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We end the section by solving a problem called the Birthday Problem .

If there are 25 people in a room, what is the probability that at least two people have the same birthday?

Let event E size 12{E} {} represent that at least two people have the same birthday.

We first find the probability that no two people have the same birthday.

We analyze as follows.

Suppose there are 365 days to every year. According to the multiplication axiom, there are 365 25 size 12{"365" rSup { size 8{"25"} } } {} possible birthdays for 25 people. Therefore, the sample space has 365 25 size 12{"365" rSup { size 8{"25"} } } {} elements. We are interested in the probability that no two people have the same birthday. There are 365 possible choices for the first person and since the second person must have a different birthday, there are 364 choices for the second, 363 for the third, and so on. Therefore,

P No two have the same birthday = 365 364 363 341 365 25 = 365 P 25 365 25 size 12{P left ("No two have the same birthday" right )= { {"365" cdot "364" cdot "363" dotsaxis "341"} over {"365" rSup { size 8{"25"} } } } = { {"365"P"25"} over {"365" rSup { size 8{"25"} } } } } {}

Since P at least two people have the same birthday = 1 P No two have the same birthday , size 12{P left ("at least two people have the same birthday" right )=1 - P left ("No two have the same birthday" right ),} {}

P at least two people have the same birthday = 1 365 P 25 365 25 = . 5687 size 12{P left ("at least two people have the same birthday" right )=1 - { {"365"P"25"} over {"365" rSup { size 8{"25"} } } } = "." "5687"} {}
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Conditional probability

Suppose you and a friend wish to play a game that involves choosing a single card from a well-shuffled deck. Your friend deals you one card, face down, from the deck and offers you the following deal: If the card is a king, he will pay you $5, otherwise, you pay him $1. Should you play the game?

You reason in the following manner. Since there are four kings in the deck, the probability of obtaining a king is 4 / 52 size 12{4/"52"} {} or 1 / 13 size 12{1/"13"} {} . And, probability of not obtaining a king is 12 / 13 size 12{"12"/"13"} {} . This implies that the ratio of your winning to losing is 1 to 12, while the payoff ratio is only $1 to $5. Therefore, you determine that you should not play.

Now consider the following scenario. While your friend was dealing the card, you happened to get a glance of it and noticed that the card was a face card. Should you, now, play the game?

Since there are 12 face cards in the deck, the total elements in the sample space are no longer 52, but just 12. This means the chance of obtaining a king is 4 / 12 size 12{4/"12"} {} or 1 / 3 size 12{1/3} {} . So your chance of winning is 1 / 3 size 12{1/3} {} and of losing 2 / 3 size 12{2/3} {} . This makes your winning to losing ratio 1 to 2 which fares much better with the payoff ratio of $1 to $5. This time, you determine that you should play.

In the second part of the above example, we were finding the probability of obtaining a king knowing that a face card had shown. This is an example of conditional probability . Whenever we are finding the probability of an event E under the condition that another event F has happened, we are finding conditional probability.

The symbol P E F size 12{P left (E \lline F right )} {} denotes the problem of finding the probability of E size 12{E} {} given that F size 12{F} {} has occurred. We read P E F size 12{P left (E \lline F right )} {} as "the probability of E size 12{E} {} , given F size 12{F} {} ."

A family has three children. Find the conditional probability of having two boys and a girl given that the first born is a boy.

Let event E size 12{E} {} be that the family has two boys and a girl, and F size 12{F} {} the event that the first born is a boy.

First, we list the sample space for a family of three children as follows.

S = BBB , BBG , BGB , BGG , GBB , GBG , GGB , GGG size 12{S= left lbrace ital "BBB", ital "BBG", ital "BGB", ital "BGG", ital "GBB", ital "GBG", ital "GGB", ital "GGG" right rbrace } {}

Since we know that the first born is a boy, our possibilities narrow down to four outcomes, BBB size 12{ ital "BBB"} {} , BBG size 12{ ital "BBG"} {} , BGB size 12{ ital "BGB"} {} , and BGG size 12{ ital "BGG"} {} .

Among the four, BBG size 12{ ital "BBG"} {} and BGB size 12{ ital "BGB"} {} represent two boys and a girl.

Therefore, P ( E F = 2 / 4 size 12{P \( E \lline F rbrace =2/4} {} or 1 / 2 size 12{1/2} {} .

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Source:  OpenStax, Applied finite mathematics. OpenStax CNX. Jul 16, 2011 Download for free at http://cnx.org/content/col10613/1.5
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