1. ## probability help

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

2. Hello, sri340!

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?
With a little Thought, the solution is quite straight forward.

We have a 5-digit number and its tens digit is odd.
There are only 2 choices for the odd digit.

The tens digit is 3: . _ _ _ 3 _
To be divisible by 4, the units digit must be 2.
And the other 3 digits can be placed in $\displaystyle 3! \:=\: 6$ ways.

The tens digit is 7:_ _ _ 7 _
To be divisible by 4, the units digit must be 2.
And the other 3 digits can be placed in $\displaystyle 3! \:=\:6$ ways.

Hence, there are: .$\displaystyle 6 + 6 \:=\:12$ such numbers.

Since there are $\displaystyle 5! = 120$ possible 5-digit numbers,
. . The probabiity is: .$\displaystyle \frac{12}{120} \:=\:\frac{1}{10}$