## Algebra 1: Common Core (15th Edition)

A)12 possible digits B)$\frac{5}{12}$ C)$\frac{1}{3}$
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)= $\frac{4}{12}$= $\frac{1}{3}$