits, you forgot xxx32 and xxx72. There are 4 * 5 = 20 ways to choose the last two digits, and they are all equally likely, so the probability that the number is divisible by 4 is (# of ways to choose last 2 digits such that the number is divisible by 4)/ways to choose last 2 digits = 6/20 = 3/10.

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