Question

Two uniform dice marked 1 to 6 are thrown together The probability that the sum is odd is

• A. $\frac{1}{4}$
• B. $\frac{1}{3}$
• C. $\frac{1}{2}$
• D. $\frac{1}{12}$

Answer 1

Answer: (C)

$\frac{1}{2}$

For the sum to be odd is, the possible combinations are $=(1,2),(1,4),(1,6),(3,2),(3,4),(3,6),(5,2),(5,4),(5,6)$ and vice-versa.
Number of favourable cases $=18$
Total number of cases $=36$
Hence, required probability $=\frac{1}{2}$