Answer
a.{12,15,18,21,24 ... 99} *Note that 99 = 3*33 & 12 = 3*4 hence the number of 2 digits positive integers that are multiple of 3 is (33-4 + 1) = 30 integer
b.two digits integer are the digits from 10 till 99 hence they are ( 99-10+1) = 90 positive integer, P(E) = $\frac{ favorable -nums}{ total -outcomes}$ = $\frac{30}{90}$ = 33.33%
c. { 12,16, ... , 96} *Note that 96 = 4*24 & 12 = 4*3 hence the number of 2 digits positive integers that are multiple of 3 is (24-3 + 1) = 22 integer.
P(E) = $\frac{ favorable -nums}{ total -outcomes}$ = $\frac{22}{90}$ = 24.44%
Work Step by Step
a.{12,15,18,21,24 ... 99} *Note that 99 = 3*33 & 12 = 3*4 hence the number of 2 digits positive integers that are multiple of 3 is (33-4 + 1) = 30 integer
b.two digits integer are the digits from 10 till 99 hence they are ( 99-10+1) = 90 positive integer, P(E) = $\frac{ favorable -nums}{ total -outcomes}$ = $\frac{30}{90}$ = 33.33%
c. { 12,16, ... , 96} *Note that 96 = 4*24 & 12 = 4*3 hence the number of 2 digits positive integers that are multiple of 3 is (24-3 + 1) = 22 integer.
P(E) = $\frac{ favorable -nums}{ total -outcomes}$ = $\frac{22}{90}$ = 24.44%
