In a single throw of two dice, find the probability of first dice always occur odd number and sum of two dice is greater than 5.

Concept:

In a random experiment, let S be the sample space and let E ⊆ S. Then, E is an event.

The probability of occurrence of E is defined as,

\(\rm P(E)=\frac{n(E)}{n(S)}\)where, n(E) = number of elements in E and n(S) = number of possible outcomes.

Calculation:

We know that in a single throw of two dice, the total number of possible outcomes is 6 × 6 = 36

Let S be the sample space. Then, n(S) = 36

Let E = first dice always occur odd number and sum of two dice is greater than 5. Then, 

E = {(1,5 ), (1, 6), (3, 3), (3, 4), (3, 5), (3, 6), (5,1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)} 

∴ n(E) =  12

So, the probability of obtaining a total of '12’  = P(E) = \(\rm \frac{n(E)}{n(S)}=\frac{12}{36} = {1\over3}\)

Hence, option (4) is correct.