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.
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.
none of these answers are correct lol
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