Download 1.5 Rules for Probability Probability Rules: For any events, E and F

1.5 Rules for Probability
Probability Rules: For any events, E and F , in a sample space S, we have
(1) 0 ≤ P (E) ≤ 1
(2) P (S) = 1
(3) P (∅) = 0
(4) P (E ∪ F ) = P (E) + P (F ) − P (E ∩ F ) (Union Rule for Probability)
Proof. Since n(E ∪ F ) = n(E) + n(F ) − n(E ∩ F ), divide both sides by n(S), we have
n(E ∪ F )
n(E) n(F ) n(E ∩ F )
=
+
−
.
n(S)
n(S)
n(S)
n(S)
We complete the proof by definition of probability.
(5) If E and F are mutually exclusive events, then P (E ∪F ) = P (E)+P (F ). (by part (4), n(E ∩F ) = 0)
(6) P (E C ) = 1 − P (E) and P (E) = 1 − P (E C ) (Complement Rule. P (S) = 1. E and E C are mutually
exclusive)
Example 1. Let E and F be two events of an experiment with sample sample space S. Suppose
P (E) = 0.6, P (F ) = 0.3 and P (E ∪ F ) = 0.7. Compute the following:
(a) P (E ∩ F ); (b) P (F C ); (c) P (E C ∪ F C ); (d) P (E C ∩ F ).
Solution. (a) P (E ∩ F ) = P (E) + P (F ) − P (E ∪ F ) = 0.2.
(b) P (F C ) = 1 − P (F ) = 0.7.
(c) P (E C ∪ F C ) = P ((E ∩ F )C ) = 1 − P (E ∩ F ) = 0.8.
(d) P (E C ∩ F ) = P (F ) − P (E ∩ F ) = 0.1.
Example 2. If P (E) = 0.6 and P (F ) = 0.3 with E and F mutually exclusive, what is P (E C ∪ F C )?
Solution. P (E C ∩ F C ) = P ((E ∪ F )C ) = 1 − P (E ∪ F ) = 1 − (P (E) + P (F )) = 0.1.
Example 3. Suppose you are given the following probability distribution for a sample space S =
{s1 , s2 , s3 , s4 , s5 .s6 }.
Outcome
s1
Probability
3
10
s2
s3
1
20
s4
s5
s6
1
12
1
10
Suppose E = {s1 , s4 , s5 }, F = {s2 , s3 }, G = {s2 , s5 }, and P (E) =
the table and then calculate the following.
a) P (F ∩ G) and P (E ∩ F )
b) P (GC ) and P (E ∪ G)
1
11
.
20
Fill in the missing probability in
c) P (F C ∪ GC )
3
1
Solution. P (s4 ) = P (E) − P (s1 ) − P (s5 ) = 11
− 10
− 12
= 16 and P (s4 ) = 1 − P (s1 ) − P (s2 ) − P (s3 ) −
20
3
.
P (s5 ) − P (s6 ) = 10
3
(a) P (F ∩ G) = P (s2 ) = 10
. P (E ∩ F ) = P (∅) = 0.
3
1
37
1
1
17
C
(b) P (G ) = 1−P (G) = 1− 10
− 12
= 60
. P (E∪G) = P ({s3 , s6 }C ) = 1−P (s3 )−P (s6 ) = 1− 20
− 10
= 20
.
7
C
C
C
C
(c) P (F ∪ G ) = P ((F ∩ G) ) = P ({s2 } ) = 1 − P (s2 ) = 10 .
Example 4. An experiment consists of flipping a fair coin and rolling a fair six-sided die.
a) Find the probability of flipping a heads and rolling an even number.
b) Find the probability of flipping a heads or rolling an even number.
c) Find the probability of rolling an even number or rolling 5.
d) Find the probability of not rolling a 5.
e) Find the probability that neither a head is flipped nor an even number is rolled.
Solution. The sample space
S = {(H, 1), (H, 2), (H, 3), (H, 4), (H, 5), (H, 6), (T, 1), (T, 2), (T, 3), (T, 4), (T, 5), (T, 6)}.
(a) The event {flipping a heads and rolling an even number} = {(H, 2), (H, 4), (H, 6)}.
3
= 14 .
So the probability is 12
(b) The event E = {flipping a heads} = {(H, 1), (H, 2), (H, 3), (H, 4), (H, 5), (H, 6)}.
The event F = {rolling an even number} = {(H, 2), (H, 4), (H, 6), (T, 2), (T, 4), (T, 6)}.
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So E ∪ F = {(H, 1), (H, 2), (H, 3), (H, 4), (H, 5), (H, 6), (T, 2), (T, 4), (T, 6)} and P (E ∪ F ) = 12
= 34 .
(c) The event G = {rolling 5} = {(H, 5), (T, 5)}.
8
= 23 .
So G ∪ F = {(H, 2), (H, 4), (H, 6), (T, 2), (T, 4), (T, 6), (H, 5), (T, 5)} and P (E ∪ F ) = 12
(d) P (GC ) = 1 − P (G) = 56 .
