Transcript probability review lecture

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Addition Rule
Multiplication Rule
Compliments
Conditional Probability
Permutation
Combinations
Expected value
Geometric Probabilities
Binomial Probabilities
Addition Rule for
Non Mutually Exclusive Events
(A or B) = P(A) + P(B) – P(A and B)
One card is drawn from a
standard deck of cards.
What is the probability that
it is red or an ace?
= P(Red) + P(Ace) – P(Both Red and Ace)
= 26/52 + 4/52 – 2/52
= 28/52
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Multiplication Rule
Finding the probability of more than one event.
The word “AND” is always used when describing the
situation.
1) P(rolling a 4 and then a 2) = 1/6 *1/6 = 2.8%
2) P(rolling 3 odd #’s) = 3/6*3/6*3/6 = 12.5%
What’s the probability of pulling out two aces in a row
Example
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continued
from a deck of 52 cards?
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P(A1 AND A2) = P(A1)P(A2|A1)
P(A1) = 4/52
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There are now 3 aces left in a 51-card pack
P(A2|A1) = 3/51
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Overall: P(A1 AND A2) = (4/52) (3/51)
= .0045
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Compliment:
P(A') = 1 - P(A)
If A is an event within the sample space S of an
activity or experiment, the complement of
A (denoted A') consists of all outcomes in S that
are not in A.
The complement of A is everything else in the
problem that is NOT in A.
Conditional Probability
and
P( A | B) 
P( A B)
P( B)
measures the probability of an event given that
another event has occurred
1% of the population has disease X.
If someone has the disease and gets tested the test
is positive every time.
If a healthy person gets tested for disease X they will
get a false positive 10% of the time.
If the lab comes back positive what will be the probability
the person actually has the disease?
P(D Ç +)
P(D +) =
P(+)
Permutations
an arrangement of items in a particular order.
n = total number of items
r = number chosen
n!
p 
n r (n  r )!
Permutations Examples
A combination lock will open when the
right choice of three numbers (from 1
to 30) is selected. How many different
lock combinations are possible assuming
no number is repeated?
30!
30!
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 30 * 29 * 28  24360
30 p3 
( 30  3)! 27!
Combinations
an arrangement of items in which order does not
matter. There are always fewer combinations
than permutations.
n = total number of items
r = number chosen
n!
C 
where 0  r  n .
n r r! ( n  r )!
Combinations Example
To play a particular card game, each player is
dealt five cards from a standard deck of 52
cards. How many different hands are
possible?
52!
52!
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52 C5 
5! (52  5)! 5!47!
52 * 51 * 50 * 49 * 48
 2,598,960
5* 4* 3* 2*1
Expected value
a weighted average of all possible values where the
weights are the probabilities of each outcome
E(X) =
x
p(x
)
å i i
all x
E ( x)  x1  P( x1 )  x2  P( x2 ) 
 xn  P( xn ).
Example: expected value
probability distribution of ER arrivals
x is the number of arrivals in one hour
X
10
11
12
13
14
P(x)
.4
.2
.2
.1
.1
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 x p( x)  10(.4)  11(.2)  12(.2)  13(.1)  14(.1)  11.3
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i 1
Geometric Distribution
want to find the number of trials for the 1st success
p = probability of success
q = 1 – p = probability of failure
X = # of trials until first success occurs
x-1
p(x) = q p
Two Ways to use the Geometric Model
#1: the probability of getting your first success on
the x trail
x-1
p(x) = q p
#2: the number of trials until the first success is
certain
1
p(x) =
p
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EXAMPLE:
On Friday’s 25% of the customers at an ATM make deposits.
What is the probability that it takes 4 customers at the ATM
before the first one makes a deposit.
✔ Two Categories:
Success: make a deposit
Failure: don’t make a deposit
✔ Probability success same for each trial
✔ Wish to find the probability of the first
x-1
The desired probability is: p(x) = q
4 1
p(4)  (.75) (.25)  0.0117
p
BINOMIAL PROBABILITY
finding the probability of a specific
number of successes
p(x) = n C x × p × q
x
n-x
n = number of trials
x = number of successes
n – x = number of failures
p = probability of success in one trial
q = 1 – p = probability of failure in one trial
EXAMPLE 2
You are taking a 10 question multiple choice test. If each
question has four choices and you guess on each question,
what is the probability of getting exactly 7 questions correct?
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p = 0.25 = guessing the correct answer
q = 0.75 = guessing the wrong answer
n = 10
x=7
p(x = 7) = 10C7 ´ 0.25 ´0.75
p(x = 6) =0.0031
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