Help me with this probability question, please?

Question

Help me with this question, please?
And explain/show how to get the result? :/
Thank you a bunch. <3

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

Thanks.

Answer 1

P(A: tenth digit is odd) = 2/5
P(B: the number is divisible by 4) = P(the number ends with 2) = 1/5
P(A and B) = (2/5)(1/5) = 2/25

Answer 2

OK, let's see here. Since any multiple of 100 is divisible by 4, the 100's, 1,000's, and 10,000's digits don't come into play, because they could be anything and it can't affect whether or not the number meets the 2 given conditions. So what are the combinations of 10's and 1's digits that result in a number meeting the requirements? Just xxx32 and xxx72.

So there's 5! ways to make the 5-digit number, and 2 * 3! ways to hit the magic combination. So the probability of getting the magic combination is:

(2 * 3!) / 5!
(2 * 3 * 2 * 1) / ( 5 * 4 * 3 * 2 * 1)
2 / ( 5 * 4)
1 / 10

Or you could think of it like this: The upper 3 digits don't even matter. There are 20 ways to make the lower 2 digits, 2 of which result in an odd tens digit and a number divisible by 4. So the probability of this is 2 out of 20, or 1 / 10.

Answer 3

A number divisible by 4 is first of all even, so the last digit must be 2, 4, or 8. To be divisible by 4, the last TWO digits must be divisible by 4, so given an odd 10's digit, our possibilities are 32, 72, 34, 74, 38, and 78. Of these, only 32 and 72 work. So we have 1 choice for last digit, 2 choices for 10's digit, and 3! choices for remaining digits. That's 3!(2) = 12 choices, compared with 5! arrangements of the digits without restriction, so the probability is 12/5! = 12/120 = 1/10.

Answer 4

hey , i am there again to help u out man.
now he tottal possible 5 digit nos are - 5*4*3*2*1
= 120.
consider the following number form -
T.T T H T U (UNITS , TENS HUNDREDS, .......]
NOW T place can be filled with 2 odd digits viz- 3&7.
now u must know that last 2 digits must be divisible by 4 in order to make whole 5 digit no divisible by 4.
therefore units place can only be 2.that s only 1way.
now the remaining 1st 3 digits can be filled with remaining 3 numbers in 3*2*1 = 6 ways.
last 2 digits can be 3 , 2 or 7 , 2.
hence total numbers are = 6*2 = 12.
probability = 12/120 = 10%