Answer
a. $\frac{1}{25}$
b. $\frac{12}{25}$
c. $\frac{2}{5}$
Work Step by Step
a. The number is even
There is only one even number (2) among a total of 25 prime numbers.
So the probability of getting the even number is $p=\frac{1}{25}$
b. The sum of the number’s digits is even
There are 12 (11 13 17 19 31 37 53 59 71 73 79 97) numbers whose sum of the number’s digits is even.
So the probability of getting such a number is $p=\frac{12}{25}$
c. The number is greater than 50
There are 10 numbers in the list that is greater than 50.
So the probability of getting such a number is $p=\frac{10}{25}=\frac{2}{5}$
