Probability: Basic concepts/Self-check assessment

Use the following quiz questions to check your understanding of the basics of probability. Note that as soon as you have indicated your response, the question is scored and feedback is provided. As feedback is provided for each option, you may find it useful to try all of the responses (both correct and incorrect) to read the feedback, as a way to better understand the concept.

Probability statements

Enter one of the following probabilities to complete each of the following statements: 0, .1, .5, .6, .99, 1

• A is an event which in fact cannot occur: P(A) = 0.
• A is an event which occurs a bit more often than it does not occur: P(A) = .6.
• A is an event which is equally likely to occur or not to occur: P(A) = .5.
• A is an event which will occur for certain: P(A) = 1.
• A is an event which occurs somewhat rarely, but will show up once in a while in a long series of trials: P(A) = .1.
• To assess dental hygiene habits in youth who wear braces, 387 12-17 year olds wearing braces were asked whether they had brushed their teeth that morning. 132 of the youth reported that they had not brushed their teeth. If event A is defined as a youth who wears braces does not brush his/her teeth in the morning, what is the empirical estimate of P(A)? .34
  • P(A) is the probability that a randomly chosen youth who wears braces has brushed his/her teeth in the morning. Use the observed data to make an estimate of this probability: P(A) = (number of times A occurred) / (total number of repetitions/trials) = 132/387 = .34

Understanding sample space and events

• The concept of a sample space is only relevant for experiments with numerical outcomes.
  • True
    • That's not quite right. The sample space lists the particular outcomes of a random phenomenon, whether they are numerical or not, e.g., the outcomes of flipping a coin are heads or tails, S = {heads, tails}
  • False
    • That's correct. The sample space lists the particular outcomes of a random phenomenon, whether they are numerical or not.
• A coin is tossed 3 times, and the number of heads is counted. What is the sample space?
  • S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}
    • That's not quite right. Note that the random phenomenon counts the number of heads in three tosses of a coin. Try again.
  • S = {H, TH, TTH, TTT}
    • That's not quite right. Note that the random phenomenon counts the number of heads in three tosses of a coin. Try again.
  • S = {1, 2, 3)
    • That's not quite right. The random phenomenon counts the number of heads in three tosses of a coin. Note that one possible outcome is no heads (TTT). Try again.
  • S = {0, 1, 2, 3}
    • That's correct. The random phenomenon counts the number of heads in three tosses of a coin.
• For a student chosen at random from the student body of a high school, record the student's gender (M or F) and whether the student's ears are pierced or not (P or N). What is the sample space?
  • S = {M, F, P, N)
    • That's not quite right. Note that for each student both the gender and whether the ears are pierced or not is recorded, so each possible outcome in the sample space is a unique combination of the two attributes. Try again.
  • S = {MP, MN, FP, FN}
    • That's correct. The possible outcomes are the set of unique combinations of the two attributes (gender and ears pierced or not).
  • S = {MF, PN}
    • That's not quite right. Note that for each student both the gender and whether the ears are pierced or not is recorded, so each possible outcome in the sample space is a unique combination of the two attributes. Try again.
  • A sample space is not relevant because the two attributes are not independent events.
    • That's not quite right. Although it is very likely that these two random phenomena (whether the student is male or female and whether their ears are pierced or not) are not independent (females are more likely to have pierced ears), independence is not relevant to the specification of the sample space. Try again.
• At an airport, a group of travelers are seated together. There are 2 people from Canada (CA), 7 from France (FR), 3 from Japan (JA), and 3 from the USA (US). Two travelers are chosen at random. Event C occurs when the first traveler chosen is from Canada. Which of the following correctly defines event C?
  • C = {US, FR, JA}
    • That's not quite right. These are three of the individual outcomes if one traveler were chosen. Note that two travelers are chosen. Try again.
  • C = {US, CA, FR, JA}
    • That's not quite right. These are the four individual outcomes if one traveler were chosen. Note that two travelers are chosen. Try again.
  • C = {(CA, FR), (CA, JA), (CA, US)}
    • That's not quite right. All of these outcomes are included in event C, but the listing is not complete. Try again.
  • C = {(CA, CA), (CA, FR), (CA, JA), (CA, US)}
    • That's correct. The traveler from Canada who is selected first could be paired wih travelers from France, Japan, USA, or another traveler from Canada.

Calculating probabilities

Four students are traveling together: one male and one female from Paris (MParis and FParis) and one male and one female from London (MLond and FLond). Among the four seat assignments on the airplane, there is one window seat, one aisle seat and two middle seats. The travelers agree to randomly choose 2 from the group for the preferred seating. The first chosen will get the aisle seat, and the second chosen will get the window seat. The two remaining will work out who sits in which middle seat.

