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Question

There are three events X, Y and Z, out of which only one must happen. If the odds are 8:3 against X, 5:2 against Y, the odds against Z must be:


A

43:34

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B

8:11

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C

7:5

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D

None of these

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Solution

The correct option is A

43:34


According to the question,P(X)P(X) = 83, P(X') = 811, P(X) = 311

Similarly P(Y') = 57, P(Y) = 27

Now, out of X,Y and Z, one and only one can happen.

P(X) + P(Y) + P(Z) = 1

So, P(Z) = 3477

P(Z) = 1-P(Z) = 4377

So, odds against Z = 43:34.


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