Download Chapter 8 Conditional and Binomial Probabilities

Chapter 8
Conditional and Binomial Probabilities
D. S. Malik
Creighton University, Omaha, NE
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Conditional Probabilities
Suppose that two distinct fair dice are rolled.
We want to …nd the probability of getting a sum of 7, when it is given
that the digit in the …rst die is greater than that of the second.
Let S be the sample space of this experiment. Then n (S ) = 36.
Because the dice are fair, each of these outcomes is equally likely.
Let E be the event that the sum of the digits of the two dice is 7, and
F be the event that the digit in the …rst die is greater than the
second. Then
E = f(6, 1), (5, 2), (4, 3), (3, 4), (2, 5), (1, 6)g
F = f(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (5, 1), (5, 2), (5, 3),
(5, 4), (4, 1), (4, 2), (4, 3), (3, 1), (3, 2), (2, 1)g.
Let G be the event that the sum of the digits in the two dice is 7 but
the digit in the …rst die is greater than the second. Then
G = f(6, 1), (5, 2), (4, 3)g = E \ F .
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It is given that the event F has occurred. Therefore, we reduce the
original sample space of 36 outcomes to the sample space F . In this
sample space, we consider the event G and the probability of G in F is
n (G )
3
1
=
= .
n (F )
15
5
This is a conditional probability because it is given that the digit in
the …rst die is greater than that of the second.
We denote it by the symbol Pr[E j F ].
Now F and G are events in S. Thus,
Pr[F ] =
n (F )
15
5
n (G )
3
1
=
= , Pr[G ] =
=
= , and
n (S )
36
12
n (S )
36
12
Pr[G ]
Pr[E \ F ]
=
=
Pr[F ]
Pr[F ]
1
12
5
12
1
= .
5
Hence,
Pr[E j F ] =
Pr[E \ F ]
.
Pr[F ]
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De…nition
Let E and F be two events in a sample space. Then the conditional
probability of E , when it is given that the event F has occurred, i.e.,
Pr[F ] 6= 0, written Pr[E j F ], is given by
Pr[E j F ] =
Pr[E \ F ]
.
Pr[F ]
Example
Let E and F be events of an experiment. Suppose that Pr[E ] = 0.6,
Pr[F ] = 0.7 and Pr[E \ F ] = 0.3. Then
Pr[E j F ] =
Pr[E \ F ]
0.3
3
=
= .
Pr[F ]
0.7
7
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Example
Let E and F be events of an experiment. Suppose that Pr[E ] = 0.65,
Pr[F ] = 0.75 and Pr[E [ F ] = 0.9. We want to determine Pr[E j F ].
We are given Pr[F ], but not Pr[E \ F ]. So …rst we …nd Pr[E \ F ]. Now
)
)
)
)
)
Pr[E [ F ] = Pr[E ] + Pr[F ] Pr[E \ F ]
0.9 = 0.65 + 0.75 Pr[E \ F ]
0.9 = 1.40 Pr[E \ F ]
0.9 1.40 = Pr[E \ F ]
0.50 = Pr[E \ F ]
Pr[E \ F ] = 0.5.
Hence,
Pr[E j F ] =
Pr[E \ F ]
0.5
50
2
=
=
= .
Pr[F ]
0.75
75
3
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Let E and F be two events of an experiment. Suppose that
Pr[F ] 6= 0 and we want to …nd Pr[E j F ].
Now Pr[E j F ] is the probability of E given that the event F has
occurred. We can think of this as a two-step experiment in which the
…rst step consists of the occurrence of the event F and the second
step consists of the occurrence of the event E . This means that we
can draw a tree diagram showing the conditional probabilities as
follows:
F
Pr[
]
Pr[
F’
]
| F]
Pr[E
E
Pr[E’
E’
F
|
Pr[E
F’
Pr[E’
| F]
F’]
E
| F’]
E’
Pr[F
E] = Pr[FE]
Pr[F
E’] = Pr[FE’]
Pr[F’
E] = Pr[F’E]
Pr[F’
E’] = Pr[F’E’]
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F
E = FE
F
E’ = FE’
F’
E = F’E
F’
E’ = F’E’
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Remark
Let E and F be events of a sample space such that Pr[F ] 6= 0. From the
previous …gure, it follows that
Pr[E j F ] + Pr[E 0 j F ] = 1.
This implies that
Pr[E 0 j F ] = 1
Pr[E j F ].
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Suppose that two cards without replacement are drawn at random
from a deck of 52 cards. Let S be the sample space of this event.
We want to …nd Pr[2nd card is a kingj 1st card is a king].
