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[tex]\times[/tex]5[tex]\times[/tex]4 = 120.
Probability of the three-digit numbers with distinct digits :
The number of 3-digit numbers with distinct digits = (6[tex]\times[/tex]5[tex]\times[/tex]4) / (3[tex]\times[/tex]2[tex]\times[/tex]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[tex]\times[/tex]6[tex]\times[/tex]2) = 24.
P(1st and 3rd digits even)= no. of 1st and 3rd digit even/Total no. of 3-digits.
⇒ 24/120
⇒ 1/5
