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Question

There are only 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

A
12:5
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B
5:12
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C
34:43
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D
43:34
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Solution

The correct option is C 43:34
Given odds against A are 8:3,
So, P(A)=38+3=311
Given odds against B are 5:2,
P(B)=25+2=27
Since the events A,B,C are mutually exclusive and totally exhaustive,
So, P(A)+P(B)+P(C)=1
311+27+P(C)=1
P(C)=3477
Odds in favour of C are 34:77
Hence, the odds against C are as (7734):34 i.e, 43:34

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