#### Answer

A)12 possible digits
B)$\frac{5}{12}$
C)$\frac{1}{3}$

#### Work Step by Step

A)There are 4 ways to pick the first digit.
There are 3 ways to pick the second digit
Therefore,there are 4 $\times$3=12 possible digits.
B)This is an overlapping event since there are numbers where 2 and 5 are both presents. From the possible 2-digit numbers,
6 contain a 2=$\frac{6}{12}$ --> P(2)
6 contains a 5:$\frac{6}{12}$--> P(5)
6 contain 2 and 5: $\frac{2}{12}$ -->P( 2 and 5)
P(2 or 5)=P(2) + P(5) + P(2 and 5)
P(2 or 5)=$\frac{6}{12}$ +$\frac{6}{12}$ +$\frac{2}{12}$=$\frac{10}{12}$
P(2 or 5)=$\frac{5}{6}$
C)The prime numbers are 13,23,31,53.Therefore,the probability is P(prime)= $\frac{4}{12}$= $\frac{1}{3}$