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1
Tree Diagrams
Example: A man takes either a bus or the subway
to work with probabilities 0.3 and 0.7 , respectively.
When he takes the bus, he is late 30% of the days.
When he takes the subway, he is late 20% of the days.
What is the probability that he is late?
Solution:
First we can write a tree diagram for the corresponding probabilities as follows:







0.3


−→










The man →










0.7


−→








0.3

−→



by bus
→



0.7

 −→


0.2

−→



by subway →



0.8

 −→
late
on time
late
on time
From above tree diagram, we can get
P({The man is late}) = 0.3(0.3) + 0.7(0.2) = 0.23.
Example: As items come to the end of a production
line, an inspector chooses items to undergo a complete
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inspection. Of all items produced 20% are defective.
50% of all defective items go through a complete inspection, and 30% of all good items go through a complete inspection. Given that an item is completely inspected, what is the probability that it is defective?
Solution: Define
D : An item defective.
C : An item is completely inspected.
Then, we have following information:
P(C|D) = 50%
P(D) = 20%
P(C|Dc) = 30%
P(Dc) = 80%



0.5



−→
inspected







0.2


−→
defective →








0.5


 −→

not inspected



An item →










0.8


−→








0.3

−→



good
→



0.7

 −→
inspected
not inspected
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From above tree diagram, we can get we have
P(D ∩ C)
P(C)
P(D)P(C|D)
=
P(C)
0.2(0.5)
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=
=
0.2(0.5) + 0.8(0.3) 34
P(D|C) =
Chapter Thirteen: Binomial Distributions
First, let us call the counting rule for combinations.
Counting Rule for Combinations
The number of different combinations of n different
objects that can be formed, taking them r at a time,
is
 
n
n!
  = Crn =
r!(n − r)!
r
Example: We want to choose 5 people from 20 people to organize a traveling group. Are there how many
different ways to choose a group?
Solution: This is just how many different combinations. So the solution is
 
