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

Answer:

0.1 or 1/10

Step-by-step 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 $= \frac{a!}{(a-n)!}$

⁵P₃ $= \frac{5!}{(5-3)!}$

⁵P₃ $= \frac{5!}{2!}$

⁵P₃ $= \frac{5*4*3*2*1}{2*1}$

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:

$P=\frac{ favorable outcomes}{equally likely possible outcomes}$

$P=\frac{6}{60}$

$P=\frac{1}{10}$