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May 1st, 2018, 12:25 AM  #1 
Senior Member Joined: Aug 2014 From: India Posts: 332 Thanks: 1  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. 
May 1st, 2018, 01:03 AM  #2 
Senior Member Joined: Sep 2015 From: USA Posts: 2,124 Thanks: 1103 
mutually exclusive means $A \cap B = \emptyset$ and thus $P[A \cap B] = 0$ 
May 1st, 2018, 01:06 AM  #3 
Senior Member Joined: Aug 2014 From: India Posts: 332 Thanks: 1 
I edited my question.

May 1st, 2018, 02:21 AM  #4 
Senior Member Joined: Aug 2014 From: India Posts: 332 Thanks: 1 
Ca you tell me what values should I substitute in P(A) and P(B)?


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$pa, cap, cup, place, substituted, theta$ 
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