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May 1st, 2018, 12:25 AM   #1
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What value is substituted in place of $P(A \cap B )$?

The odds in favour of events A and B are 1:3 and 1:2 respectively. What is the probability that at least one of the two would occur if A and B are mutually exclusive events?

Solution: Given $P(A) = \dfrac{1}{1+2} = \dfrac {1}{3}$

$P(B) = \dfrac{1}{1+3}$

A and B are mutually exclusive $(A \cup B = \theta)$

$P (A \cup B ) = P(A) + P(B) - P(A \cap B )$

= $\dfrac {1}{3} + \dfrac{1}{4} = \dfrac{7}{12}$

What value is substituted in place of $P(A \cap B)$?

Last edited by Ganesh Ujwal; May 1st, 2018 at 01:05 AM.
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May 1st, 2018, 01:03 AM   #2
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mutually exclusive means $A \cap B = \emptyset$

and thus $P[A \cap B] = 0$
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May 1st, 2018, 01:06 AM   #3
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I edited my question.
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May 1st, 2018, 02:21 AM   #4
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Ca you tell me what values should I substitute in P(A) and P(B)?
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