Two dice are thro

given: two dices are thrown

formula:

total possible outcomes are ⁶C₁⁶C₁

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

Therefore, probability of event E’ is

P(E’) =1-P(E)

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

Therefore, probability of event E’ is

P(E’) =1-P(E)

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

Therefore, probability of event E’ is

P(E’) =1-P(E)

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