Two dice are thrown. Find

Solution:

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 $=\frac{5}{36}$

We know that, If odds in favor of the occurrence an event are a:b, then the probability of an event to occur is $\frac{a}{a+b}$

Now we got $\frac{\mathrm{a}}{\mathrm{a}+\mathrm{b}}=\frac{5}{36}$

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 $6=\frac{6}{36}$

$=\frac{1}{6}$

We know that,

If odds in favor of the occurrence an event are a:b, then the probability of an event to occur is $\frac{a}{a+b}$

Now we got $\frac{\mathrm{a}}{\mathrm{a}+\mathrm{b}}=\frac{1}{6}$

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 $5: 1$