Question

The probability

Answer 1

Probability of distinct 3-digits is 1/6.

Probability of both the first digit and the last digit are even numbers is 1/5.

Step-by-step explanation:

Given set of numbers {1,2,3,5,8,9} = 6

Total number of 3-digits can be formed = 6$\times$5$\times$4 = 120.

Probability  of the three-digit numbers with distinct digits :

The number of 3-digit numbers with distinct digits = (6$\times$5$\times$4) / (3$\times$2$\times$1) = 20.

P(distinct 3-digits) =  no. of distinct 3-digits / Total no.of 3-digits

⇒ 20/120

⇒ 1/6

Probability that both the first digit and the last digit of the three-digit number are even numbers :

The even numbers in the set are 2 and 8 = 2 possibilities.

The number of 3-digits with 1st and 3rd digit are even = (2$\times$6$\times$2) = 24.

P(1st and 3rd digits even)= no. of 1st and 3rd digit even/Total no. of 3-digits.

⇒ 24/120

⇒ 1/5