Answer :

**given:** two dices are thrown

**formula:**

total possible outcomes are ^{6}C_{1}^{6}C_{1}

therefore n(S)=6^{2}=36

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

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

n(E)= ^{3}C_{1}=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)= ^{4}C_{1}=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)= ^{5}C_{1}=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

Rate this question :

A bag contains 6 RD Sharma - Mathematics

In a lottery, a pRD Sharma - Mathematics

In a hand at WhisRD Sharma - Mathematics

A bag contains 5 RD Sharma - Mathematics

Two unbiased diceRD Sharma - Mathematics

From a deck of 52RD Sharma - Mathematics

A bag contains 6 RD Sharma - Mathematics

In shutting a pacRD Sharma - Mathematics

Two unbiased diceRD Sharma - Mathematics

In a single throwRD Sharma - Mathematics