HW4- Probability
(solutions)
Due: Thursday April 7, 2011
HW #4 - SOLUTIONS
Topics: Fundamentals of probability, combining probabilities, the law of large numbers,
expected value, counting and probability
I. Probability Fundamentals [15]
1. Probability. What is the probability of randomly meeting someone who is born between
2:00pm and 3:00pm? Assume that a person is equally likely to be born at any time of day.
[5 points]
If someone is equally likely to be born at any time of day, then there are 24 hours in a
day, and only one hour in the event “being born between 2:00pm and 3:00pm”, so the
probability of being born between 2:00pm and 3:00pm is
2. Probability. Emily has a large jar with 30 red gumballs and 50 green gumballs. What
is the probability that she draws out a red ball? What is the probability that she
draws out a green ball? [4 points]
Recall that for equally likely outcomes
So for this event, each ball that could be drawn is one possible outcome.
3. Probability. What is the theoretical probability of not rolling a 6 with a fair die?
[3 points]
Recall that the probability of an event not happening is just 1 minus the probability of
that event happening:
.
4.
Empirical probability. A doctor diagnosed pneumonia in 250 patients, and 225 of them
actually had pneumonia. What is the probability that the next patient diagnosed with
pneumonia will actually have pneumonia? [3 points]
Recall that the empirical probability can be calculated by using the number of observed
occurrences of an event out of the total possible.
IGE104: Logic and Mathematics for Daily Living (3-2010)
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HW4- Probability
(solutions)
Due: Thursday April 7, 2011
II. Combining probabilities [25]
For questions 4 - 6 a) determine whether the events are independent or dependent [2 points]
and b) find the probability of the event [3 points]
5. Probability with And. Rolling two 6’s followed by two 1’s on four tosses of a fair die.
a) Since the die is fair, each die roll is independent of any previous rolls. So the event of
rolling a 6 is independent of the event of rolling a second 6, which is independent of
rolling a 1 and independent of rolling another 1.
b) recall that
Since the
, and
then
6. Probability with And. Drawing three Aces in a row from a standard deck of 52 playing
cards when the drawn cards are not returned to the deck.
a) Since the cards are drawn without replacement, the events are not independent.
Once the first Ace is drawn from the deck, there are only three Aces left, out of a total of
51 cards. Once the second Ace is drawn, there are only two Aces left out of a deck of 50
cards.
b) Recall that for non-independent events
So:
7. Probability with And. Randomly meeting 3 new international students in a row on a
campus where 1 in 12 students is an international student. [3 points]
IGE104: Logic and Mathematics for Daily Living (3-2010)
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HW4- Probability
(solutions)
Due: Thursday April 7, 2011
8. Probability with Or. Find the probability of randomly meeting either a man or an
American in a group composed of 40 English women, 30 English men, 20 American
women, and 10 American men. [5 points]
Gender
Count of people in the
group Men Women Total
American
10
20
30
Nationality
English
30
40
70
Total
40
60
100
From the table above we can see that the total number of men, for example) is 40.
Recall that
so for this example we have:
9. Probability with At Least. If the probability of rain on a single day is 0.2, what is the
probability of getting rain at least once in 6 days? [5 points]
Recall that the probability of an event occurring at least once is just one minus the event
never occurring. So
Since the
, then
. So:
III. The Law of Large Numbers & Expected Value [30 points]
10. Insurance Claims. An insurance policy sells for $800. Based on past data, an average 1
in 50 policy holders will file a $10,000 claim, an average of 1 in 100 policyholders will
file a $20,000 claim, and an average 1 in 250 policyholders will file a $50,000 claim.
What is the expected profit of the company of if it sells 10,000 policies? [6 points]
The expected profit = expected revenue – expected cost.
From the question above we can make the following table of empirical probabilities.
Using the formula for expected value
X = $10,000
X = $20,000
X = $50,000
Probability that a
policyholder will file
claim X
Expected number of
200
100
40
people out of 10,000
Expected Cost
$2,000,000
$2,000,000
$2,000,000
IGE104: Logic and Mathematics for Daily Living (3-2010)
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HW4- Probability
(solutions)
Due: Thursday April 7, 2011
Expected Cost = (expected #people who claim $10,000) x $10,000
+ (expected #people who claim $20,000) x $20,000
+ (expected #people who claim $50,000) x $50,000
= [Pr(claim of $10,000) x 10,000] x $10,000
+ [Pr(claim of $20,000) x 10,000] x $20,000
+ [Pr(claim of $50,000) x 10,000)] x $50,000
=[ 0.02 x 10,000] x $10,000
+ [0.01 x 10,000] x $20,000
+ [0.004 x 10,000] x $50,000
= (200 x $10,000) + (100 x $20,000) + (40 x $50,000)
= $2,000,000 + $2,000,000 + $2,000,000
= $6,000,000
Expected Revenue = amount received per policy x # of policies
= $800 x 10,000 = $8,000,000
Expected Profit = $8,000,000 - $6,000,000 = $2,000,000
11. Gambler’s Fallacy and dice. Suppose you roll a die with the following rules: You win
$1 if the die comes up with an even number and you lose $1 if it comes up odd.
a. Suppose you get 45 even numbers in your first 100 rolls. How much money
have you won or lost? [2 points]
I have won $45 and I have lost $55, so overall I have lost $10.
b. On the second 100 rolls, your luck improves and you roll 47 even numbers.
