Download FM Lial 9th 7.5 Notes Sp10

7.5 Notes O’Brien Sp10
FM 9th ed Lial
7.5
Conditional Probability; Independent Events
I.
Introduction
Suppose that someone rolls a single fair die, out of your sight, and then you are asked
“What is the probability that a 2 was rolled?”
Since you know the sample space for the experiment is S = {1, 2, 3, 4, 5. 6} and there is only one way to
roll a two, you reply “The probability that a 2 was rolled is 16 .”
Now suppose the experiment is repeated, but this time the person rolling the die tells you that an even
number has been rolled, and then you are asked “What is the probability that a 2 was rolled?”
Knowing an even number has been rolled, you mentally reduce the sample space to S r = {2, 4, 6} and
reply “In this case, the probability that a 2 was rolled is 13 .”
When we are asked to find the probability of an event E, given that event F has already occurred, we are
being asked to find a conditional probability. The mathematical notation for this type of probability is
P E F which is read as “the probability of E given F” or “the probability of event E given that event F has


already occurred.”
In many cases, a conditional probability can be computed in a straightforward way by considering a
reduced sample space, Sr.
Example 1
If two fair dice are rolled, find the probability that a double was rolled, given that a sum of 8 was rolled.
[like 6]
Knowing that a sum of 8 was rolled, the sample space can be reduced from
S = {(1, 1), (1, 2), …, (6, 5), (6, 6)} to Sr = {(2, 6), (3, 5), (4, 4), (5, 3), (6, 2)}. Of these outcomes,
only one is a double, so P(double | sum of 8) = 15 .
Example 2
Two cards are randomly drawn, in succession, without replacement, from a standard deck of 52 cards.
a.
What is the probability that the second card drawn is a heart, given that the first was a heart?
[7]
Knowing the first card drawn was a heart, we can reduce the sample space for the second drawing to
12 hearts and 39 non-hearts. Therefore,
P(2nd card is heart | 1st card was heart) =
b.
12
51
=
4
17
What is the probability that the second card is a face card, given that the first card was a jack?
[9]
Knowing the first card was a jack, we can reduce the sample space for the second drawing to
11 face cards and 40 non-face cards. Therefore,
P(2nd card is face card | 1st card was a jack) =
11
51
For some experiments it may be difficult, inconvenient, or even impossible to find a reduced sample space.
In such cases, the following formula can be use to calculate conditional probabilities.
II.
Conditional Probability Formula
The conditional probability of event E given event F is
PE  F
, provided P(F) ≠ 0.
P E F 
PF


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7.5 Notes O’Brien Sp10
FM 9th ed Lial
Example 3
The Midtown Bank has found that most customers at the tellers’ windows either cash a check or make a deposit.
The following table indicates the transactions for one teller for one day.
Cashed Check (C)
Didn’t Cash Check (C')
Totals
Made Deposit (D)
60
20
80
Didn’t Make Deposit (D')
30
10
40
Totals
90
30
120
Express each probability in words and find its value.
a.
P(C | D) is the probability that a customer cashed a check, given that s/he made a deposit.
P(C | D) 
b.
3
60
P(C  D)
=
=
80
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P(D)
P(D' | C) is the probability that a customer did not make a deposit given that s/he cashed a check.
P(D | C) 
III.
[33]
[34]
1
30
P(D  C)
=
=
3
90
P(C)
Product Rule of Probability
For events E and F, P(E ∩ F) may be found using either of the following formulas.
P(E ∩ F) = P(E) ∙ P(F | E)
or
P(E ∩ F) = P(F) ∙ P(E | F)
The product rule gives a method for finding the probability that events E and F both occur.
Example 4
Two cards are randomly drawn, in succession, without replacement, from a standard deck of 52 cards.
Find the probability that an ace and a 4 are drawn.
[12]
There are two mutually exclusive ways to get an ace and a 4: ace followed by 4 and 4 followed by ace.
Since there are 4 aces and 4 fours in a deck of 52 cards, for the first draw, P(ace) = P(4) =
4
52
=
1
13
.
Once a card has been drawn, 51 cards remain. If an ace was drawn first, 4 fours remain, so P(4 | ace) =
If a 4 was drawn first, four aces remain, so P(ace | 4) =
4
51
4
51
.
. Therefore,
P(ace ∩ 4) = P(ace followed by a 4) + P(4 followed by an ace) =
P(ace) ∙ P(4 | ace) + P(4) ∙ P(ace | 4) =
IV.
1
13
4 +
∙ 51
1
13
4 =
∙ 51
8
663
A Probability Tree Diagram
A natural way to visually display probabilities of events that occur in an order, such as a first draw followed by a
second draw, is a probability tree diagram.
The first branching shows the possible outcomes of the first draw along with their corresponding probabilities.
The second branching shows the possible outcomes & corresponding probabilities of the second draw GIVEN the
outcome of the first-draw. These are conditional probabilities.
To find the probability of the final outcome of each two-branch sequence, we multiply the probability of the first
branch times the conditional probability of the second branch.
For the tree diagram shown below, we can find P(F) by adding P(E ∩ F) & P(E' ∩ F) and we can find P(F')
by adding P(E ∩ F') & P(E' ∩ F'); i.e.,
P(F) = P(E ∩ F) + P(E' ∩ F)
and
P(F') = P(E ∩ F') + P(E' ∩ F').
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7.5 Notes O’Brien Sp10
FM 9th ed Lial
Example 5
A bicycle factory runs two assembly lines, A and B. 60% of the factory’s bikes come off of assembly line A and
the rest come off line B. 95% of line A’s products pass inspection while only 85% of line B’s products pass
inspection.
a.
Find the probability that one of the factory’s bikes
did not pass inspection and came off line A. [41]
P(A ∩ PI') = P(A) ∙ P(PI' | A) = .60  .05 = .03
b.
Find the probability that one of the factory’s bikes
did not pass inspection and came off line B. [42]
P(B ∩ PI') = P(B) ∙ P(PI' | B) = .40  .15 = .06
c.
Find the probability that one of the factory’s bikes
did not pass inspection. [43]
P(PI') = P(A ∩ PI') + P(B ∩ PI') = .03 + .06 = .09
V.
Independent Events
A.
Definition of Independent Events
Events E and F are independent events if
P(F | E) = P(E)
or
P(E | F) = P(E).
In words, E and F are independent events if the probability of F given E is the same as the
probability of F or the probability of E given F is the same as the probability of E.
Knowledge that E occurred gives no information about the probability of the occurrence or
non-occurrence of F, and vice-versa.
B.
The Product Rule for Independent Events
Events E and F are independent events if and only if P(E ∩ F) = P(E) ∙ P(F).
Caution:
Do not confuse independent events and mutually exclusive events.
E and F are mutually exclusive events if E ∩ F = Ø and P(E ∩ F) = 0.
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7.5 Notes O’Brien Sp10
FM 9th ed Lial
M
M'
Totals
C
.035
.004
.039
C'
.452
.509
.961
Totals
.487
.513
1.000
Example 6
The adjacent table shows frequencies for red-green color blindness.
M represents the event “person is male”.
C represents the event “person is color-blind”.
a.
Find P(M).
[52]
P(M) = .487
b.
Find P(C).
[53]
P(C) = .039
c.
Find P(M ∩ C).
[54]
P(M ∩ C) = .035
d.
Are events C and M dependent? What does this mean?
[59]
P(M) ∙ P(C) = .487  .039 = .018993
Since P(M) ∙ P(C) ≠ P(M ∩ C), C and M are not independent.
This means having red-green color blindness and being male are dependent events.
Red-green color blindness does not occur as frequently among women as it does among men.
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