(e) P (E C ∩ F C ) = P ((E ∪ F )C ) = 1 − P (E ∪ F ) = 14 .
Example 5. The table below gives the number of students of each classification who are majoring and
not in business in a class of 110 students.
Freshmen
Sophmores
Juniors
Seniors
Total
Business
10
17
20
12
59
Non-Business
8
3
15
25
51
Total
18
20
35
37
110
A student is randomly selected from this class. Find the probability of the following events.
a) The student is not a junior.
2
b) The student is a lower classman (freshman or sophmore).
c) The student is an upper classman non-business major.
d) The student is a business major or a sophmore.
35
7
Solution. (a) Let the event A = {The students is a junior}. So P (A) = n(A)
= 110
= 22
. P (AC ) =
n(S)
1 − P (A) = 15
.
22
(b) Let the event B = {The student is a lower classman}. So n(A) = 18 + 20 = 38. Then, P (B) =
38
= 19
.
110
55
(c) Let the event C = {The student is a upper classmain and non-business major}. Then, n(C) =
4
40
= 11
.
15 + 25 = 40. Then, P (C) = 110
(c) Let the events D = {The student is a business major} and E = {The student is a sophmore}. So
n(D) = 59, n(E) = 20, n(D ∩ E) = 17. Then n(D ∪ E) = n(D) + n(E) − n(D ∩ E) = 62. Thus,
62
P (D ∪ E) = 110
= 31
.
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Odds: The odds of an event E are defined to be the ratio of P (E) to P (E C ), or
P (E)
.
P (E C )
Example 6. Given the probability that it will rain tomorrow is 0.3, what are the odds that it will rain
tomorrow? What are the odds that is will not rain tomorrow?
Solution. Let the event E = {It rains tomorrow}. So the odds that it will rain are
0.3
3
P (E)
=
= .
1 − P (E)
0.7
7
The odds that it will not rain are
P (E C )
7
= .
P (E)
3
Obtaining Probability from Odds: Suppose that the odds for an event E occurring is given as
or a : b, then
a
P (E) =
.
a+b
a
b
Example 7. The odds of Whipper Snapper (a horse) winning at hte Kentucky Derby are 5 to 3. What
is the probability that Whipper Snapper will win? What is the probability that Whipper Snapper will not
win?
Solution. Let the event E = {Whipper Snapper will win}. So
P (E)
5
= .
1 − P (E)
3
3
Then,
P (E) =
and
5
5
= .
5+3
8
3
P (E C ) = 1 − P (E) = .
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1.6: Conditional Probability
Example 8. Throw a six-sided dice and observe the number of dots on the top face.
a) Let the event E = {the number 4 or 3}. So
P (E) =
2
1
= .
6
3
b) Assume that we know the number we observe is always even (Maybe using some high-techies). What
is the probability we observe the number 3 or 4?
Solution. Since there are only three outcomes F = {the number 2 or 4 or 6}. We could only observe
4. So the reasonable probability should be one out of three, i.e. 1/3. Consider the general probability
framework we discussed before. The sample space has been changed! The new sample space is F
and the desired event should be E ∩ F . Therefore, our probability under the condition F is
P (E|F ) =
1
n(E ∩ F )
= .
n(F )
3
Conditional Probability: Let E and F be two events in a sample space S. The probability that E
occurs given that F has occurred is defined to be
P (E|F ) =
n(E ∩ F )
n(E ∩ F )/n(S)
P (E ∩ F )
=
=
.
n(F )
n(F )/n(S)
P (F )
Example 9. A pair of fair six-sided dice is rolled. One is red and the other is green.
a) What is the probability that the sum is equal to 5 given that a 2 is rolled?
b) What is the probability that a 2 is rolled given that the sum is less than 5?
Solution. Note that n(S) = 36. Let the event E = {the sum is equal to 5} = {(1, 4), (2, 3), (3, 2), (4, 1)}
and the event F = {2 is rolled} = {(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (1, 2), (3, 2), (4, 2), (5, 2), (6, 2)}.
So E ∩ F = {(2, 3), (3, 2)} and
(a)
n(E ∩ F )
2
P (E|F ) =
= .
n(F )
11
4
(b)
P (F |E) =
n(E ∩ F )
2
1
= = .
n(E)
4
2
Example 10. Let P (E) = 0.7, P (F ) = 0.3, and P (E ∩ F ) = 0.1. Find
a) P (E|F ); b) P (F C |E); c) P (F |E C ); d) P (E ∪ F |F ); e) P (E C ∩ F |E).
Solution. a)
P (E|F ) =
P (E ∩ F )
0.1
1
=
= .
P (F )
0.3
3
b)
P (F C |E) =
P (E) − P (E ∩ F )
0.7 − 0.1
6
P (F C ∩ E)
=
=
= .
P (E)
P (E)
0.7
7
c)
P (F |E C ) =
F ∩ EC
P (F ) − P (F ∩ E)
0.3 − 0.1
2
=
=
=
.
P (E C )
1 − P (E)
1 − 0.7
3
d)
P (E ∪ F |F ) =
P ((E ∪ F ) ∩ F )
P (F )
=
= 1.
