| 1 | a. This statement is an example
of a personal probability because it is based on an individual's belief
of his/her chances of getting the flu. b. The relative frequency interpretation applies here because this probability is based on repeated observations of the frequency of flu cases each winter. |
| 2 | a. Assuming that the traits
of having red hair and having blond hair are mutually exclusive, Rule 2
can be applied. So the probability of having red hair should be 23% - 14%
= 9%. b. The probability that a child is living with one parent is the probability that a child is living with just his or her mother or just his or her father. Because these two possibilities are mutually exclusive, the probabilities can be added using Rule 2. The answer is 0.247 .c. The probability of a birth resulting in more than one child is 0.0231 + 0.0008 = 0.0239. So, the probability that a birth will result in only one child is 1 - 0.0239 = 0.9761. |
| 3 |
a. This contradicts Rule 1. There are only two possibilities:
a person either is wearing a seat belt or is not. So the two probabilities
should sum to 1 |
| 4 | a. The probability of not
being born on Friday the 13th is 1 - 1/214 = 213/214, by Rule 1. b. No. For instance, the probability of being born on Friday the 13th if you are born in a year with two is roughly 2/365, whereas if there are three in the year it would be 3/365. c. "The probability of being born on Friday the 13th is 1/214" means that over the long run (many years) 1 in 214 people will be born on a Friday the 13th. |
| 5 | Explanation 1 is closer to what
it means. The number of heads may be far from half for a small number of
tosses, so Explanation 2 is incorrect. |
| 6 | Not all events are repeatable. An
example is the probability that the Earth will be uninhabitable by the year
2010. |
| 7 | a. Because 1/2 the cards
are red and 1/2 black, in the long run 1/2 the cards you pick will be black
and 1/2 will be red. So whether you guess black or red you have a probability
of 1/2 of being right by chance. b. It is a relative-frequency probability because it is based on knowledge of the physical situation, which can be repeated numerous times. c. This would be a personal probability because it is based not on physical assumptions nor repeated observations, but on a person's belief. d. In this case the value of 0.60 would be a relative frequency probability because it was arrived at by repeated observations of her "guessing" ability. |
| 8 | a. You could either go through
the whole phone book and count the portion of people listed who had your
first name or you could repeatedly randomly select an entry and note the
portion of the time that the first name matched yours. b. By either method the resulting probability would be a relative-frequency probability. The resulting probability would be based either on physical assumptions (how many listings you know to have your first name) or on repeated observations. |
| 9 | a. By Rule 1, it must be
1 - 0.9 = 0.1, or 10%. b. By Rule 3, the probability that customer A and customer B will not pay their bills in one week is (0.1)(0.1)= 0.01, or 1%. The assumption has been made that customer A not paying in one week does not influence whether or not customer B pays in a week. Yes, it seems reasonable to assume that there will, in general, be no connection between these two events. |
| 10 |
No. To say the probability is 5/10, you are assuming that the events are mutually exclusive, but you could easily get an interesting piece of mail on two of these days or all of them. (You would need to subtract the probability that you received an interesting piece of mail on two or more of these days.)
|
| 11 | We would have to assume
independence, so that if a person belongs to one of these organizations
it would not make them more or less likely to belong to the other. This
assumption is probably not valid because older people are more likely to
belong to AAA, and only those over 50 are allowed to belong to AARP. |
| 12 | Alternative A must have a higher
probability than B because bank tellers who are active in the feminist movement
are a subset of all bank tellers. |
| 13 | This is a relative-frequency probability.
