There are three events A, B, C one of which must and only one can happen; The odds are 8 to 3 against A, 5 to 2 against B; find the odds against C.

Explanation

According to the question

`(P(A'))/(P(A))​=8/3`

`​P(A)=3/(11)`​ and
`P(A′)=8/(11)​`

`(P(B'))/(P(B))​=5/2​=P(B)=2/7​ `and `P(B′)=5/7`​

Now out of A,B and C, one and only one can happen.

`P(A)+P(B)+P(C)=1`

`P(C)=(34)/(77)​`

So odd against `C=P(C′)P(C)​=(43)/(34)`

...