P (F )
P (F )
e)
P ((E C ∩ F ) ∩ E)
P (∅)
P (E ∩ F |E) =
=
= 0.
P (E)
P (E)
C
Product Rule: If E and F are two events in a sample space S with P (E) > 0 and P (F ) > 0,
then
P (E ∩ F ) = P (F )P (E|F ) = P (E)P (F |E).
Example 11 (ex. 38). Two machines turn out all the products in a factory, with the first machine
producing 40% of the product and the second 60%. The first machine produces defective products 2% of
the time and the second machine 4% of the time.
a) What is the probability that a defective product is produced at this factory and it was made on the
first machine?
b) What is the probability that a defective product is produced at this factory?
c) Given a defective product, what is the probability it was produced on the first machine?
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Solution. Let E1 = {Products from the first machine} and
E2 = {Products from the second machine}.
We also let F1 = {Defective products from the first machine} and
F2 = {Defective products from the second machine}.
So P (E1 ) = 0.4, P (E2 ) = 0.6, P (F1 |E1 ) = 0.02 and P (F2 |E2 ) = 0.04.
a) Let E = {Defective products at this factory}. Then
P (E ∩ E1 ) = P (E|E1 )P (E1 ) = 0.02 ∗ 0.4 = 0.008.
b)
P (E) = P (F1 ) + P (F2 ) = P (F1 ∩ E1 ) + P (F2 ∩ E2 ) = P (E1 )P (F1 |E1 ) + P (E2 )P (F2 |E2 ) = 0.032.
c)
P (E1 |E) =
P (E1 ∩ E)
P (F1 )
P (F1 |E1 )P (E1 )
0.4 × 0.02
1
=
=
=
= .
P (E)
P (E)
P (E)
0.032
4
Independent Events: We say that two events E and F are independent if the outcome of the one
does not affect the outcome of the other. So the conditional probability P (E|F ) does not depend on
P (F ) (The event E happens no matter whether the event F has happened), that is P (E|F ) = P (E).
Similarly P (F |E) = P (F ).
Note that if P (E) > 0 and P (F ) > 0,
P (E) = P (E|F ) =
P (E ∩ F )
,
P (F )
P (F ) = P (F |E) =
P (E ∩ F )
.
P (E)
and
So P (E ∩ F ) = P (E)P (F ). This is independent events theorem.
Mutually exclusive vs independence.
Example 12. Determine if the givens events E and F are independent.
a) P (E) = 0.5, P (F ) = 0.6, and P (E ∩ F ) = 0.4
b) P (E ∩ F C ) = 0.3, P (E ∩ F ) = 0.2, and P (E C ∩ F ) = 0.2
Solution. (a) P (E)P (F ) = 0.3 6= P (E ∩ F ). They are not independent.
(b) P (E) = P (E ∩ F C ) + P (E ∩ F ) = 0.5. P (F ) = P (F ∩ E C ) + P (E ∩ F ) = 0.4. So P (E)P (F ) =
0.2 = P (E ∩ F ). They are independent.
Example 13. Assume the events E and F are independent with P (E) = 0.3 and P (F ) = 0.6. Find
P (E ∪ F ).
6
Solution.
P (E ∪ F ) = P (E) + P (F ) − P (E ∩ F ) = P (E) + P (F ) − P (E)P (F ) = 0.3 + 0.6 − 0.3 × 0.6 = 0.72.
Example 14. A fair coin is flipped three times. Let E be the event “at most one head” and F the
event “at least one head and at least one tail.” Are there two events independent?
Solution. Let E = {At least one head} and F = {At least one head and at least one tail}. Then
P (E) = 84 = 21 and P (F ) = 68 = 43 . P (E ∩ F ) = 38 . So P (E)P (F ) = 38 = P (E ∩ F ). Thus, they
are independent.
A set of events {E1 , E2 , · · · , En } is said to be independent if, for any k of these events, the probability
of the intersection of these k events is the product of the probabilities of each of the k events. Thus
must hold for any k = 2, 3, · · · , n.
For example, if we want to verify the three events E, F and G are independent, we need to verify
P (E ∩ F ) = P (E)P (F ),
P (F ∩ G) = P (F )P (G),
P (G ∩ E) = P (G)P (E),
P (E ∩ F ∩ G) = P (E)P (F )P (G).
Example 15. If E, F and G are independent,
a) Show that E C and F are independent.
Proof.
P (E C )P (F ) = P (F ) − P (E)P (F ) = P (F ) − P (E ∩ F ) = P (E C ∩ F ).
b) Show that E C , F and G are independent.
Proof. By definition,
P (E C ∩ F ) = P (E C )P (F ),
P (F ∩ G) = P (F )P (G),
by part (a)
P (G ∩ E C ) = P (G)P (E C ),
similar argument by part (a)
C
C
P (E ∩ F ∩ G) = P (E )P (F )P (G) = (1 − P (E))P (F )P (G)
= P (F )P (G) − P (E)P (F )P (G)
= P (F ∩ G) − P (E ∩ F ∩ G) = P (E C ∩ F ∩ G).
7