Let F be the event that the …rst card is a king and E be the event
that the second card is a king. Then F \ E is the event that the …rst
card is a king and the second card is a king.
Because two cards are drawn without replacement,
n (S ) = 52 51 = 2652.
Now F is the event that the …rst card is king. So the second card can
be any one of the remaining 51 cards. Because the number of kings is
4, n (F ) = 4 51 = 204.Hence,
Pr[F ] =
4 51
4
1
=
= .
52 51
52
13
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Next, F \ E = E \ F is the event that the …rst card is a king and the
second card is a king. Because there are 4 kings and cards are drawn
without replacement, it follows that
n (E \ F ) = n (F \ E ) = 4 3 = 12. Thus,
Pr[E \ F ] =
1
4 3
=
.
52 51
221
Next we …nd Pr[E j F ]. This is a conditional probability. The sample
space reduces to F . In the set F , the number of cards such that the
second card is a king is 4 3 = 12. Thus,
Pr[E j F ] =
4 3
4 3
3
1
=
=
= .
n (F )
4 51
51
17
You can also draw a tree diagram of this experiment as shown in the
next slide:
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|F
P r[E
F]
Pr[
Pr[
F
’]
=4
=4
/ 52
8/5
2
51
] = 3/
E
F
Pr[E’
F’
|
P r[ E
Pr[E’
| F] =
48/51
F’ ] =
| F’]
4/51
= 47/
51
E’
(4/52)(3/51) = 1/221
(4/52)(48/51) = 16/221 F
(48/52)(4/51) =16/221
E
E’
F
(48/52)(47/51) = 188/221
E = FE
E’ = FE’
F’
E = F’E
F’
E’ = F’E’
Remark
When drawing the tree diagram of conditional probabilities you must
correctly identify the steps of the experiment and the outcomes. For a
good number of problems, these steps can be easily identi…ed.
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Example
An urn contains 5 red, 6 brown, and 3 white marbles. Two marbles
without replacements are drawn. What is the probability that the second
marble is brown, given that the …rst marble is white?
This can be considered a two-step experiment and we can draw the tree
diagram as shown in the following diagram:
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First
Marble
Second Marble
(5/14)(4/13) = 10/91
R R = RR
R
4/13
6/13
R
3/13
5/14
W
R
5/13
6/14
5/13
B
3/13
B
5/13
6/13
(5/14)(3/13) = 15/182
R
B = RB
R
W = RW
B
R = BR
(6/14)(5/13) = 15/91 B
B = BB
(6/14)(5/13) = 15/91
(6/14)(3/13) = 9/91
W = BW
R (3/14)(5/13) = 15/182 W
R = WR
B
2/13
Pr[2nd marble brown | 1st marble is white]
(5/14)(6/13) = 15/91
B
W
3/14
W
B
W
(3/14)(6/13) = 9/91
(3/14)(2/13) = 3/91
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W
B = WB
W
W = WW
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Theorem
Let E and F be two events of an experiment.
(i) Suppose that Pr[F ] 6= 0. Then
Pr[E \ F ] = Pr[F ] Pr[E j F ].
(ii) Suppose that Pr[E ] 6= 0. Then
Pr[E \ F ] = Pr[E ] Pr[F j E ].
De…nition
Let E and F be two events of an experiment. Then E and F are called
independent if
Pr[E \ F ] = Pr[E ] Pr[F ].
Theorem
Let E and F be two independent events of an experiment.
(i) If Pr[F ] 6= 0, then
Pr[E j F ] = Pr[E ],
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Example
Let E and F be events.
(i) Suppose that Pr[E ] = 0.45, Pr[F ] = 0.48, and Pr[E \ F ] = 0.216. Now
Pr[E ] Pr[F ] = 0.45 0.48 = 0.216 = Pr[E \ F ].
Thus, E and F are independent.
(ii) Suppose that Pr[A] = 0.55, Pr[B ] = 0.62, and Pr[A \ B ] = 0.35. Now
Pr[A] Pr[B ] = 0.55 0.62 = 0.341 6= 0.35 = Pr[A \ B ].
Thus, A and B are not independent.
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Example
To exercise regularly, a survey of 100 students shows that 40% play
basketball, 70% jog, and 82% play either basketball or jog. Let B be the
event that a randomly selected student plays basketball and J be the event
that a randomly select student jogs.
Note that Pr[B ] = 0.40, Pr[J ] = 0.70, and Pr[B [ J ] = 0.82. Now
)
)
)
)
)
Pr[B [ J ] = Pr[B ] + Pr[J ] Pr[B \ J ]
0.82 = 0.40 + 0.70 Pr[B \ J ]
0.82 = 1.10 Pr[B \ J ]
0.82 1.10 = Pr[B \ J ]
0.28 = Pr[B \ J ]
Pr[B \ J ] = 0.28.
