Q. 22

# Two dice are thro

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

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