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PROBABILITY- INDEPENDENT EVENTS

PROBABILITY- INDEPENDENT EVENTS

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HOW DRAWING A DIAMOND AND DRAWING AN ACE ARE INDEPENDENTS EVENTS AND WHY NOT OTHERS ?

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365 Team
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Hey Mohd,
 
The idea here is that we’re drawing a single card and we’re trying to determine how knowing 1 thing about the card reduces/increases/doesn’t change the likelihood of the other event as well. 
 
Hence, we’re looking for two events, A and B, such that P(A) = P(A|B) (and of course P(B) = P(B|A)). 
If A = drawing a diamond, then P(A) = 13/52 = 1/4 = 0.25, since there are 13 diamonds in the deck. Similarly, P(A|B) = 1/4 = 0.25, since out of the 4 Aces, precisely 1 is a diamond. To be absolutely certain, we can check P(B) and P(B|A). 
P(B) = 4/52 = 1/13, since exactly 4 of the 52 cards in the deck are Aces. 
P(B|A) = 1/13, since just 1 of the 13 diamonds is an Ace. 
 
Thus, P(B) = P(B|A) and P(A) = P(A|B). If we go through the other options, we’ll see that such a relationship doesn’t hold true for any of them. 
 
Hope this helps!
Best, 
365 Vik
 

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