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Defining Probabilities: Random Variables
• Examples:
– Out of 100 heart catheterization procedures performed
at a local hospital each year, the probability that more
than five of them will result in complications is
__________
– Drywall anchors are sold in packs of 50 at the local
hardware store. The probability that no more than 3 will
be defective is
__________
– In general,
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___________
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Discrete Random Variables
• Example:
– Look back at problem 3, page 46. Assume someone
spends $75 to buy 3 envelopes. The sample space
describing the presence of $10 bills (H) vs bills that
are not $10 (N) is:
_____________________________
– The random variable associated with this situation, X,
reflects the outcome of the choice and can take on
the values:
_____________________________
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Discrete Probability Distributions
• The probability that there are no $10 in the
group is
P(X = 0) = ___________________
(recall results from last time)
• The probability distribution associated with the
number of $10 bills is given by:
x
0
1
2
3
P(X = x)
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Another Example
• Example 3.3, pg 66
P(X = 0) =
_____________________
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Discrete Probability Distributions
• The discrete probability distribution function (pdf)
– f(x) = P(X = x) ≥ 0
– Σx f(x) = 1
• The cumulative distribution, F(x)
– F(x) = P(X ≤ x) = Σt ≤ x f(t)
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Probability Distributions
• From our example, the probability that no more
than 2 of the envelopes contain $10 bills is
P(X ≤ 2) = F(2) = _________________
• The probability that no fewer than 2 envelopes
contain $10 bills is
P(X ≥ 2) = 1 - P(X ≤ 1) = 1 - F(1) = ________________
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Another View
• The probability histogram
0.45
0.4
0.35
f(x)
0.3
0.25
0.2
0.15
0.1
0.05
0
0
1
2
3
x
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Your Turn …
• The output of the same type of circuit board from two assembly
lines is mixed into one storage tray. In a tray of 10 circuit boards, 6
are from line A and 4 from line B. If the inspector chooses 2 boards
from the tray, show the probability distribution function associated
with the selected boards being from line A.
 6  4 
  
0 2
1* 6
f (0)  P ( X  0)     
 0.133
45
10 
 
 2
 6  4 
  
1 1
6*4
f (1)  P ( X  1)     
 0.533
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10 
 
 2
 6  4 
  
2 0
15 * 1
f ( 2)  P ( X  1)     
 0.333
45
10 
 
 2
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x
0
P(x)
1
2
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Continuous Probability Distributions
• Examples:
– The probability that the average daily temperature in
Georgia during the month of August falls between 90
and 95 degrees is
__________
– The probability that a given part will fail before 1000
hours of use is
__________
– In general,
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__________
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Understanding Continuous
Distributions
• The probability that the
average daily
temperature in Georgia
during the month of
August falls between 90
and 95 degrees is
-5
-3
-1
1
3
5
• The probability that a
given part will fail before
1000 hours of use is
0
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10
15
20
25
30
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Continuous Probability Distributions
• The continuous probability density function (pdf)
f(x) ≥ 0, for all x ∈ R

 f ( x )dx  1

b
P (a  X  b)   f ( x )dx
a
• The cumulative distribution, F(x)
x
F ( x )  P ( X  x )   f (t )dt

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Probability Distributions
• Example: Problem 7, pg. 73
f(x) =
{
x,
2-x,
0,
0<x<1
1≤x<2
elsewhere
1st – what does the function look like?
a) P(X < 120) = ___________________
b) P(50 < X < 100) = ___________________
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Your turn
• Problem 14, pg. 73
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Additional useful information* …
Joint probability distributions
• Example 1 (discrete): the joint probability mass function
(see defn. 3.8, pg. 75)
In the development of a new receiver for a digital
communication system, each received bit is rated as
acceptable, suspect, or unacceptable, depending on the
quality of the received signal, with the following probabilities:
P(acceptable) = P(x) = 0.9
P(suspect) = P(y) = 0.08
P(unacceptable) = P(z) = 0.02
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* Section 3.4 in your book (Optional)
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If we let X denote the number of acceptable bits and Y denote the
number of suspect bits, then the joint probability associated with
the number of acceptable and suspect bits in 4 transmitted bits is
denoted by:
fXY(x, y), where
fXY(x, y) ≥ 0
∑x ∑y fXY(x, y) = 1
fXY(x, y) = P(X = x, Y = y)
So, the probability of exactly two acceptable bits and exactly 1
suspect bit in the first 4 bits is
fXY(2, 1) = P(X = 2, Y = 1)
Assuming independence, we can determine the probability of a
particular combination, say aasu, as:
P(aasu) = P(a)*P(a)*P(s)*P(u) = 0.9*0.9*0.08*0.02
= 0.0013
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Recognizing that aasu is just one of several possible combinations
of 4 bits, we next need to determine the number of possible
permutations of 2 acceptable and 1 suspect bit in 4 tested bits.
That is (from theorem 2.6, pg. 36),
4!
 12
2!* 1!* 1!
So,
fXY(2, 1) = P(X = 2, Y = 1) = 12(0.0013) = 0.0156
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Joint probability distributions
• Example 2 (continuous): the joint density function (see
definition 3.9, pg. 76)
If X denotes the time until a computer server connects to
your machine (in milliseconds) and Y denotes the time until
the server authorizes you as a valid user (in milliseconds.)
Each measures the wait from a common starting time and X
< Y. Assume that the joint probability density function for X
and Y is given as:
fXY(x, y) = 6 x 10-6exp(-0.001x – 0.002y)
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for x < y
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The probability that X < 1000 and Y < 2000 is:
1000 2000
P ( X  1000 ,Y  2000 )  
0
x
 0.001x
 6  10
dy e
dx
   e
0  x

4 
1000  0.002 x
e

e
6
0.001x


 6  10
e
dx
  0.002 
0 

6
1000  2000
 f XY ( x, y )dxdy
0.002 y
1000
 0.003  e 0.003 x  e  4 e 0.001x dx
0
 1  e 3   4  1  e 1 
e 
  0.915
 0.003 
 0.001 
 0.003 


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[Note: we can verify that this density function integrates to 1 as
follows:
 

 6 0.001x 0.002 y

dy dx
 f XY dydx     6  10 e

0 x

 6  10
6
 
 6  10
6


 0.001x
0.002 y

dy e
dx
 e
0 x

0.002 x
e
  0.002
0
 0.003  e
0
0.003 x
 0.001x
e
dx


1
dx  0.003 (
) 1
0.003
]
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For further study …
• Read section 3.4
• Solve selected problems on pp. 84-86
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