It means that over the long run the portion of times luggage will be temporarily
lost is 1/176. |
| 14 |
a. This probability was most likely determined by observing the
relative frequency. Flights from New York to San Francisco were observed
and their punctuality over a long period of time was recorded. |
| 15 | a. Whatever the birth month
of the first person, the probability that the second person was born in
that same month is 1/12. The probability of a different month is therefore
1 - 1/12 = 11/12. You could also notice that for the second person, 11 out
of the 12 months would not match the first person's birth month, so the
probability of a different month is 11/12. b. Two months have been taken by the first two people, so there are ten left. If the third person was born in any of those ten months, their birth month would be different. Therefore, the probability is 10/12. c. If parts a and b both happen, then all three were born in different months. The probability of this happening can be found by multiplying the probabilities: (11/12)(10/12) = 110/144 or 0.76. d. If part a and part b did not both happen this would mean that at least two people have the same birth month. In part c we found the probability of the complement of this to be 110/144. Therefore, we know that the probability that parts a and b do not both happen is 1 - 110/144 = 34/144 = 0.236. Hence, the probability that at least two of the people were born in the same month is 34/144 or 0.236. |
| 16 | Here is an example provided by a
student: I think that the probability that I will finish my math homework
by the due date is 70%. I also believe that I will receive an A with a probability
of 60%. I will need to turn in the assignment on time in order to get an
A, so getting an A is a subset of turning in the assignment. This prediction
is coherent because 60% is smaller than 70%. |
| 17 | a. These are not likely
to be independent because spouses' political views are very likely to influence
each other, and spouses are likely to belong to the same political party.
b. The temperature plays a role in whether or not it snows so these can not be independent. c. These will be independent because the numbers are chosen randomly each week and knowledge of one week's numbers will not give you any knowledge of the second week's numbers. d. These events are probably independent because the Dow Jones Industrial Average is not likely to be affected by earthquakes and earthquakes are clearly not affected by the Dow Jones Industrial Average. You might argue that they are not independent because a major earthquake in a city like New York, Los Angeles or San Francisco would probably play havoc with the stock market. This is an acceptable answer if you provide an explanation. |
| 18 | a. The probability that you
do not find money is 1 - 0.1 = 0.9. b. The probability of finding money for the first time on the third try would be (0.9)(0.9)(0.1) = 0.081. c. This is the accumulated probability at the third try, which is the sum of the individual probabilities for tries 1, 2 and 3. It is thus (0.1) + (0.9)(0.1) + (0.9)(0.9)(0.1) = 0.1 + 0.09 + 0.081 = 0.271. |
| 19 | a. 99/100 b. (99/100)(1/100) = .0099. |
| 20 | a. The probability of making
it 70 years without being struck by lightning would be (684,999/685,0000)70.
b. If the average lifetime is 70 years, then Krantz's probability should be 1-(684,999/685,000)70, or one minus the probability in part a. c. These probabilities do not apply to any particular person. Your chances depend on things like where you live and what you do during storms. d. About (260,000,000) (1/685,000), or approximately 380 people. |
| 21 | a. The expected commute time
is (20 min)(0.2) + (15 min)(0.8) = 16 minutes. b. No, the commute times are always either 15 or 20 minutes. This is just an average over the long run. |
| 22 | a. You lose $3 if the births are
all girls, so the probability that you lose $3 is (0.49)(0.49)(0.49) = 0.117649.
Your friend loses $3 if they are all boys, so the probability is (0.51)(0.51)(0.51)
= 0.132651. b. Expected value = (+$1)(0.51) + (-$1)(0.49) = 0.51 - 0.49 = 0.02 or 2 cents for each birth. c. Making 2 cents on each birth, after 1,000 births you make about ($0.02)(1,000) = $20. |
| 23 | It costs $1 to play, so the possible
outcomes are $19, $1 and -$1. The expected value is ($19)(0.012) + ($1)(0.137)
+ (-$1)(0.851) = -$0.486. So your expected loss would be about 49 cents
for each dollar played. |
| 24 | Your expected grade point average
would be (4.0)(0.3) + (3.0)(0.7) = 3.3. You would not necessarily get this
in any one quarter; expected value gives you only a long-term average. |
| 25 | The possible number of parents is
2 with probability 0.72, 1 with combined probability for the mother and
father of 0.25, and 0 with probability 0.03. The expected value for the
number of parents is (2)(0.72) + (1)(0.25) = 1.69. This is not particularly
meaningful for this example because we cannot interpret 1.69 parents. |
| 26 | The best examples are for discrete
data in which the expected value is also the mode. For instance, suppose
you are waiting tables in a restaurant and want to find your expected value
for tips at any given table. If you know that you will receive either a
$2, $3 or $4 tip and you assess your probability of getting each as 1/3,
then the expected value will be a $3 tip, which is also a reasonable possible value. |