Answer:
<h2>Hope it helps you......</h2>
Really easy i just don't feel like doing it, 10 extra points, (find x)
1 answer:
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Probability of distinct 3-digits is 1/6.
Probability of both the first digit and the last digit are even numbers is 1/5.
<u>Step-by-step explanation</u>:
Given set of numbers {1,2,3,5,8,9} = 6
Total number of 3-digits can be formed = 65
4 = 120.
<u>Probability of the three-digit numbers with distinct digits</u> :
The number of 3-digit numbers with distinct digits = (65
4) / (3
2
1) = 20.
P(distinct 3-digits) = no. of distinct 3-digits / Total no.of 3-digits
⇒ 20/120
⇒ 1/6
<u>Probability that both the first digit and the last digit of the three-digit number are even numbers</u> :
The even numbers in the set are 2 and 8 = 2 possibilities.
The number of 3-digits with 1st and 3rd digit are even = (26
2) = 24.
P(1st and 3rd digits even)= no. of 1st and 3rd digit even/Total no. of 3-digits.
⇒ 24/120
⇒ 1/5