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Section 11.5 Bayes’ Theorem and Ch 12-13 Lecture note review for test 4
P ( B ) P ( A | Bk )
, where k  1, 2, 3, ......m . The conditional probability
P( Bk | A)  m k
 P( Bi ) P( A | Bi )
i 1
P( Bk | A) is often called the posterior probability of Bk .
1. The Belgian 20-framk coin (B20), the Italian 500-lire coin (I500), and the Honk
Kong 5-dollar (HK5) are approximately the same size. Coin purse one (C1)
contains six of each of these coins. Coin purse two (C2) contains nine B20s, six
I500s, and three HK5s. A fair four-sided die is rolled. If the outcome is {1}, a
coin is selected randomly from C1. If the outcomes belong to {2, 3, 4}, a coin is
selected randomly from C2. Find
a) P (I 500) , the probability of selecting an Italian coin
b) P(C1 | I 500) , the conditional probability that the coin selected from C1,
given that it was an Italian coin.
Solution: We have P (C1)  1 / 4 and P(C 2)  3 / 4 . We know that (you may draw a
tree diagram) P( I 500 | C1)  6 / 18  1 / 3 , P( B 20 | C1)  6 / 18  1 / 3 and
P( HK 5 | C1)  6 / 18  1 / 3 , and you can find for C2 in the same way.
a)
P( I 500)  P( I 500 | C1) P(C1)  P( I 500 | C 2) P(C 2)  1 / 3 *1 / 4  1 / 3 * 3 / 4  4 / 12  1 / 3
P(C1) P( I 500 | C1) 1 / 4 *1 / 3

 1/ 4
P( I 500)
1/ 3
2. (Homework problem 1) The Belgian 20-framk coin (B20), the Italian 500-lire
coin (I500), and the Honk Kong 5-dollar (HK5) are approximately the same size.
Coin purse one (C1) contains six of each of these coins. Coin purse two (C2)
contains nine B20s, six I500s, and three HK5s. A fair four-sided die is rolled. If
the outcome is {1}, a coin is selected randomly from C1. If the outcomes belong
to {2, 3, 4}, a coin is selected randomly from C2. Find
b) P(C1 | I 500) 
a) P(B 20) , the probability of selecting a Belgian coin (Answer: 11/24)
b) P(C1 | B 20) , the conditional probability that the coin selected from C1,
given that it was a Belgian coin. (Answer: 2/11)
3. A package, say A, of 24 crocus bulbs contains eight yellow, eight white, and eight
purple crocus bulbs. A package, say B, of 24 crocus bulbs contains six yellow, six
white and 12 purple crocus bulbs. One of the two packages is selected at random.
a) If three bulbs from this package were planted and all three yielded white
flowers, compute the conditional probability that the package A was
selected.
b) If the three bulbs yielded one yellow, one white, and one purple flower,
compute the conditional probability that the package B was selected.
Solution: Suppose WWW means three white bulbs, then
C (6, 3)
C (8, 3)
20
56
, P(WWW | A) 
, also
P(WWW | B) 


C (24,3) 2024
C (24,3) 2024
1
P( A)  P( B) 
2
a)
P( A) P(WWW | A)
P( A | WWW ) 
P( A) P(WWW | A)  P( B) P(WWW | B)
1 / 2 * 56 / 2024
56


1 / 2 * 56 / 2024  1 / 2 * 20 / 2024 76
b)
P( B) P(YWP | B)
P( B | YWP ) 
P( A) P(YWP | A)  P( B) P(YWP | B)
6 * 6 *12
1/ 2 *
54
2024