How much money have you won or lost over 200 rolls? [2 points]
During the second set of 100 rolls, I have won $47 and I have lost $53, so in sum, over
the entire 200 rolls, I have lost $16
c. Your luck continues to improve, and you roll 148 even numbers in your next
300 rolls. How much money have you won or lost over your total of 500 rolls?
[2 points]
Over the next 300 rolls, I have lost $148 and won $152, so overall, the past 500 rolls, I
have lost $30.
d. How many even numbers would you have to roll in the next 100 rolls to break
even (earn a total of 0 from the game)? Is this likely? [4 points]
In order to break even, I will need to gain $30. So in the next 100 rolls, I will need to roll
30 even numbers more than odd numbers, so in the next 100 rolls, I will need to roll 65
even numbers and 35 odd numbers. This is not likely since the theoretical probability of
an even number is 50% on a fair die, so we would expect 50 even rolls from a fair die
e. What were the percentages of even numbers after 100, 200 and 500 rolls?
Explain how these illustrate the Law of Large Numbers. [4 points]
100
200
500
number of even rolls
45
45 + 47 = 92
45 + 47 + 148 = 240
percentage of even
45%
46%
48%
This illustrates the Law of Large Numbers which says that as the number of observations
increases, the empirical probability nears the theoretical probability. Thus as the number
of rolls increases, the percentage nears 50%, which is the theoretical probability.
IGE104: Logic and Mathematics for Daily Living (3-2010)
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HW4- Probability
(solutions)
Due: Thursday April 7, 2011
12. Expected Value. You are playing the game of Monopoly in which you move according
to the sum of a dice roll. What is the expected value of the sum of a dice roll when there
are two dice? [6 points]
There are two approaches to solving this problem. The first is to calculate the expected
value of the sum of the two dice. The second is to find the sum of the expected value of
one die.
1) To find the expected value of the sum of a dice roll, we first need to find the different
values possible for the sum of two dice. Assuming we are rolling two fair dice, each
combination below is equally likely.
First Die
sum of dice
probability
1
2
3
Second Die
4
5
6
1
2
2
3
3
1/6
3
1/6
6
1/6
6
1/6
1/6
1/6
8
1/6
10
10
10
1/6
1/6
1/6
1/6
9
1/6
1/6
9
9
1/6
9
1/6
8
1/6
1/6
8
1/6
1/6
1/6
8
1/6
7
8
1/6
7
7
1/6
7
7
1/6
1/6
7
1/6
6
6
6
6
1/6
1/6
5
1/6
5
5
5
5
1/6
4
4
4
4
1/6
11
1/6
11
1/6
1/6
12
1/6
1/6
To find the probability of rolling a given sum, we simply sum up the different
probabilities for each combination that yields that sum. Alternatively, we could count
the different number of combinations out of 36, the total possible number of
combinations.
Sum of dice
2
3
4
5
6
7
8
9
10
11
12
probability
The expected value is just the sum of the probability of a sum times its value.
2) The second method for calculating this is to see that ExpVal(First Die + Second Die) =
ExpVal(First Die) + ExpVal(SeconDie)
The expected value of a single die can be calculated by finding the probability for each
possible value of a die.
Die roll 1
2
3
4
5
6
probability
Thus the ExpVal(Die)
So the ExpVal(First Die + Second Die) = 3.5 + 3.5 = 7.
IGE104: Logic and Mathematics for Daily Living (3-2010)
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HW4- Probability
(solutions)
Due: Thursday April 7, 2011
13. Expected Wait. Suppose you arrive at a bus stop randomly, so all arrival times are
equally likely. The bus arrives regularly every 30 minutes without delay (say on the hour
and on the half hour). What is the expected value of your waiting time? Explain. [4
points]
If the bus arrives regularly every 30 minutes then when you arrive at the bus stop
you can wait anywhere between 0 and 29 minutes. The table below gives the
number of minutes you will have to wait if you arrive at a given time during the hour
X (the hour doesn’t matter in determining how long you have to wait). Since all
arrival times are equally likes, the probability that you will need to wait a certain
number of minutes, say 5 is just 2 (the number of arrival times that result in waiting
5 minutes) divided by 60 (the number of possible arrival times), or 1/30.
