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Section 5.1

Discrete Probability

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LaPlace’s definition of probability

• Number of successful outcomes divided by the number of possible outcomes

• This definition works when all outcomes are equally likely, and are finite in number

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Finite Probability

• Experiment: a procedure that yields one of a given set of possible outcomes

• Sample space: set of possible outcomes

• Event: a subset of the sample space

• LaPlace’s definition, stated formally, is: The probability p of an event E, which is a subset of a finite sample space S of equally likely outcomes is p(E) = |E|/|S|

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Example 1

• What is the probability that you will draw an ace at random from a shuffled deck of cards?

• There are 4 aces, so |E| = 4

• There are 52 cards, so |S| = 52

• So p(E) = |4|/|52|, or 1/13

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Example 2

• What is the probability that at randomly-selected integer chosen from the first hundred positive integers is odd?

• S = 100, E = 50

• So p(E) = 1/2

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Example 3

• What is the probability of winning the grand prize in the lottery, if to win you must pick 6 correct numbers, each of which is between 1 and 40?

• There is one winning combination

• The total number of ways to choose 6 numbers out of 40 is C(40,6) = 40!/(34!6!)

• So your chances of winning are 1/3,838,380

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Example 4

• What is the probability that a 5-card poker hand does not contain the ace of hearts?

• If all hands are equally likely, the probability of a hand NOT containing a particular card is the quotient of:– probability of picking 5 cards from the 51

remaining: C(51,5) and– probability of picking any 5 cards from entire deck:

C(52,5)

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Example 4

• C(51,5) = 51!/5!46!

• C(52,5) = 52!/5!47!

• So C(51,5)/C(52,5) = (51!/5!46!)(5!47!/52!)

• Through cancellation, we get: 47/52 or ~.9

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Example 5

• There are C(52,5) = 52!/(47!5!) = 2,598,960 possible hands of 5 cards in a deck of 52

• What is the probability of getting 4 of a kind in a hand of 5?

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Example 5

• Using the product rule, the number of ways to get 4 of a kind in a hand of 5 is the product of:– the number of ways to pick one kind: C(13,1)

– the number of ways to pick 4 of this kind from the total number in the deck of this kind: C(4,4)

– the number of ways to pick the 5th card: C(48,1)

• So the probability of being dealt 4 of a kind is:(C(13,1)C(4,4)C(48,1))/C(52,5) = 13*1*48/2,598,960 or .00024

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Example 6

• What is the probability of a 5-card poker hand containing a full house (3 of one kind, 2 of another)?

• By the product rule, the number of hands containing a full house is the product of:– ways to pick 2 kinds in order: P(13,2): order matters

because 3 aces, 2 tens 3 tens, 2 aces

– ways to pick 3 out of 4 of first kind: C(4,3)

– ways to pick 2 out of 4 of second kind: C(4,2)

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Example 6

• P(13,2) * C(4,3) * C(4,2):– P(n,r) = n(n-1) * … * (n-r+1); since 13-2 = 11,

P(n,r) = 13 * 12– C(4,3) = 4!/3!1! & C(4,2) = 4!/2!2! = 24/6 & 24/4– So result is 156 * 4 * 6 = 3744

• Because there are 2,598,960 possible poker hands, the probability of a full house is 3744/2598960 = ~.0014

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Example 7

• What is the probability that a 5-card poker hand contains a straight (5 consecutive cards, any suit)?

• Assuming ace is always high, the 5 cards could start with any of: {2,3,4,5,6,7,8,9,10}

• So there are C(9,1) or 9 ways to start• There are 4 suits, so there are 4 cards of each kind,

so there are C(4,1) or 4 ways to make each of the 5 choices

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Example 7

• Putting this information together with the product rule, there are 9 * 45 = 9,216 different possible hands containing a straight

• Since there are 2,598,960 hands possible, the probability of a hand containing a straight is: 9,216/2,598,960 = ~.0035

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Complement of an event

• Let E be an event in sample space S. The probability of E, the complementary event of E is: p(E) = 1 - p(E)

• Note that |E| = |S| - |E|

• So p(E) = (|S| - |E|)/|S| = 1-|E|/|S| = 1- p(E)

• Sometimes it’s easier to find the probability of a complement than the probability of the event itself

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Example 8

• A sequence of 10 bits is randomly generated. What is the probability that at least one bit is 0?– E: at least 1 of 10 bits is 0– E: all bits are 1s– S: all strings of 10 bits

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Example 8

• Since p(E) = 1-p(E) and there are 210 possible bit combinations, but only 1 containing all 1s:

• p(E) = 1/210 = 1-1/1024 = 1023/1024, or 99.9% probability that at least one bit is 0 in a random 10-bit string

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Probability of Union of 2 Events

• The probability of the union of 2 events is the sum of the probabilities of each of the events, less the probability of the intersection of the 2 events:

• p(E1 E2) = p(E1) + p(E2) - p(E1 E2)

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Probability of Union of 2 Events: proof of theorem

• Recall the formula for the number of elements in the union of 2 sets:

• |E1 E2| = |E1| + |E2| - |E1 E2|

• So, p |E1 E2| = |E1 E2|/|S| =(|E1| + |E2| - |E1E2|)/|S| = |E1|/|S| + |E2|/|S| - |E1E2|/|S| = p(E1)+p(E2)-p(E1E2)

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Example 9

• What is the probability that a positive integer selected at random from the set of positive integers not exceeding 100 is divisible by either 2 or 5?

• E1 = selected integer divisible by 2; |E1|=50

• E2 = selected integer divisible by 5; |E2|=20

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Example 9

• So E1 E2 is the event that the number is divisible by either 2 or 5 and

• E1 E2 is the event that the number is divisible by both (divisible by 10): |E1E2|=10

• So p(E1 E2) = p(E1) + p(E2) - p(E1 E2) = 50/100 + 20/100 - 10/100 = 60/100 = 3/5

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Probabilistic reasoning

• … is determining which of 2 events is more likely• The Monty Hall 3-door puzzle is an example of such

reasoning:– select one of 3 doors, one of which has the GRAND

PRIZE!!!!! behind it

– once selection is made, Monty opens one of the other doors (knowing it is a loser)

– then he gives you the option to switch doors -should you?

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Solution to Monty Hall 3-door puzzle

• Initial probability of selecting the grand prize door is 1/3

• Monty always opens a door the prize is NOT behind

• The probability you selected incorrectly is 2/3 (since you only had a 1/3 chance of a correct selection)

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Solution to Monty Hall 3-door puzzle

• If you selected incorrectly, when Monty selects another door (without prize), the prize must be behind the remaining door

• You will always win if you chose wrong the first time, then switch

• So by changing doors, you probability of winning is 2/3

• Always change doors!

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Section 4.4

Discrete Probability

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