How drawing a diamond and drawing an ace are independent events? How other options are not independent? Question is from Bayesian Inference, Quiz after lesson # 6, Question # 2

Hello, Tanzeela! For starters, this question assumes drawing a single card out of a standard deck containing 52 cards. With that out of the way, let's go back to the definitions. For events A and B to be independent from one another, then P(A|B) = P(A), as well as P(B|A) = P(B). Let event A be the card being an Ace and event B - the card being a diamond. Then, P(A) = 4/52 = 1/13 since there are 4 aces in the deck and P(A|B) = 1/13 since there is a single Ace in the entire suit (diamonds). Similarly, P(B) = 13/52 = 1/4, since 13 of the 52 cards are diamonds; and P(B|A) = 1/4 since only one of the 4 Aces is a diamond. Hence, P(A|B) = P(A) and P(B|A) = P(B), so the two events are independent. Hope this helps! Best, 365 Vik

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Tanzeela Arshad

Last answered:

11 Apr 2022

Posted on:

11 Apr 2022

Independent and Dependent events

in Probability / Course Introduction

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Viktor Mehandzhiyski

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Posted on:

11 Apr 2022

Independent and Dependent events

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