Time of Arrival
#
minutes
Probability
to wait Probability
x Value
X:30
X:00
0
1/30
0
X:29
X:59
1
1/30
1/30
X:28
X:58
2
1/30
2/30
X:27
X:57
3
1/30
3/30
X:26
X:56
4
1/30
4/30
X:25
X:55
5
1/30
5/30
X:24
X:54
6
1/30
6/30
X:23
X:53
7
1/30
7/30
X:22
X:52
8
1/30
8/30
X:21
X:51
9
1/30
9/30
X:20
X:50
10
1/30
10/30
X:19
X:49
11
1/30
11/30
X:18
X:48
12
1/30
12/30
X:17
X:47
13
1/30
13/30
X:16
X:46
14
1/30
14/30
X:15
X:45
15
1/30
15/30
X:14
X:44
16
1/30
16/30
X:13
X:43
17
1/30
17/30
X:12
X:42
18
1/30
18/30
X:11
X:41
19
1/30
19/30
X:10
X:40
20
1/30
20/30
X:9
X:39
21
1/30
21/30
X:8
X:38
22
1/30
22/30
X:7
X:37
23
1/30
23/30
X:6
X:36
24
1/30
24/30
X:5
X:35
25
1/30
25/30
X:4
X:34
26
1/30
26/30
X:3
X:33
27
1/30
27/30
X:2
X:32
28
1/30
28/30
X:1
X:31
29
1/30
29/30
Expected Value(#minutes to wait)
14.5
So the expected value of the number of minutes to wait is 14.5
IGE104: Logic and Mathematics for Daily Living (3-2010)
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HW4- Probability
(solutions)
Due: Thursday April 7, 2011
IV. Counting and Probabilities [20]
For questions 13 – 15 count the number of possibilities using the appropriate counting
technique: arrangements with repetition, permutations, or combinations. [3 points each]
14. Counting. How many license plates can be made of the form XXX-YYYY where X is a
letter of the English alphabet and Y is a numeral 0-9?
There are 26 possible values for X and 10 possible values for Y so using the
multiplication rule, the total number of license plates possible (if order matters, and we
can repeat the same letter/digit in the same question) is:
15. Counting. How many different three- letter “words” can be formed from the genetic
alphabet ACGT?
There are 26 possible values for the three different spaces in a given three-letter
“word” . Note that in this example, the order of the letters matters, and there can be
repeating letters. Thus we will use the multiplication rule.
16. Counting. The debate team has 15 members but only 4 can compete. How many
different teams of 4 can be formed?
In this example we are choosing a group of 4 out of a team of 15. Since we are dealing
with people here, we cannot select the same person twice to be in the group that is
competing (this means no repetition). All we are interested is in finding different teams
of 4, regardless of the order that we select them in. So selecting a competition team of
Alice, Joe, Bob, and Kate is the same as selecting Joe, Bob, Alice, and Kate. Thus we will
use our formula for combinations, which says that we have
possible groups of 4 from a team of 15.
17. Counting. Luigi’s Pizza Parlor advertises 56 different three-topping pizzas. How many
individual toppings does Luigi actually use? Ramona’s Pizzeria advertises 36 different
two-topping pizzas. How many toppings does Ramona actually use? (Hint: in this
problem you are given the total number of combinations, and you must find the number
of toppings that are used) [4 points]
If Luigi’s has 56 different 3 topping pizzas, then
where
IGE104: Logic and Mathematics for Daily Living (3-2010)
, so we have to solve for n
Page 7 of 9
HW4- Probability
(solutions)
Due: Thursday April 7, 2011
So, what we are solving for is
If we look at the number 336, we can see that
, so it must be true that
. At Luigi’s there are 8 different pizza toppings.
Repeating this same procedure for Ramona Pizzeria, we get that
, so
So, what we are solving for is
If we look at the number 72, we can see that
must be true that
. At Ramona’s there are 9 different pizza toppings.
, so it
18. Probability. What is the probability of choosing five numbers that match five randomly
selected balls when the balls are numbered 1 through 40? [2 points]
Recall that
To find the Pr(Choosing the correct 5 numbers), we can think of a single group of five
numbers as one combination (the order doesn’t matter). So the number of outcomes in
our event is just one. Then we look at the total possible number of outcomes and we
are effectively choosing a group of five (numbers on the selected balls) out of a possible
40 (possible numbers on all the balls). So using our combinations formula we have that
19. Probability. What is the probability of guessing the top three winners (in order) from a
group of eight finalists in a soccer tournament? [ 2 points]
To find the Pr(Choosing the top three winners), we can think of a single group of three
winners as 3! outcomes, since the order matters. So the number of outcomes in our
event is 6. Then we look at the total possible number of outcomes and we are
effectively choosing a group of three top winners out of a possible 8 finalists. So using
our permutations formula we have that
IGE104: Logic and Mathematics for Daily Living (3-2010)
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HW4- Probability
(solutions)
Due: Thursday April 7, 2011
20. Probability. What is the probability of at least two people sharing the same birthday in a
group of 12 people (assume 365 days in a year)? [3 points]
Recall that Pr(at least 1 in E) = 1 – Pr(none in E)
So, what we are trying to find is the probability that no two people share the same birthday
in a group of 12 people. Thus the Pr(no 2 share the same birthday) = Pr(12 different
birthdays) =
BONUS [10]
Suppose you are on a game show with three numbered closed doors. Behind one of the
doors there is a car, and behind the other two doors there is a goat. You do not know what
is behind each door. Whichever door you pick you get to keep what is behind it.
You pick door #1 and the host of the show, Vanna White, opens door 2 to show you that
it has a goat behind it. She asks you if you want to switch your choice to door 3. Should
you switch your door choice?
IGE104: Logic and Mathematics for Daily Living (3-2010)
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