stats exam 2 finding probability of events
when finding "at least one of" which rules should you use
multiplication and complement rule
Can events be both disjoint and independent?
no
when two events are independent of each other, which rule do we use
rule 5 P(Aand B)=0
is P(A and B) larger or smaller than either one of the individual properties
smaller since it is a more specific (restrictive) event
T or F In probability and in its applications, we are frequently interested in finding out the probability that a certain event will not occur
t
t or f In probability, "OR" means either one or the other or both.
t
When the events are independent, which two rules give the same results
the Complement Rule strategy and the General Addition Rule
A 2011 poll by the Pew Research Center for People and the Press estimated that 62% of U.S. adults favor the death penalty for persons convicted of murder, 31% oppose it, with the remaining 7% undecided. What is the probability that two randomly chosen U.S. adults support the death penalty for persons convicted of murder?
.384 .62*.62
In the previous section we learned about disjoint events, which are events that can never happen together. This means that if A and B are disjoint, then P(A and B) must be: a. 1 b. 0 c. p(A) + p(B)
0 If events A and B are disjoint, they can never happen together. In other words the event "A and B" can never occur, and thus P(A and B)=0
Suppose that Jim is applying to two colleges: College A, an "Ivy League" school, and College B, a state university. Based on his credentials and the requirements of the two colleges, Jim estimates his chances with the following probabilities: i. Probability that he will be admitted to college A is 0.10. ii. Probability that he will be admitted to college B is 0.75. iii. Probability that he will be admitted to both colleges is 0.05. What is the probability that Jim will be admitted to at least one of the two colleges?
0.80 P(A or B) = 0.10 + 0.75 − 0.05 = 0.80
A six-sided cube is rolled. What is the probability that the number is odd or less than 4? Event A: Numbers on a six-sided cube are odd: 1, 3, 5 Event B: Numbers on a six sided cube are less than 4: 1, 2, 3 a. 1/2 b. 2/3 c. 5/6
2/3 P(A) = 3/6 P(B) = 3/6 P(A and B) = 2/6 bc the numbers 1 and 3 are repeated P(A or B) = P(A) + P(B) - P(A and B)
When A and B are not disjoint (general addition rule) P(A or B) means
P(A occurs or B occurs or both events occur).
One of the following choices is smaller than the other two. Which is it? P(B) P(B or M) P(B and M)
P(B and M)
One of the following choices is larger than the other two. Which is it? P(B) P(B or M) P(B and M)
P(B or M)
2nd basic probability rule
P(S) = 1; that is, the sum of the probabilities of all possible outcomes is 1.
3rd basic probability rule
P(not A) = 1 - P(A); that is, the probability that an event does not occur is 1 minus the probability that it does occur (complement rule)
For safety reasons, four different alarm systems were installed in the vault containing the safety deposit boxes at a Beverly Hills bank. Each of the four systems detects theft with a probability of 0.99independently of the others. What is the probability that when a theft occurs, all four systems will detect it? a. (0.99)^4 b. (0.99) * 4 c. (0.01)4 d. 4 * (0.01) * (0.99)3 e. 4 * (0.99) * (0.01)3
a. (0.99)^4
In each of the following situations, choose the correct sample space (S) for the random experiment that is described. (a) A pair of dice is rolled, and the sum of the dots on the two faces that come up is recorded: a. S={2,3,4,5,6,7,8,9,10,11,12} b. S={1,2,3,4,5,6,7,8,9,10,11,12} c. S={2,4,6,8,10,12}
a. S={2,3,4,5,6,7,8,9,10,11,12} When rolling two dice, the possible outcomes are: (1,1) (1,2) (1,3) (1,4) (1,5) (1,6) (2,1) (2,2) ... (6,6). Thus, the sums can be 2-12.
Recall the estimate by the Pew Research Center that 62% of U.S. adults favor the death penalty for murder. The same report gave a much lower estimate for the percentage of U.S. college graduates supporting the death penalty in cases of murder. According to census data from 2000, roughly 28% of U.S. adults have a college degree. What is the probability that a randomly selected U.S. adult has a college degree and favors the death penalty? Let A be the event that a U.S. adult has a college degree. Let B be the event that this person supports the death penalty. We want to find P(A and B). Which answer is correct, Answer 1 or 2? a. Answer 1: P(A and B) = P(A) * P(B) = 0.62(0.28) = 0.1736 We can use the Multiplication Rule for Independent Events because the events are independent. Having a college degree does not determine a person's views on the death penalty. b. Answer 2: We do not have enough information to answer the question. We cannot use the Multiplication Rule for Independent Events because these events are dependent. Having a college degree affects the likelihood that a person supports the death penalty.
b. Answer 2: We do not have enough information to answer the question. We cannot use the Multiplication Rule for Independent Events because these events are dependent. Having a college degree affects the likelihood that a person supports the death penalty. -A and B are dependent events because if A occurs (we have selected a person with a college degree), then the probability of B is affected (a smaller chance that the person supports the death penalty.) So we cannot use the Multiplication Rule for Independent Events in this situation.
