Probability please help!

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

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

thanks

none of these answers are correct lol

You are all wrong!!!! Its 3/10

hahaha is RIGHT!!!!!!! ur all worng!!
:) :(

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.

A number is divisible by 4 if and only if the number formed by its last two digits is divisible by 4. Using the given digits, we find that the only two-digit numbers that are divisible by 4 are 24, 28, 32, 48, 72, and 84.

There are $5 \cdot 4 = 20$ ways to choose the last two digits, and they are all equally likely, so the probability that the number is divisible by 4 is $6/20 = \boxed{3/10}$.

*Facepalm* it is 3/10

Answer - 3/10
Solution: The way to find out if a number is divisible by 4 is to look at the last 2 digits. If they are divisible by 4, then the whole number is divisible by 4. This give us 6 possible endings: 24, 28, 32, 48, 72, and 84. However, there aren't just 6 of these numbers.
87324 and 37824 both end in 24, but they are different numbers. Since there are three digits that you can choose for the beginning of the number, you have 3*2*1 = 6 different numbers per number ending. Therefore, we get 6*6 = 36 total solutions. Then, we have to find the total amount of numbers that could be made. We get 5*4*3*2*1 = 120 because we can't repeat numbers. If we put our correct solutions over the total amount of numbers, we get
36/120 = 3/10

The answer is 3/10. There are 5! (120) different ways to arrange the five digits. If a number is divisible by four the last 2 digits are also divisible by four. There are 6 ways to have the last two digits divisible by 4: 32, 72, 24, 84, 28, and 48. We have to keep in mind that for every pair of last two digits, there are 6 ways to arrange the 3 digits in front. Since there are 6 pairs of 2 digit numbers that can be divisible by four, there are 36 ways to make a 5 digit number divisible by 4 with the numbers provided. But, they are asking for the probability of getting a number that is divisible by 4. That would be 36/120 = 3/10. Thanks for looking at my answer!

The answer is that you guys are posting HW probs and answers here. Not good, guys.

It’s 1/10 bots

Yes, this is cheating