8*8*8
6 * 6 *12 118
1/ 2 *
 1/ 2 *
2024
2024
4. (Homework problem 2) A package, say A, of 24 crocus bulbs contains eight
yellow, eight white, and eight purple crocus bulbs. A package, say B, of 24 crocus
bulbs contains six yellow, six white and 12 purple crocus bulbs. One of the two
packages is selected at random.
a) If three bulbs from this package were planted and all three yielded purple
flowers, compute the conditional probability that the package B was
selected. (Answer: P(B|PPP) = 55/69)
b) If the three bulbs yielded one yellow, one white, and one purple flower,
compute the conditional probability that the package A was selected.
(Answer: P(A|YWP) = 32/59)
5. (Homework problem 3) Given two urns, suppose urn I contains 4 black and 7
white balls. Urn II contains 3 black, 1 white, and 4 yellow balls. Select an urn and
then select a ball.
a) What is the probability that you obtain a black ball? (Answer: 65/176)
b) What is the probability that you obtain a ball from urn II, given that the ball is a
black ball? (Answer: 33/65)
6. (Homework problem 4) An absence minded nurse is to give Mr. Brown a pill
each day. The probability that the nurse forgets to administer the pill is 2/3. If he
receives the pill, the probability that Mr. Brown will die is 1/3. If he does not get
the pill, the probability that he will die is ¾. Mr. Brown died. What is the
probability that the nurse forgot to give Mr. Brown the pill? (Answer: 9/11)
Solution hints: B = the nurse forgets to give pill, A = do not forget, E = Mr.
Brown dies. Now P(B) = 2/3, P(E|A) = 1/3, P(E) = P(A)P(E|A)+P(B)P(E|B), find
P(B|E).
Ch12-13:
1. Let the p.m.f. of X be defined by f ( x)  x / 9, x  2, 3, 4 . Draw a) a bar graph
for this p.m.f. b) a probability histogram for this p.m.f.
Solution: To draw bar graph and histogram find f(2) = 2/9, f(3) = 3/9, f(4) = 4/9,
then plot x along horizontal line and f(x) along vertical line. Review from your
note.
2. Given the p.m.f. f ( x)  x / c, x  2, 3, 4 . Determine c and rewrite the p.m.f.
Find P(2  x  3)
Solution: Find f(2) = 2/c, f(3) = 3/c, f(4) = 4/c, and then set 2/c + 3/c + 4/c=1 and
find c= 9. P(2  x  3) = P(2) + P(3) = 2/9 + 3/9 = 5/9
3. Let the random experiment be the cast of a pair of unbiased dice, each having
three faces numbered 1, 2, 3, and let the random variable X be the product of
the dice. Determine p.m.f. f(x) of X and draw the bar graph of f(x). Find
P(2  x  3)
Solution: The sample space is 1,1 1,2
1,3
2,1 2,2
2,3
3,1 3,2
3,3
Now P(X = 1) = 1/9 P(X = 2) = 2/9
P(X = 3) = 2/9
P(X = 6) = 2/9
P(X = 9) = 1/9
The p.m. f is defined by
1 / 9, x  1, 4, 9
f ( x)  
2 / 9, x  2, 3, 6
Now you can draw the bar graph.
P(2  x  3)  P(2)  P(3)  2 / 9  2 / 9  4 / 9
P(X = 4) = 1/9
4. According to a survey, 70% of all students at a large university suffer from
math anxiety. Three students are randomly selected from this university. Let X
be the number of students in the sample who suffer from math anxiety.
Develop the probability distribution of X.
Solution: Draw the tree diagram and find the distribution
X
0
1
2
3
P(X)
.027
.189
.441
.343
5. Let X have the p.m.f. f ( x)  x / 10, x  1, 2, 3, 4 , find E(X) and Var(X).
Solution: E(X) = 1(1/10) + 2(2/10) + 3(3/10) + 4(4/10) = 3
Var(X) = 12 (1 / 10)  2 2 (2 / 10)  32 (3 / 10)  4 2 (4 / 10)  9  1
6. Given the following distributions, find mean  and variance  2 . (Look at text
book p#471, # 16)
3
a) f ( x)   (1 / 4) x (3 / 4) 3 x , x  0,1, 2, 3
 x
 4
b) f ( x)   (1 / 2) 4 , x  0,1, 2, 3, 4
 x
7. Given the p.d.f. f ( x)  x 3 / 4, 0  x  c . Find c. Find P(1  x  3) .
c
 x4 
c4
1 c  2
Solution:  x / 4 dx  1     1 
16
16


0
0
c
3
2
 x4 
P(1  x  3)   x / 4 dx     1  1 / 16  15 / 16
 16  1
1
8. A normal distribution is given with a mean of 75 and standard deviation of 5.
Find a) P(65  x  85) b) P(65  x) c) P ( x  85)
Solution: a) P(65  x  85) =95.45%
b) P(65  x)  .5  .4772  97.72%
c) P ( x  85) = .5 + .4772 = 97.72
2
3
Draw normal curve and verify the results.