Question

The digits 4 comma 5 comma 6 comma 7 comma and 8 are randomly arranged to form a​ three-digit number. ​ (Digits are not​ repeated.) Find the probability that the number is even and greater than 800.

Answer 1

0.1 or 1/10

Explanation:

The digits 4, 5, 6, 7 and 8 are randomly arranged to form a three digit number, where the digits are not repeated.

This is question of permutation.

Imagine this sum as; there are 3 boxes(blank spaces for digits) and 5 different fruits(digits) are to be put in these boxes, where a box can hold a maximum of only 1 fruit. The number of such permutations are: ⁵P₃

By formula (a! is factorial a):

ᵃPₙ
`= (a!)/((a-n)!)`

⁵P₃
`= (5!)/((5-3)!)`

⁵P₃
`= (5!)/(2!)`

⁵P₃
`= (5*4*3*2*1)/(2*1)`

⁵P₃= 60

This is the total count of possible numbers that can be formed.

Now, for a number to be greater than 800 and even; first digit should necessarily be 8. Last digit can be 4 or 6. Using these conditions, there are 6 possibilities. 854, 864, 874, 846, 856, 876 are the numbers.

The probability that number is even and greater than 800 is:

`P=( favorable outcomes)/(equally likely possible outcomes)`

`P=(6)/(60)`

`P=(1)/(10)`