Download Chapter(5)

Discrete Probability
Distributions
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Introduction
5-1 Probability Distributions
5-2 Mean , Variance, Standard Deviation ,and Expectation
5-3 The Binomial Distribution
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5-1:Probability Distributions
A random variable is a variable whose values are
determined by chance.
Classify variables as discrete or continuous.
 A discrete probability distribution consists of the values
a random variable can assume and the corresponding
probabilities of the values.
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For example 1: S ={T , H}
X= number of heads “H”
X= 0 , 1
:‫خطوات الحل‬
n(S)= (2)^2=4 ‫ ايجاد الحجم الكلي‬.1
x ‫ ايجاد‬.2
‫= عدد ظهور الحدثة \ الحجم الكلي‬P(X) ‫ ايجاد‬.3
P(x=0) = P(T) =
P(x=1) = P(H)=
Probability Distribution Table
X
P(X)
0
1
Total=
1
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For example 2: S={TT , HT , TH , HH}
X= number of heads “H”
X= 0 , 1 , 2
P(x=0) = P(TT) =
.
=
P(x=1) = P(HT) + P(TH) =
P(x=2) = P(HH)=
.
.
.
+
=
=
Probability Distribution Table
X
P(X)
0
1
2
Total=
1
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For example 3: Tossing three coins
T TTT
T
T
H
H TTH
T THT
H
THH
T HTT
T
H
H
H HTH
T
HHT
H
HHH
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S={TTT , TTH , THT , HTT , HHT , HTH , THH , HHH}
X= number of heads
X= 0 , 1, 2 , 3
P(x=0) = P(TTT) =
..
=
P(x=1) = P(TTH) + P(THT) + P(HTT)
=
..
+
..
+
..
=
P(x=2) = P(HHT) + P(HTH) + P(THH)
=
..
+
P(x=3) = P(HHH)=
..
.. =
+
..
=
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Probability Distribution Table
Number of heads (X)
Probability P(X)
0
1
2
3
Total= 1
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Example 5-2: Tossing Coins
Represent graphically the probability distribution for the
sample space for tossing three coins .
X
0
1
2
3
P(X)
Solution :
P(X)
0
1
2
3
X
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Example 5-3:
The baseball World Series is played by the winner of the National League
and the American League. The first team to win four games, wins the
world sreies.In other words ,the series will consist of four to seven games,
depending on the individual victories. The data shown consist of the
number of games played in the world series from 1965 through 2005.The
number of games (X) .Find the probability P(X) for each X ,construct a
probability distribution, and draw a graph for the data.
x
4
5
Number of games played
8
7
6
7
9
16
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P(for 4 games) =
P(for 5 games) =
= 0.200
= 0.175
P(for 6 games) =
P(for 7 games) =
= 0.225
= 0.400
Probability Distribution Table
X
P(X)
4
0.200
5
0.175
6
7 Total
0.225 0.400 = 1
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P(X)
0.40
0.30
0.20
0.10
4
5
6
7
X
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The sum of the probabilities of all events in a
sample space add up to 1.
∑ p(x) = 1
Each probability is between 0 and 1, inclusively.
0 ≤ P(x) ≤ 1
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Example 5-4: Determine whether each distribution is a probability
distribution.
X
P(X)
0
5
10
15
20
√
X
P(X)
0
-1
2
1.5
4
0.3
6
0.2
×
X
P(X)
1
2
3
4
√
X
P(X)
2
0.5
3
0.3
7
0.4
×
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5-2: Mean, Variance, Standard
Deviation, and Expectation
Mean
The mean of a random variable with a discrete
probability distribution .
µ = X1 . P(X1) + X2 . P(X2) + X3 . P(X3)+ … + Xn . P(Xn)
µ = ∑ X . P(X)
Where X1, X2 , X3 ,…, Xn are the outcomes and P(X1), P(X2),
P(X3), …, P(Xn) are the corresponding probabilities.
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Example : Children in Family
In a family with two children ,find the mean of the number
of children who will be girls.
gb
b
g
Solution:
g gg
Probability Distribution Table
b bb
b
X
g bg
0 1 2 Total
:‫خطوات الحل‬
P(X)
=1
µ= ∑X . P(X)= 0 .
n(S)= (2)^2=4 ‫ايجاد الحجم الكلي‬
x ‫ايجاد‬
‫= عدد ظهور الحدثة \ الحجم الكلي‬P(X) ‫ايجاد‬
mean‫ثم ايجاد الوسط الحسابي‬
+1.
+2.
.1
.2
.3
.4
=1
Example 5-5, 5-6 : see page 266
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Example 5-7: Tossing Coins
If three coins are tossed ,find the mean of the number of
heads that occur.
Solution:
Probability Distribution Table
Number of heads (X)
0
1
2
3
+2.
+3.
= 1.5
Probability P(X)
µ= ∑X . P(X)= 0 .
+1.
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Example 5-8: No. of Trips of 5 Nights or More
The probability distribution shown represents the number of
trips of five nights or more that American adults take per year.
(That is, 6% do not take any trips lasting five nights or more,
70% take one trip lasting five nights or more per year, etc.)
Find the mean.
Solution :
   X  P  X   0  0.06   1 0.70   2  0.20 
 3  0.03  4  0.01
 1.2 trips
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Variance and Standard Deviation
The formula for the variance of a probability distribution is
Variance:
Standard Deviation:
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Example 5-9: Rolling a Die
Compute the variance and standard deviation for the
probability distribution in Example 5–5.
Solution :
 2    X 2  P  X     2
 2  12  16  22  16  32  16  42  16
 5   6    3.5 
