Ch5.3 Conditional Probability and Independence
-----------------------------------------------------------------------------------------------------------Topics:
§5.3: Conditional probability
Multiplicative law
Independent events
Calculating probabilities
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Conditional Probability
 Consider the distribution of blood type in a certain population:
Blood Type
Total # of
people
TO
70
P( TO )=70/200=0.35
TA
50
P( TA ) =50/200=0.25
TB
50
P( TB ) =50/200=0.25
TAB
30
P( TAB ) =30/200=0.15
Total
200

Blood Type
Male
Female
TO
37
33
Total # of
people
70
TA
27
23
50
TB
26
24
50
TAB
16
106
14
94
30
200
Total
 Marginal probability
 Row margins:
P( TO ), P( TA ), P( TB ), P( TAB )
 Column margins: P( Male ) =106/200=0.53
P( Female ) =94/200=0.47
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We can jointly consider “Gender” and “Blood Type” (i.e., joint probability)
P( TO and Male ) = 37/200=0.185
P( TAB and Female ) = 14/200=0.07

We can also concentrate on “Male” only (i.e., conditional probability):
Among males, the probability of being type “O” =37/106=0.349
Such probability is denoted by
P[ To |Male]
Ex. P( TA | Male) =27/106=0.254
P( TAB | Female) =14/94=0.149
 Conditional distribution is all about reducing the space of interest from the
Sample space to the conditioning event
 P( A | B ) =
# of simple events in A and B
P( A  B)
=
# of simple events in B
P( B)
(if each simple event is equally likely to happen)
 Conditional Probability:
The conditional probability of A given that event B has occurred is
P( A | B) 
P( A  B)
P( B)
Ex. A student is randomly selected from a class where 35% of the class is lefthanded and 50% are sophomores. We further know that 5% of the class
consists of left-handed sophomores. Given that a randomly selected student
is a sophomore, what is the probability that he/she is left-handed?
What we know:
Define A = event that a randomly selected student is left-handed,
B = event that a randomly selected student is a sophomore.
Then P(A)=0.35, P(B)=0.5, P( A  B)  0.05.
What we want: P(A|B)
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Solve: By the definition,
P( A | B) 
P( A  B) 0.05

 0.1
P( B)
0.5
Ex. A certain system can experience 2 difference types of defects. Let Ai , i=1,2
denote the event that the system has a defect of type i. Suppose that
P A1   0.15, P A2   0.10 , and P A1 and A 2   0.08 . If the system has a type 1
defect, what is the probability that it has a type 2 defect?
What we know:
P( A1 )  0.15, P( A2 )  0.10, P( A1  A2 )  0.08
What we want:
P( A2 | A1 ).
Solve: By the definition,
P( A2 | A1 ) 
P( A1  A2 ) 0.08

 0.53.
P( A1 )
0.15
 Multiplicative Law:
P( A  B)  P( A) P( B | A)  P( B) P( A | B)
Ex. Six balls in a basket: 2 white, 3 blue and 1 yellow. Two balls are drawn
randomly one after the other. What is the probability that the first ball is
white, and the second ball is white?
What we know:
Define A = the event the first selected ball is white,
B = the event the second selected ball is white.
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What we want: P( A  B).
Solve: Since P(A) is easy to find (2/6) and P(B|A) is easy to find (1/5), we use the
fomula:
P( A  B)  P( A) P( B | A)  (2 / 6)(1/ 5)  1/15  0.067
Alternative way to solve this problem: The Tree Diagram
(Please read the text book on page 205)
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Ex. A store sells 2 different brands of DVD players. Of its DVD player sales, 60% are
brand A (less expensive) and 40% are brand B. Each manufacturer offers a 1-yr
warranty on parts and labor. It is known that 25% of brand A’s DVD players require
warranty repair work, whereas 10% for brand B.
(a) What is the probability that a randomly selected purchaser has bought a brand A
DVD player that will need repair while under warranty?
What we know:
Define A = event that a randomly selected purchaser bought a brand A DAD
player.
B = event that a randomly selected purchaser bought a brand B DAD
player.
W = event that purchased DVD play requires repair work.
So P(A) = 0.6, P(W|A) = 0.25, P(W|B) = 0.1
What we want: P( A  W )
Solve:
P( A  W )  P( A) P(W | A)  0.6  0.25  0.15
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(b) Given that the probability that a randomly selected purchaser has a DVD player
that will need repair while under warranty is 0.19 (can be obtained from the
given conditions). Now if a customer returns to the store with a DVD player that
need warranty repair work, what is the probability that it is brand A? Brand B?
What we know:
P(W)=0.19
What we want: P(A|W) & P(B|W)
Solve:
P( A | W ) 
P( A  W ) 0.15

 0.79
P(W )
0.19
We can find P( B  W )  P( B) P(W | B)  0.4  0.1  0.04 . So
P( B | W ) 
P( B  W ) 0.04

 0.21
P(W )
0.19
Or alternatively,
P(B|W)=P(A’|W) = 1 – P(A|W) = 1-0.79 = 0.21
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Independent events:
(1) Events A and B are independent events if the occurrence of either one
event is not affected by the occurrence of the other event. That is,
P A | B  P A or PB | A  PB
In this case,
P( A  B)  P( A) P ( B )
(2) For k independent events A1 , A2 ,..., Ak ,
P A1 and A2 and .... and Ak   P A1 P A2  P Ak 
Ex. A system consists of 3 components, as shown in the picture. The
components work or fail independently of one another, and each component
works with probability 0.9. What is the probability that the entire system will
work?
(a)
1
2
3
What we know:
Define Ai = event that component i (=1,2,3) works. P( Ai )  0.9 .
What we want: P(the system works)
Solve: Since the system works if and only if every component works. So
P(system works)= P( A1  A2  A3 )  P( A1 ) P( A2 ) P( A3 )  0.9  0.9  0.9  0.729
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(b)
2
1
3
What we know: Define Ai = event that component i (=1,2,3) works.
P( Ai )  0.9 .
What we want: P(the system works)
Solve: Define the mini-system consisting of components 1 and 2 as S1 . The
system works if and only if either S1 works or component 3 works. That is
P(system works)=P( S1  A3 ) =P( S1 ) + P( A3 ) – P( S1  A3 ).
But S1  A1  A2 . By independence, P( S1 )=0.9*0.9=0.81.
P( S1  A3 )=P( A1  A2  A3 ) =0.729 (from (a)). Therefore,
P(system works)=P( S1  A3 ) =0.81+ 0.9– 0.729=0.981.
Ex. Two independent services A and B are used to deliver a document. The
probability of on-time delivery for service A is 0.8, and for service B is 0.7.
What is the probability that the document being delivered on time?
What we know:
Define A = event that the document delivered by service A will be on-time.
B = event that the document delivered by service B will be on-time.
Then A and B are independent and P(A) = 0.8, P(B) = 0.7
What we want: P( A  B) (since the document will be delivered on-time as
long as one service is on-time
Solve: By the additive rule and the fact that A and B are independent events,
we have
P( A  B)  P( A)  P( B)  P( A  B)  P( A)  P( B)  P( A) P( B) =0.8+0.7-0.8*0.7=0.94
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