# Independent and Dependent events

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