The sample space needs to reflect that the order in which the two are chosen matters, the first chosen gets the aisle seat and the second chosen gets the window seat. Each pair is listed once, and then in reverse order.

S = {(MParis, FParis), (FParis, MParis), (MParis, MLond), (MLond, MParis), (MParis, FLond), (Flond, MParis), (FParis, MLond), (MLond, FParis), (FParis, FLond), (FLond, FParis), (MLond, FLond) (FLond, MLond)}

• Each of the outcomes is equally likely. The probability that the two chosen students are from the same city is .33. (round to two decimal places)

Calculating probabilities

• Advanced Placement test scores are reported on a 5 point scale. The table below lists probabilities for each score across all tests, for the 2011 testing year.. Given the information in the table, the probability that a randomly chosen 2011 test-taker received a score of 5 is .12.

Distribution of AP test scores

Test Score	5	4	3	2	1
Probability	?	.21	.25	.18	.24

Calculating probabilities

In the 2012 Frankenstorm which hit the northeast region of the US, residences in Middlesex County, New Jersey experienced a number of problems:

--72% of households experienced a power outage (label this event E)

--23% of households experienced internal flooding (label this event F)

--81% of households experienced at least one of the two problems: power outage, flooding, or both (label this event A)

• What is the probability that a Middlesex Co household chosen at random did not experience a power outage?
  • .72
    • That's not quite right. This is the probability that a Middlesex Co. household experienced a power outage, P(E) = .72. The question asks for P(not E). Try again.
  • .28
    • That's correct. The question asks for P(not E). Using the complement rule, P(not E) = 1 - P(E) = 1 - .72 = .28.
  • .23
    • That's not quite right. This is the probability that a Middlesex Co. household experienced a flooding, P(F) = .23. The question asks for P(not E). Try again.
  • .77
    • That's not quite right. This is the probability that a Middlesex Co. household did not experience flooding, P(not F) = .77. The question asks for P(not E). Try again.
• The complement of event A, "not A", is the event that
  • a household experienced both problems.
    • That's not quite right. This response includes some of the outcomes in event A: a household experienced at least one of the two problems, meaning it experienced a power outage, flooding or both a power outage and flooding. The question asks for the complement of A; what is not in A. What else could have occurred that is not part of event A? Try again.
  • a household experienced only one of the two problems.
    • That's not quite right. This response includes some of the outcomes in event A: a household experienced at least one of the two problems, meaning it experienced a power outage, flooding or both a power outage and flooding. The question asks for the complement of A; what is not in A. What else could have occurred that is not part of event A? Try again.
  • a household experienced neither of the two problems.
    • That's correct. The event A (a household experienced at least one of the two problems) can occur in one of three ways: a household experienced a power outage (E), a household experienced flooding (F), or a household experienced both a power outage (E) and flooding (F). The only way that A can not occur (event not A) is if a household experienced neither a power outage (E) or flooding (F).
  • a household experienced at least one of the two problems.
    • That's not quite right. This response simply restates event A: a household experienced at least one of the two problems, meaning it experienced a power outage, flooding or both a power outage and flooding. The question asks for the complement of A; what is not in A. What else could have occurred that is part of event A? Try again.
• Referring to the previous series of problems, the probability of "Not A" is
  • .19
    • That's correct. P(not A) = 1 - P(A) = 1 - .81 = .19
  • 1.9
    • This answer is obviously wrong because a probability must be a number between 0 and 1 (inclusive). The question asks for P(not A). As P(A) is provided in the description above, use the complement rule to find P(not A). Try again.
  • .28
    • That's not quite right. This response is the P(not E). The question asks for the P(not A). As P(A) is provided in the description above, use the complement rule to find P(not A). Try again.
  • .05
    • That's not quite right. The question asks for the P(not A). As P(A) is provided in the description above, use the complement rule to find P(not A). Try again.

Calculating probabilities

How old are college students? The probabilities in the table below were calculated from NCES 2010 counts of fall enrollment in degree-granting institutions in the US, by age..

• What is the probability that a randomly selected college student is 30 years old or older?
  • .087
    • That's not quite right. .087 is the probability that a randomly selected student is 30 to 34 years old. The event "is 30 years old or older" is composed of two events in the table. Try again.
  • .274
    • That's correct. P("is 30 years old or older") = P("30 to 34 years old" or "35 years old or older") = P("30 to 34 years old") + P("35 years old or older") = .087 + .187 = .274. We can add these probabilities because the events "30 to 34 years old" and "35 years old or older" are disjoint. These events are disjoint because each individual in the table is included in only one age category.
  • Can't tell because the events "30 to 34 years old" and "35 years old or older" are not disjoint.
    • That's not quite right. The events "30 to 34 years old" and "35 years old or older" are disjoint because an individual is included in only one age category.

