Question

If a, b and c are three numbers (not necessarily different) chosen randomly and with replacement from the set $\{1,2,3,4,5\}$, the probability that (ab + c) is even, is

• A. $\frac{50}{125}$
• B. $\frac{59}{125}$
• C. $\frac{64}{125}$
• D. $\frac{75}{125}$

Answer 1

The correct option is B $\frac{59}{125}$

Given, a,b,c as three numbers from the set {1,2,3,4,5}
No.of even numbers in the set $1,2,3,4,5=2$
No.of odd numbers in the set $1,2,3,4,5=3$
Number of ways of selecing 3 numbers from the set with replacement$=5\ast 5\ast 5=125$
Inorder to get even number for $ab+c$ as even,refer below table.No.of ways of selecting a,b,c as odd numbers,$S1=3\ast 3\ast 3=27$
No.of ways of selecting a as odd and b,c as even numbers,$S2=3\ast 2\ast 2=12$
No.of ways of selecting b as odd and a,c as even numbers,$S3=3\ast 2\ast 2=12$
No.of ways of selecting a,b,c as even numbers,$S4=2\ast 2\ast 2=8$
Total number of ways of selecting a,b,c such that ab+c as even=$S1+S2+S3+S4=27+12+12+8=59$
$\therefore Therequiredprobability=\frac{59}{125}$