2
1
6
2
  2.9
,   1.7
2
1
6
2
Variance
standard deviation
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Example 5-10: Selecting Numbered Balls
A box contains 5 balls .Two are numbered 3 , one is numbered
4 ,and two are numbered 5. The balls are mixed and one is
selected at random . After a ball is selected , its number is
recorded . Then it is replaced . If the experiment is repeated
many times , find the variance and standard deviation of the
numbers on the balls.
Solution :
Number on each ball (X)
3
4
5
Probability P(X)
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Number on each ball (X)
3
4
5
32=9
42=16
52=25
Probability P(X)
X2
Step 1 :
µ= ∑X . P(X)= 3 .
Step 2 :
+4.
+5. =4
Variance
standard deviation
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Days
0
probability 0.06
1
0.7
2
0.2
3
0.03
4
0.01
A) Mean = 1.23, variance= 0.4171
B) Mean = 0.645, variance = 1.23
C) Mean =1.23, variance= 1.93
D) Mean =1.93, variance = 1.23
X
0
1
2
4
6
P(x)
0.2
0.1
K
0.3
0.2
What the value K would be needed to complete the probability
distribution?
A) 0.15
B) 0.2
C) -0.25
D) -0.2
What is the probability value of x less than or equal 2
1- X is a discrete random variable. The mean of its
probability distribution is 3, and the standard
deviation is 4. Find ∑X2 P(X).
2- A box contains 6 balls. One is numbered 2, three are
numbered 3 and two are numbered 4. Construct a
probability distribution for the numbers on the balls.
Expectation
The expected value, or expectation, of a discrete
random variable of a probability distribution is the
theoretical average of the variable.
The expected value is, by definition, the mean of
the probability distribution.
EX    X PX
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One thousand tickets are sold at $1 each for a color television
valued at $350. What is the expected value of the gain if you
purchase one ticket?
Solution :
Win
Lose
Gain(X)
$349
-$1
Probability P(X)
An alternate solution :
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Example : Winning Tickets
One thousand tickets are sold at $1 each for four prizes of
$100, $50, $25, and $10. After each prize drawing, the
winning ticket is then returned to the pool of tickets. What is
the expected value if you purchase two tickets?
$98
Gain X
Probability
P(X)
$48
$23
$8
-$2
2
2
2 992
2
1000 1000 1000 1000 1000
Solution :
2
2
2
E  X   $98  1000
 $48  1000
 $23  1000
992
2
 $8  1000
  $2   1000
 $1.63
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An alternate solution :
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Example 5-13:
Six balls numbered 1,2,3,5,8 and 13 are placed
in a box. A ball is selected at random, and its
number is recorded and then it is replaced.
Find the expected value of the numbers that
will occur.
1-If a player rolls one die and when gets a number greater
than 4, he wins 12$, the cost to play the game is 5$. What
is the expectation of the gain?
A) 2$
B) -1$
C) -2$
D) 1$
2- One thousand tickets are sold at $3 each for a PC value
at $1600. What is the expected value of the gain if a
person purchases two ticket?
EX (page:273)
Exercises 5-1 page 263-264
7-8-9-10-11-12-13-14-15-16-17-18
Exercises 5-2 page 272-273
1-2-10-15-18
5-3:The Binomial Distribution
Mean, Variance and Standard deviation for
The Binomial Distribution
 Many types of probability problems have only
two possible outcomes or they can be reduced to
two outcomes.
 Examples include: when a coin is tossed it can
land on heads or tails, when a baby is born it is
either a boy or girl, etc.
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The binomial experiment is a probability experiment
that satisfies these requirements:
1. Each trial can have only two possible
outcomes—success or failure.
2. There must be a fixed number of trials.
3. The outcomes of each trial must be
independent of each other.
4. The probability of success must remain the
same for each trial.
Example 5-15: see page 276-277
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Notation for the Binomial Distribution
P(S) :The symbol for the probability of success
P(F) :The symbol for the probability of failure
p
:The numerical probability of success
q
:The numerical probability of failure
P(S) = p and P(F) = 1 – p = q
n
:The number of trials
X
:The number of successes
Note that X = 0, 1, 2, 3,...,n
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In a binomial experiment, the probability of exactly
X successes in n trials is
n!
X
n X
P X  
 p q
 n - X ! X !
or
P X  
n
Cx
number of possible
desired outcomes
 p q
X
n X
probability of a
desired outcome
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The outcomes of a binomial experiment and the
corresponding probabilities of these outcomes
are called a binomial distribution
A coin is tossed 3 times. Find the probability of getting
exactly two heads.
n=3
Solution :
x= 2
p=
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Example 5-17: Survey on Doctor Visits
A survey found that one out of five Americans say he or she
has visited a doctor in any given month. If 10 people are
selected at random, find the probability that exactly 3 will
have visited a doctor last month.
n = 10
Solution :
x= 3
P X  
n!
 p X  q n X
 n - X ! X !
10!  1 
P  3 
 
