Answer :

Total outcomes are (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6),

(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6),

(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6),

(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6) ,

(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6) ,

(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)

Total cases where sum will be 6 is (1, 5), (2, 4), (3, 3), (4, 2), (5, 1) i.e. 5

Probability of getting sum 6

We know that, If odds in favor of the occurrence an event are a:b, then the probability of an event to occur is

Now we got

So, a = 5 and a+b = 36 i.e. b = 31

Therefore odds in the favor of getting the sum as 6 is 5:31

Conclusion: Odds in favor of getting the sum as 6 is 5:31

(ii) Total outcomes are (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6),

(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6),

(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6),

(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6) ,

(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6) ,

(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)

Total cases where sum will be 7 is (1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1) i.e. 6

Probability of getting sum 6

We know that,

If odds in favor of the occurrence an event are a:b, then the probability of an event to occur is

Now we got

So, a = 1 and a+b = 6 i.e. b = 5

Therefore odds in the favor of getting the sum as 7 is 1:5

Odds against getting the sum as 7 is b:a i.e. 5:1

Conclusion: Odds against getting the sum as 7 is 5:1

Rate this question :

A natural number RD Sharma - Mathematics

In a single throwRD Sharma - Mathematics

One of the two evRD Sharma - Mathematics

A card is drawn aRD Sharma - Mathematics

One of the two evRD Sharma - Mathematics

The probability tRD Sharma - Mathematics

If the probabilitRD Sharma - Mathematics

Without repetitioRD Sharma - Mathematics

6 boys and 6 girlRD Sharma - Mathematics