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 May 1st, 2018, 12:25 AM #1 Senior Member   Joined: Aug 2014 From: India Posts: 355 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: 2,430 Thanks: 1315 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: 355 Thanks: 1 I edited my question. May 1st, 2018, 02:21 AM #4 Senior Member   Joined: Aug 2014 From: India Posts: 355 Thanks: 1 Ca you tell me what values should I substitute in P(A) and P(B)? Tags $pa, cap, cup, place, substituted, theta$ Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post Ganesh Ujwal Trigonometry 4 April 29th, 2018 05:08 AM wirewolf Trigonometry 4 March 1st, 2016 11:23 AM SourPatchKid Complex Analysis 0 April 21st, 2015 04:15 PM mared Geometry 1 June 15th, 2014 09:49 AM mauro125 Algebra 3 February 22nd, 2014 04:57 PM

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