given: two dices are thrown

Formula: P(E) = \(\frac{favorable\ outcomes}{total\ possible\ outcomes}\) 

total possible outcomes are ⁶C₁ 6C₁ 

therefore n(S)=6² = 36 

(i) let E be the event that total sum is 4 on dice 

E= {(1,3) (3,1) (2,2)} 

n(E)= ³C₁ = 3

P(E) = \(\frac{n(E)}{n(S)}\)

P(E) = \(\frac{3}{36}=\frac{1}{12}\) 

Therefore, probability of event E’ is 

P(E’) =1-P(E)

P(E’) = \(1-\frac{1}{12}=\frac{11}{12}\)

Odds in favour of getting sum as 4 is P(E):P(E’) =1:11 

(ii) let E be the event that total sum is 5 on dice 

E= {(1,4) (2,3) (3,2) (4,1)} 

n(E)= ⁴C₁ = 4

P(E) = \(\frac{n(E)}{n(S)}\)

P(E) = \(\frac{4}{36}=\frac{1}{9}\) 

Therefore, probability of event E’ is 

P(E’) = 1 - P(E)

P(E’) = \(1-\frac{1}{9}=\frac{8}{9}\)

Odds in favour of getting sum as 5 is P(E):P(E’) =1:8 

(iii) let E be the event that total sum is 6 on dice 

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

n(E)= ⁵C₁ = 5

P(E) = \(\frac{n(E)}{n(S)}\)

P(E) = \(\frac{5}{36}\) 

Therefore, probability of event E’ is 

P(E’) =1-P(E)

P(E') = \(1-\frac{5}{36}=\frac{31}{36}\) 

Odds against of getting sum as 6 is P(E’):P(E) =31:5