Question

there are three events A B and C one of which must only one can happen the odds are 8 to 3 against A, 5 to 2 against B , then odds against C is

Answer 1

Answer: 43:34.

Odds against A are 8:3
=> P(A) = 3÷ 8+3 = 3÷11

Odds in favor of B are 2:5.
=> P(B) = 2÷2+5 = 2÷ 7

P(A ∪ B ∪ C)= 1(∵ A,B, and C are mutually exhaustive.) 
=> P(A)+P(B) + P(C) =1 (∵ A,B and C are mutually exhaustive. )
=> 3÷ 11 + 2÷7 + P(C)=1
=> P(C) = 34÷77 

Therefore,
Number of favarable outcomes for the event C = 34
Number of unfavarable outcomes for the event C = 77-32
                                                                               = 43
Therefore, Odds against C= 43:34.

Hope it helps u!

Answer 2

Answer:

43/34

Step-by-step explanation:

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