1198/DCP1206 Probability, Fall 2015
23-Oct-2015
Homework 2 Solutions
Instructor: Prof. Wen-Guey Tzeng
Scribe: Amir Rezapour
1. The theaters of a town are showing seven comedies and nine dramas. Marlon has
seen five of the movies. If the first three movies he has seen are dramas, what is the
probability that the last two are comedies? Assume that Marlon chooses the shows
at random and sees each movie at most once.
Answer.
Reduce the sample space: Marlon chooses from six dramas and seven comedies two
atrandom.
What is the probability that they are both comedies? The answer is
7
21
2
2 / 7 = 0.269.
2. A number is selected at random from the set {1, 2, ..., 10, 000} and is observed to be
odd. What is the probability that it is (a) divisible by 3; (b) divisible by neither 3
nor 5?
Answer.
(a) The reduced sample space is S = {1, 3, 5, 7, 9, ..., 9999}. There are 5000 elements
in S. Since the set {5, 7, 9, 11, 13, 15, ..., 9999} includes exactly 4998/3 = 1666 odd
numbers that are divisible by three, the reduced sample space has 1667 odd numbers
that are divisible by 3. So the answer is 1667/5000 = 0.3334.
(b) Let O be the event that the number selected at random is odd. Let F be the
event that it is divisible by 5 and T be the event that it is divisible by 3. The desired
probability is calculated as follows.
P (F c T c |O) = 1 − P (F ∪ T |O) = 1 − P (F |O) − P (T |O) + P (F T |O)
= 1 − 1000/5000 − 1667/5000 + 333/5000 = 0.5332
(1)
2
3. An urn contains five white and three red chips. Each time we draw a chip, we look at
its color. If it is red, we replace it along with two new red chips, and if it is white, we
replace it along with three new white chips. What is the probability that, in successive
drawing of chips, the colors of the first four chips alternate?
Answer.
3/8 × 5/10 × 5/13 × 8/15 + 5/8 × 3/11 × 8/13 × 5/16 = 0.0712.
2
4. From an ordinary deck of 52 cards, cards are drawn one by one, at random and
without replacement. What is the probability that the fourth heart is drawn on
the tenth draw? Hint: Let F denote the event that in the first nine draws there
1-1
are exactly three hearts, and E be the event that the tenth draw is a heart. Use
P (F E) = P (F )P (E|F ).
Answer.
(13)(39)
P (F )P (E|F ) = 3 52 6 × 10/43 = 0.059
2
(9)
5. A person has six guns. The probability of hitting a target when these guns are properly
aimed and fired is 0.6, 0.5, 0.7, 0.9, 0.7, and 0.8, respectively. What is the probability
of hitting a target if a gun is selected at random, properly aimed, and fired?
Answer.
1
6 (0.6)
2
+ 16 (0.5) + 61 (0.7) + 16 (0.9) + 61 (0.7) + 16 (0.8) = 0.7
6. Suppose that 10 good and three dead batteries are mixed up. Jack tests them one
by one, at random and without replacement. But before testing the fifth battery he
realizes that he does not remember whether the first one tested is good or is dead.
All he remembers is that the last three that were tested were all good. What is the
probability that the first one is also good?
Answer.
Reducing the sample space, we have that the answer is 7/10.
2
7. A stack of cards consists of six red and five blue cards. A second stack of cards consists
of nine red cards. A stack is selected at random and three of its cards are drawn. If
all of them are red, what is the probability that the first stack was selected?
Answer.
Since the choice of the boxes are random, so pick each box by probability 1/2. We
[(6)/(11
]×1/2
3)
have [ 6 /3 11 ]+[1×(1/2)]
= 4/37
2
(3) ( 3 )
8. An urn contains five red and three blue chips. Suppose that four of these chips are
selected at random and transferred to a second urn, which was originally empty. If a
random chip from this second urn is blue, what is the probability that two red and
two blue chips were transferred from the first urn to the second urn?
Answer.
5
3
2 (2)(2)
4 . ( 8)
4
(5)(3)
(5)(3)
(5)(3)
(5)
0. 84 + 14 . 3 8 1 + 24 . 2 8 2 + 43 . 1 8 3
(4)
( 4)
(4)
( 4)
= 0.571
(2)
2
9. A fair die is rolled twice. Let A denote the event that the sum of the outcomes is odd,
and B denote the event that it lands 2 on the first toss. Are A and B independent?
Why or why not?
Answer.
1-2
A={(1,2), (1,4), (1,6), (2,1), (4,1), (6,1), (2,3), (2,5), (3,2), (5,2), (3,4), (3,6), (4,3),
(6,3), (4,5), (5,4), (5,6), (6,5)}
B={(2,1), (2,2), (2,3), (2,4), (2,5), (2,6)}
P (AB) = (3/6)(3/18) = (1/12); so A and B are independent.
2
10. Suppose that an airplane passenger whose itinerary requires a change of airplanes in
Ankara, Turkey, has a 4% chance of losing each piece of his or her luggage independently. Suppose that the probability of losing each piece of luggage in this way is 5%
at Da Vinci airport in Rome, 5% at Kennedy airport in New York, and 4% at OHare
airport in Chicago. Dr. May travels from Bombay to San Francisco with one piece
of luggage in the baggage compartment. He changes airplanes in Ankara, Rome, New
York, and Chicago. (a) What is the probability that his luggage does not reach his
destination with him? (b) If the luggage does not reach his destination with him,
what is the probability that it was lost at Da Vinci airport in Rome?
Answer.
(a) Let E be the event that Dr. Mays suitcase does not reach his destination with
him. We have
P (E) = (0.04) + (0.96)(0.05) + (0.96)(0.95)(0.05) + (0.96)(0.95)(0.95)(0.04) = 0.168
or simply, P (E) = 1(0.96)(0.95)(0.96) = 0.168.
(b) Let D be the event that the suitcase is lost in Da Vinci airport in Rome. Then,
by Bayes formula,
P (D|E) =
P (D)
(0.96)(0.05)
=
= 0.286.
P (E)
0.168
(3)
2
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