Question

Find the probability that using the digits, 5,2,9 and 6, a 4-digit number can be formed by by using each number once, that the number is divisible by 4?

Answer 1

Total no. of numbers that can be formed using 5,2,9, 6=\[4\times 3\times 2\times 1=24\]
For a number to be divisible by 4 , the last two digits should be divisible by 4
Let's see which all combinations satisfy this
56, 52, 92, 96.
When we have last two digits fixed
the first two digits can only be from the other two digits so \(2\times 1=2\) numbers
56, 52, 92, 96.
Each of these will make two possibilities of numbers
so total 8
Probability=\(\frac{8}{24}=\frac{1}{3}\)

Answer 2

There are 24 possible no. of four digit nos that can be formed.
Now, only those no.os which have their last two digits divisible by 4 are divisible by 4.
So, numbers ending with 52,56,92,96 are divisble by 4. There are a total of 8 nos ending with those numbers(2 for each).
So probability=8/24