Age of college student	Probability
14 - 17 years old	.010
18 and 19 years old	.196
20 and 21 years old	.193
22 to 24 years old	.175
25 to 29 years old	.152
30 to 34 years old	.087
35 years old and older	.187

Understanding disjoint events

A teacher needs 2-3 volunteers for a classroom job. He decides to randomly select children from among the children in the class until a boy and a girl are chosen, or until there are 3 children chosen. The sample space of possible outcomes is S = {GB, BG, GGB, BBG, GGG, BBB}.

--Event A: the children chosen are all the same gender.

--Event C: three children are chosen.

--Event G: only one girl is chosen.

Indicate whether each of the following events is disjoint or not disjoint.

• Events A and C: not disjoint
• Events A and G: disjoint
• Events C and G: not disjoint

Understanding disjoint and independent events

• If A and B are disjoint, then P(A and B) is
  • 0
    • That's correct. If events A and B are disjoint then it's impossible for them both to occur "jointly", as would be needed for a greater than 0 probability for P(A and B)
  • 1
    • That's not quite right. The P(A and B) = 1 indicates that A and B always happen at the same time. But if A and B are disjoint, then in fact they never happen together. Try again.
  • P(A) + P(B)
    • That's not quite right. Note that the question asks for P(A and B) not P(A or B). When A and B are disjoint, P(A or B) = P(A) + P(B).

Understanding disjoint and independent events

About 18% of public schools in the US require students to wear school uniforms. Suppose a US school is randomly selected from among all possible US schools. Two events are defined:

--A: the selected school requires students to wear school uniforms.

--B: the selected school does not require students to wear school uniforms.

• Are events A and B disjoint or not disjoint? disjoint
• Are events A and B independent or dependent? dependent

Now suppose two US schools are randomly selected from among all possible US schools. Again, two events are defined:

--C: the first selected school requires students to wear school uniforms.

--D: the second selected school does not require students to wear school uniforms.

• Are events C and D disjoint or not disjoint? not disjoint
• Are events C and D independent or dependent independent

Calculating probabilities

All human blood can be typed as O, A, B or AB. The distribution of the types varies a bit among groups of people. The table below displays the probabilities of human blood types in the United States. Two people are selected simultaneously and at random from all people in the United States.

• What is the probability that both have blood type O? .19 (rounded to two decimals)
  • Let event F = "the first person has blood type O". Let event G = "the second person has blood type O". The probability requested is P(F and G). Since the two people are chosen at random from a large population, F and G are independent and we can use the multiplication rule for independent events. P(F and G) = P(F)*P(G) = .44*.44 = .1936 (rounded to .19)
• What is the probability that both have the same blood type? .38 (rounded to two decimals)
  • The probability requested is interpreted as P(both O or both A or both B or both AB). Using the multiplication rule for independent events, we calculate the probability that both people have type O, both people have type A, both people have type B and both people have type AB. Because each of these 4 events is disjoint, we can use the addition rule for disjoint events to sum them to find P(both have same type). P(both O or both A or both B or both AB) = P(type O)*P(type O) + P(type A)*P(type A) + P(type B)*P(type B) + P(type AB)*P(type AB) = [(.44)*(.44)] + [(.42)*(.42)] + [(.10)*(.10)] + [(.04)*(.04)] = .3816 (rounded to .38)

Blood Type	US probability
O	.44
A	.42
B	.10
AB	.04

Calculating probabilities

About 18 percent of public schools in the US require students to wear school uniforms. A much higher percentage of private schools require students to wear school uniforms. About 24% of US schools are private schools. What is the probability that a randomly selected US school requires students to wear uniforms and is a private school? To help us work this out, consider

Event A: a school is a private school

Event B: a school requires students to wear uniforms.

• Which of the following statements is correct?
  • We can use the multiplication rule for independent events to calculate the probability: P(A and B) = P(A)*P(B) = (.24)*(.18) = .04.
    • That's not quite right. In this scenario, the two events (A: a school is a private school and B: a school requires students to wear uniforms) are dependent. Private school are more likely to require students to wear uniforms; knowing whether or not the school is private affects the probability that the school requires school uniforms. It is not appropriate to use the multiplication rule for independent events in this situation.
  • We do not have enough information to answer the question. We cannot use the multiplication rule for independent events because the events are dependent. Knowing whether or not the school is private affects the probability that the school requires school uniforms.
    • That's correct. A and B are dependent events because if A is true (the selected school is private) then the probability of B is affected (it is more likely that the school requires students to wear uniforms). We cannot use the multiplication rule for independent events in this situation.