# What is the probability of getting face card from deck of 52 cards?

 37. P(if the series lasts 7 games, Atlanta will win). We seek P(Atlanta wins | series last 7 games). This is a conditional probability statement and we apply the conditional formula here. P(Atlanta wins | series last 7 games) = P(Atlanta wins AND series last 7 games)/P(series lasts 7 games) There are 40 different win-lose sequences that lead to a 7-game series. In 20 of these sequences, Atlanta wins 4 games to 3 and in the other 20, the Yankees win 4 games to 3. Each of the 20 ways that Atlanta wins the series in 7 games has the same probability: (0.4)^4(0.6)^3 = 0.0055296. So P(Atlanta wins AND series last 7 games) = 20*(0.4)^4(0.6)^3 = 20*0.0055296 = 0.110592. Each of the 20 ways that the Yankees win the series in 7 games has the same probability: (0.6)^4(0.4)^3 = 0.0082944. So P(Yankees win AND series last 7 games) = 20*(0.6)^4(0.4)^3 = 20*0.0082944 = 0.165888. Because the event "Yankees win in 7" and the event "Braves win in 7" are mutually exclusive, we add the two results to determine P(series last 7 games). This is 0.165888 + 0.110592 = 0.27648 = P(series last 7 games). This says, by the way, that under these conditions and assumptions, about 25% of the time the series will last 7 games. We can now compute the desired conditional probability: P(Atlanta wins | series last 7 games) = P(Atlanta wins AND series last 7 games)/P(series lasts 7 games) = (0.110592)/(0.165888 + 0.110592) = 0.4. Does this seem like a surprising result? Given that we get to game 7, we can think of the series boiling down to one game, and the probability that atlanta wins that one game is just P(B) = 0.4. Another way to think about (0.110592)/(0.165888 + 0.110592) is top think of it as a weighted outcome: P(result A and condition #1) compared to the sum of the probabilities of all possible results under condition #1. Here, there were just two results possible under the condition that we get to game 7: Either the Braves win or the Yankees win. Other situations may extend that. Suppose under a certain condition, 10 different things - - all mutually exclusive - - could happen. Then the probability that the first of those different things does happen under the given condition is just the probability that that first thing happens and the condition happens compared to the sum of the 10 probabilities for the 10 different things that could happen with the condition in place. If we have two events, A and B, the probability of A or B is the probability of A added to the probability of B less the probability of both A and B, or: P(A or B) = P(A) + P(B) - P(A and B). This equation is often called the law of addition. For example, what is the probability of drawing a card from a standard deck of cards and getting a face card or a club? The probability of drawing a club is 13/52. The probability of drawing a face card is 12/52. However, three of the clubs are also face cards, so there are only 22 of the 52 cards that are either clubs or face cards. If we simply add 13/52 and 12/52, we count three cards twice. We should only count them once, and hence we subtract out the 3/52. Our answer is then 13/52 + 12/52 - 3/52. What is the probability of getting both A and B? A formula for that is called the law of multiplication: P(A and B) = P(A)*P(B|A) = P(B)*P(A|B) The P(A|B) is called a conditional probability and is read as "the probability of A given B." What is the probability of getting a face card and a club? The formula says that we need two things, the probability of getting a face card (12/52) and the probability of getting a club given that we have a face card (1/4). If we put this into the formula, we get: P(face card and club) = (12/52)*(1/4)=3/52. If we do it as P(club and face card) = (13/52)*(3/13)=3/52, we get the same answer. If P(A and B) = 0 then A and B are mutually exclusive events, meaning that if one happens, the other cannot. If P(A|B)=P(A) then A and B are independent events, that is, the probability of getting an A does not depend on B. There are many statistical results that depend on independence.