Thus,
Pr[B ] Pr[J ] = 0.40 0.70 = 0.28 = Pr[B \ J ].
Hence, the events B and J are independent.
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Exercise: Let E and F be events of a sample space such that
Pr[E ] = 0.56, Pr[F ] = 0.45, and Pr[E \ F ] = 0.20. Find the
following probabilities.
(a) Pr[E j F ]
(b) Pr[E 0 \ F ]
Pr[F j E ]
(e) Pr[E \ F 0 ]
Solution: a.
Pr[E j F ] =
(c) Pr[E 0 j F ]
(f) Pr[F 0 j E ]
(d)
0.20
4
Pr[E \ F ]
=
= .
Pr[F ]
0.45
9
b.
Pr[E 0 \ F ] = Pr[F \ E 0 ] = Pr[F ]
= 0.45
0.20 = 0.25.
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Pr[E \ F ]
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Pr[E ] = 0.56, Pr[F ] = 0.45, and Pr[E \ F ] = 0.20.
Pr [E 0 \F ]
0.25
Solution: c. Pr[E 0 j F ] = Pr [F ] = 0.45
= 59 .You can also …nd
Pr[E 0 j F ] by using the fact that Pr[E 0 j F ] = 1 Pr[E j F ]
and part (a). That is,
Pr[E 0 j F ] = 1
d.
Pr[F j E ] =
Pr[E j F ] = 1
5
4
= .
9
9
Pr[F \ E ]
0.20
5
=
= .
Pr[E ]
0.56
14
e.
Pr[E \ F 0 ] = Pr[E ]
f.
Pr[F 0 j E ] =
Pr[E \ F ] = 0.56
0.20 = 0.36.
Pr[F 0 \ E ]
0.36
9
=
= .
Pr[E ]
0.56
14
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Exercise:
Solution:
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Exercise: On the weekend at a shopping complex food plaza, 65% of
the customers buy lunch. Of those who buy lunch, 20% also
buy ice cream for a dessert. Of those who do not buy lunch,
55% buy ice cream for a dessert.
a. What is the probability that a randomly selected customer
does not buy ice cream, given that the customer does not
buy lunch.
b. What is the probability that a randomly selected customer
buys ice cream.
Solution: Let L be the set of customers who buy lunch and C be the
set of customers who buy ice cream. Then from the given
information
Pr[L] = 0.65, Pr[C j L] = 0.20, and Pr[C j L0 ] = 0.55.
Also, Pr[L0 ] = 1 Pr[L] = 1 0.65 = 0.35.
a. We want to …nd Pr[C 0 j L0 ]. To …nd Pr[C 0 j L0 ], we use
the fact that Pr[C 0 j L0 ] = 1 Pr[C j L0 ]. Thus,
Pr[C 0 j L0 ] = 1
Pr[C j L0 ] = 1
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0.55 = 0.45.
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Solution: b. We want to …nd Pr[C ]. Note that C is the set of
customers who buy ice cream. This is the same as the set of
customers who buy lunch and ice cream, or the customers
who do not buy lunch, but buy ice cream. That is
C = (C \ L ) [ (C \ L0 ).
So next we …nd Pr[C \ L]. Now
Pr[C j L] =
Pr[C \ L]
.
Pr[L]
This implies that
Pr[C \ L] = Pr[C j L] Pr[L] = (0.20) 0.(65) = 0.13.
Similarly,
Pr[C \ L0 ] = Pr[C j L0 ] Pr[L0 ] = (0.55) 0.(35) = 0.1925.
Hence,
Pr[C ] = Pr[C \ L] + Pr[C \ L0 ] = 0.13 + 0.1925 = 0.3225.
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The previous exercise can also be answerd using a tree diagram as follows:
0.65
0.35
0.20
| L ]=
Pr[C
L
Pr[C’
| L ]=
0.80
C
C’
(0.65)(0.20) = 0.13
(0.65)(0.80) = 0.52
5
= 0.5
C (0.35)(0.55) = 0.1925
| L’ ]
C
[
r
P
L’
Pr[C’
(0.35)(0.45) = 0.1575
| L’ ]
= 0.4 C’
5
L
C = LC
L
C ’= LC’
L’
C = L’C
L’
C’ = L’C’
a. Pr[C 0 j L0 ] = 0.45.
b.
Pr[C ] = Pr[fLC , L0 C g] = Pr[LC ] + Pr[L0 C ]
= 0.13 + 0.1925 = 0.3225
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