Q. 355.0( 2 Votes )

Two dice are thro

Answer :

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


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