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

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

Follow this topic

MOHD AZEEM ANSARI

Last answered:

08 Oct 2020

Posted on:

03 Sept 2020

PROBABILITY- INDEPENDENT EVENTS

1 answers ( 0 marked as helpful)

Viktor Mehandzhiyski

Posted on:

08 Oct 2020

PROBABILITY- INDEPENDENT EVENTS

Submit an answer