20
20!
 =
= 15504.
5!(20
−
5)!
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An experiment is called a binomial experiment if it
satisfies following conditions:
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(1) There are a fixed number n of trials or subexperiments.
(2) The n trials are all independent.
(3) Each trial has exactly two possible outcomes: success and failure.
(4) The probability of success is p which satisfies 0 <
p < 1 and which is the same for all the trials.
Binomial distribution:
Let X denote the count of successes in the binomial
experiment. Then X has following binomial distribution:
The distribution table of r.v X
x
0
1
2
··· n
P(X = k) C0np0q n C1np1q n−1 C2np2q n−2 · · · Cnnpnq 0
where q = 1−p.pThen, X has mean µ = np and standard
deviation σ = np(1 − p).
Remark: If we consider that n different positions, each
time we choose k different positions to put successes.
Then, this is just the total number of different combinations we can get and this is where the coefficient Ckn
comes from.
Example: A couple plans to have three children. There
are 8 possible arrangements of girls and boys. For example, GGB means the first two children are girls and
the third child is a boy. All 8 arrangements are equally
likely.
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(a) Write down all 8 arrangements of the sexes of three
children. What is the probability of any one these arrangements.
(b) Let X be the number of girls the couple has. What
is the probability that X = 2?
(c) Find the distribution of X.
Solution:
(a) All different 8 arrangements are as follows:
1
2
3
4
5
6
7
8
BBB BBG BGB BGG GBB GBG GGB GGG
They are equally likely. So each has probability 1/8.
Since X is the number of girls the couple has,
P(X = 2) = P({BGG, GBG, GGB})
= P({BGG}) + P({GBG}) + P({GGB}) = 3/8.
(c) Similarly we can find that
P(X = 1) = P({BBG, BGB, GBB})
= P({BBG}) + P({BGB}) + P({GBB}) = 3/8,
P(X = 0) = P({BBB}) = 1/8,
and
P(X = 3) = P({GGG}) = 1/8.
Thus, we have the distribution table of X.
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Value of X
0
1
2
3
Probability 1/8 3/8 3/8 1/8
Now if we use binomial distribution, then, p = 1/2 and
n = 3, we can get
1 1
P(X = 1) = C13( )1( )2 = 3/8
2 2
and
Value of X
0
1
2
3
Probability 1/8 3/8 3/8 1/8
Example: A couple plans to have 5 children. Each new
baby has equal probability to be a boy or girl. Let X
be the number of girls the couple will have. What is
the probability that X = 2?
Solution:
1 1
P(X = 2) = C25( )2( )3 = 5/16.
2 2
Example: Consider an experiment of randomly drawing a chip 8 times successively with replacement from
a box containing two black chips and three white chips.
The chips are identical except their colors. Let X be
the number of black chips observed in the 8 drawings.
(a) Find the probability P(X = 5).
(b) Find its mean and variance.
Solution: (a) According to the binomial distribution
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formula, we have
P(X = 5) =
5
3 8−5
8 2
C5 ( ) ( )
5
5
= 0.124.
(b) Its mean is equal to
µ = 8(2/5) = 16/5
and its variance is equal to
48
25
Example: Consider a binomial experiment of flipping
a biased coin twenty times successively. Let A be the
event of observing a head and Ā be the event of observing a tail.
Suppose that
σ 2 = 8(2/5)(3/5) =
4
1
P(Ā) =
5
5
Let X be number of heads observed in the twenty flippings.
(a) Find the probability P(X = 8).
(b) Find its mean and standard deviation.
Solution: (a) According to the binomial distribution
formula, we have
P(A) =
P(X = 8) =
8
4 20−8
20 1
C8 ( ) ( )
5
5
(b) Its mean is equal to
µ = 20(1/5) = 4
= 0.022.
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and its standard deviation is equal to
r
p
16
σ = 20(1/5)(4/5) =
5
Example: A factory employs 2 thousand workers, of
whom 30% are Hispanic. If the 15 members of the union
executive committee were chosen from the workers at
random,let X be the number of Hispanics on the committee. Find P(X = 3).
Solution:
This is not a binomial model because it is drawing
sample without replacement.
1400
C3600 × C12
.
P(X = 3) =
2000
C15
When the number of trials n is very big, it is very
difficult to calculate the probability
P(X = k) = Cknpk (1 − p)n−k .
However, we have following approximation method.
Normal approximation to binomial distributions
Let X denote the count of successes in the binomial
experiment with n trials and success probability p. If
np ≥ 10 and n(1 − p) ≥ 10, thenp
the distribution of X is
approximately Normal, N (np, np(1 − p)).
More precisely, we have following formula. For given
constant integers a, b, use following formulae to find
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approximate probabilities:
(I)
(II)
a − np
P(X ≥ a) = P(Z ≥ √
)
npq
b − np
P(X ≤ b) = P(Z ≤ √
)
npq
Example:
Consider the binomial experiment with
n = 10, 000 and the probability of success p = 0.3. Let X
be the number of successes in the 10, 000 trials. Find
the approximate probability P(X ≤ 3100).
Solution: (1) np = 10, 000(0.3) = 3, 000 > 10 and nq =
10, 000(0.7) = 7, 000 > 10.
p
√
(2) µ = np = 3, 000, σ = npq = 10, 000(0.3)(0.7) = 45.82
(3) To find the probability P(X ≤ 3, 100), according to
(II), we have
b − np
)
P(X ≤ b) = P(Z ≤ √
npq
3, 100 − 3, 000
)
45.82
= P(Z ≤ 2.19)
= 0.9857
P(X ≤ 3, 100) = P(Z ≤
Example: Suppose that early statewide election returns indicate totals of 33, 00 votes for candidate A versus 27, 00 for candidate B, and that these early returns
can be regarded as a random sample selected from the
population of all 10, 000, 000 eligible voters in the state.
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Let X be the number of votes for A.
(a) If the statewide vote will be split 50 − 50, find the
expected number of votes for A in the sample of 60, 000
early returns.
(b) Find the standard deviation of X.
(c) Find the probability that (X > 30, 100).
Solution:
(a) µ = np = 60, 000 × 0.5 = 30, 000.
√
√
(b) σ = npq = 15, 000 = 122.47449
P
60,000
(c) P(X > 30, 0100) = 1 − 30,100
(0.5)i(0.5)60,000−i
i=0 Ci
However, we can find an approximate probability as
follows:
P(X > 30, 100) = 1 − P(X ≤ 30, 100)
30, 100 − 30, 000
= 1 − P(Z ≤
)
122.47
= 1 − P(Z ≤ 0.8165) = 0.208