Transcript document

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 randomlyselected 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| - |E1E2|)/|S|
= |E1|/|S| + |E2|/|S| - |E1E2|/|S| = p(E1)+p(E2)-p(E1E2)
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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):
|E1E2|=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
- ends -
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