(i) We know that,

Probability of occurrence of an event

 = \(\frac{Total\,no.of\,Desired\,outcomes}{Total\,no.of\,outcomes}\) 

Outcomes are (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), 

(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), 

(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), 

(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6) , 

(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6) , 

(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)

Total no. of outcomes are 36 

In that only (1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (4, 1) are our desired outputs as there sum is less than 6 

Therefore no. of desired outcomes are 10 

Therefore, the probability of getting a sum less than 6

= \(\frac{10}{36}\) 

= \(\frac{5}{18}\) 

Conclusion: Probability of getting a sum less than 6, when two dice are rolled is  \(\frac{5}{18}\) 

(ii) We know that, 

Probability of occurrence of an event

 = \(\frac{Total\,no.of\,Desired\,outcomes}{Total\,no.of\,outcomes}\) 

In (a, b) if a=b then it is called a doublet 

Total doublets are (1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6) 

In (a, b) if a=b and if a, b both are odd then it is called a doublet 

Odd doublets are (1, 1), (3, 3), (5, 5) 

Outcomes are (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), 

(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), 

(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), 

(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6) , 

(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6) , 

(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6) 

Total no. of outcomes are 36 and desired outcomes are 3 

Therefore, probability of getting doublet of odd numbers

= \(\frac{3}{36}\) 

= \(\frac{1}{12}\) 

Conclusion: Probability of getting doublet of odd numbers, when two dice are rolled is  \(\frac{1}{12}\) 

(iii) We know that,

Probability of occurrence of an event

= \(\frac{Total\,no.of\,Desired\,outcomes}{Total\,no.of\,outcomes}\) 

Outcomes are (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), 

(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), 

(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), 

(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6) , 

(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6) , 

(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6) 

Total no. of outcomes are 36 

Desired outputs are (1, 1), (1, 2), (1, 4), (1, 6), (2, 1), (2, 3), (2, 5), (3, 2), (3, 4), (4, 1), (4, 3), (5, 2), (5, 6), (6, 1), (6, 5) 

Total no. of desired outputs are 15 

Therefore, probability of getting the sum as a prime number

= \(\frac{15}{36}\) 

= \(\frac{5}{12}\) 

Conclusion: Probability of getting the sum as a prime number, when two dice are rolled is  \(\frac{5}{12}\)