Two dice are thrown. Find the odds in favour of getting the sum
(i) 4 (b) 5
(iii) What are the odds against getting the sum 6?
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