I don't understand how to calculate the non-mutually exclusive formula in conditional probability?

Think of a deck of cards.

Let P(A) be the probability of a King.  4/52 or 1/13.

Let P(B) be the probability of a Heart. 13/52 or 1/4.

If we do:

P(A or B) = P(A) + P(B)

we get:

P(A or B) = 4/52 + 13/52

P(A or B) = 17/52

And we can list out the 17 cards that were used to make this set:

K♥, K♦, K♣, K♠, A♥, 2♥, 3♥, 4♥, 5♥, 6♥, 7♥, 8♥, 9♥, 10♥, J♥, Q♥, K♥

If you are paying attention, we have a problem.  We counted the King of Hearts twice.  That's your P(A and B)  The probability of a card being both a King and a Heart, of which there is only 1 in the deck.  Subtracting this out will get rid of that duplicate and we end up with:

P(A or B) = P(A) + P(B) - P(A and B)

P(A or B) = 17/52 - 1/52

P(A or B) = 16/52 or 4/13

There are 16 unique cards that fit "King or Heart" in a deck of cards.

Does this example explain why you have to subtract out the "AND" condition?

The P( A and  B)  is the Probability of BOTH A and B at the same time.

EX.  P ( red card or  King) = P(red) + P(king) - P(red and king)  ... assuming you draw only one card from a normal , shuffled deck of 52 cards

P(red  or  king) =  26/52  + 4/52  - 2/52      there are 26 red cards in 52 card deck  ... 4 kings in a deck ... and 2 red kings in the deck

P(  red or king ) = ( 26 + 4 - 2 )/52 = 28/52  = 14/26 = 7/13

you have a prob. of 7 /13  you will choose a red card or a king from a 52 card shuffled deck ... it's a little better than 1/2 the time !

don't forget to choose a Best answer for the help..