What is the probability of getting a face card if a card is drawn from a well shuffled 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.

Text Solution

Solution : When a card is drawn from a well - shuffled deck of 52 cards, the number of possible outcomes is 52. (i) Let `E_(1)` be the event of getting a face card. Since there are 16 face cards, we have `n(E_(1)) = 16`. `therefore P(E_(1)) = (n(E_(1)))/(n(S)) = 16/52 = 4/13.` `rArr` odds in favour of getting a face card `= (P(E_(1)))/({1 - P(E_(1))}) = ((4//13))/((1 - 4/13)) = 4/9.` (ii) Let `E_(2)` be the event of getting a spade. Then, `n(E_(2)) = 13`. `therefore P(E_(2)) = (n(E_(2)))/(n(S)) = 13/52 = 1/4` `rArr` odds against getting a spade `= ({1 - P(E_(2))})/(P(E_(2)))` `= ((1 - 1/4))/((1//4)) = 3/1.`