7!3!  5 
3
p=
7
4
    0.201
5
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Example 5-18: Survey on Employment
A survey from Teenage Research Unlimited (Northbrook, Illinois) found
that 30% of teenage consumers receive their spending money from parttime jobs. If 5 teenagers are selected at random, find the probability that
at least 3 of them will have part-time jobs.
n=5
Solution :
5!
3
2
P  3 
  0.30    0.70   0.132
2!3!
P  4 
5!
4
1
  0.30    0.70   0.028
1!4!
5!
5
0
P  5 
  0.30    0.70   0.002
0!5!
x= 3,4,5
p=0.30
q=1-0.30 =0.70
P  X  3   0.132
0.028
0.002
 0.162
Q: find the probability that at most 3 of them will have part-time jobs.
Mean, Variance and Standard deviation
for the binomial
The mean , variance and SD of a variable that the
binomial distribution can be found by using the
following formulas:
Mean:   np
Variance:   npq
2
Standard Deviation:   npq
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Example 5-22: Tossing A Coin
A coin is tossed 4 times. Find the mean, variance and standard
deviation of number of heads that will be obtained.
Solution :
n=4
p=
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Example 5-23: Rolling a die
A die is rolled 360 times , find the mean , variance and
slandered deviation of the number of 4s that will be rolled .
Solution :
n = 360
p=
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A coin is tossed 72 times. The standard deviation for the number of
heads that will be tossed is
A) 18
B) 4.24
C) 6
D) 36
A student takes a 6 question multiple choice quiz with 4 choices for
each question. If the student guesses at random on each question, what
is the probability that the student gets exactly 3 questions correct?
A) 0.088
B) 0.0512
C) 0.132
D) 0.022

A student takes a 6 question multiple choice quiz with 4
choices for each question. If the student guesses at
random on each question, what is the probability that the
student gets exactly 3 questions wrong?

How many times a die is rolled when the mean of the
numbers greater than 4 that will be rolled = 20?

If 7% of calculators are defective, find the mean of the
number of defective calculators for a lot of 4200
calculators?
Which of the following is a binomial experiment?
A) Asking 100 people if they swim.
B) Testing five different brands of aspirin to see which brands are
effective.
C) Asking 60 people which brand of soap they buy.
D) Drawing four balls without replacement from a box contains 5
white balls, 7 blue balls and one green ball.
Exercises 5-3 page 283-284-285
1-5-6-11-13-17-20-26
Review Exercises page 304-305
1-2-3-8-16-17
Chapter Quiz page 306-307
All except 19-20 and 25 to 33