Question

The digits 3, 4, 5, 6 and 7 are randomly arranged to form a three-digit number. Digits are not repeated. What is the probability that the number is even and greater than 700?

Answer 1

`12/120=1/10=10%`

Explanation:

There are `5! = 120` ways to randomly arrange the 5 digits.

To satisfy the conditions, we want the first digit to be 7. We want the last digit to either be 4 or 6. With the 7 forced to be the first digit, we have 4 digits to account for.

To get an even digit, we have a population of 2 and we are choosing 1:

`"^nC_k=(n!)/((k!)(n-k)!)` with `n="population", k="picks"`

`"^2C_1=(2!)/((1!)(2-1)!)=2! =2`

We can now account for the remaining 3 digits, which is `3! = 6`.

Therefore, the number of ways to get an even number greater than 700 is:

`2xx6=12`

and the probability of randomly arranging numbers into this arrangement is:

`12/120=1/10=10%`