According to the information that comes with a certain prescription drug, when taking this drug, there is a 20% chance of experiencing nausea (N) and a 50% chance of experiencing decreased sexual drive (D). The information also states that there is a 15% chance of experiencing both side effects. What formula will give you the probability of experiencing neither of these side effects? a. 1 − P("not N" and "not D") b. P("not N" and "not D") c. 1 − P(N and D) d. P(N or D) e. P(N and D)
b. P("not N" and "not D")
A flight has been overbooked; however, there are 2 seats available—one in business class and one in first class. The ground crew decides to upgrade 2 of the coach (regular class) passengers so that 2 more passengers will be able to get on the flight. The crew has identified 4 passengers, 2 males and 2 females, who are traveling by themselves and who have been loyal frequent fliers on the airline. They decide to choose 2 of those passengers at random for the upgrade. The first chosen will be upgraded to first class, and the second chosen will be upgraded to business class. We'll denote the 2 males and 2 females (as before) with M1, M2, F1 and F2. a. S = { (M1,M2) (M1,F1) (M1,F2) (M2,F1) (M2,F2) (F1,F2) } b. S = { (M1,M2) (M2,M1) (M1,F1) (F1,M1) (M1,F2) (F2,M1) (M2,F1) (F1,M2) (M2,F2) (F2,M2) (F1,F2) (F2,F1) }
b. S = { (M1,M2) (M2,M1) (M1,F1) (F1,M1) (M1,F2) (F2,M1) (M2,F1) (F1,M2) (M2,F2) (F2,M2) (F1,F2) (F2,F1) } This sample space reflects the fact that the order in which the two are chosen matters. For example, the outcome (M1,M2) indicates that M1 was chosen to be upgraded to first class, and M2 was chosen to be upgraded to business class, while (M2,M1) indicates the reverse assignment.
A person in a casino decides to play blackjack until he wins a game, but he will not play more than 3 games. Let W denote a win and L denote a loss. What is the sample space for this random experiment? a. S = {WWW, WWL, WLW, , WLL, LWW, LWL, LLW, LLL} b. S = {W, LW, LLW, LLL} c. S = {W, WL, WWL, WWW}
b. S = {W, LW, LLW, LLL} The person stops playing when he wins or if he plays three games without winning.
According to a study published by the National Institute of Health, during intercourse condom breakage rates fall between 0.4% and 2.3%. Suppose that the breakage rate of Brand X is 2%. A Brand X Pleasure Pack contains 12 condoms. What is the probability that at least one of the condoms in the pack will break during intercourse? a. 0.02 b. 0.016 c. 0.215
c. 0.215 P(B) = .02 P(no B) = .98 1-P(no B)^12 (bc 12 condoms) 1 - (0.98)^12
According to the information that comes with a certain prescription drug, when taking this drug, there is a 20% chance of experiencing nausea (N) and a 50% chance of experiencing decreased sexual drive (D). The information also states that there is a 15% chance of experiencing both side effects. What is the probability of experiencing nausea or a decrease in sexual drive? a. 0.10 b. 0.40 c. 0.55 d. 0.70 e. 0.85
c. 0.55 build a table and add P(D not N) + P(N not D) + P(N and D)
A person in a casino decides to play 3 games of blackjack. Let W denote a win and L denote a loss. Define the event A as "the person wins at least one game of blackjack." What are the possible outcomes for this event? a. {WWL, LWL, LLW} b. {WWW, WWL, WLW, WLL, LWW, LWL, LLW, LLL} c. {WWW, WWL, WLW, WLL, LWW, LWL, LLW} d. {W, LW, LLW}
c. {WWW, WWL, WLW, WLL, LWW, LWL, LLW} Event A is all outcomes in the sample space except LLL.
The CSU system reported the following probabilities for their student body in 2013. The probability of their student being a male between 17 and 19 is P(A) = 0.11. The probability of their student being a female between 20 and 24, P(B) = 0.30. What is P(A and B) given that year? a. 0.033 b. 0.410 c. 0.377 d. 0
d. 0 If two events are disjoint then they cannot happen together so P(A and B) = 0.
For safety reasons, four different alarm systems were installed in the vault containing the safety deposit boxes at a Beverly Hills bank. Each of the four systems detects theft with a probability of 0.99 independently of the others. The bank, obviously, is interested in the probability that when a theft occurs, at least one of the four systems will detect it. What is the probability that when a theft occurs, at least one of the four systems will detect it? a. (0.99)^4 b. (0.01)^4 c. 1 − (0.99)^4 d. 1 − (0.01)^4
d. 1 − (0.01)^4 We want to find the probability that "at least one system detects the theft," which is the complement of "none detects." P(at least one detects) = 1 − P(none detects) = 1 − (0.01)4
Consider the following two events: A—a randomly chosen person has blood type A B—a randomly chosen person is a woman. are A and B disjoint
no, A and B can both occur at the same time, a woman that has blood type A can be chosen
A family has 4 children, two of whom are selected at random. Let B1 be the event that one child has blue eyes, and B2 be the event that the other chosen child has blue eyes. are these independent events
no. since we know that eye color is hereditary, so whether or not one child is blue-eyed will increase or decrease the chances that the other child has blue eyes, respectively.
Two people are selected simultaneously and at random from a very large population, and their blood type is checked. A—person 1 has blood type O B—person 2 has blood type O disjoint or non disjoint? independent or dependent?
non disjoint Since it is possible to choose two people with blood type O, the events are not disjoint. independent since the two people were selected simultaneously and at random from a large population, whether or not event A occurs (whether or not one of the people chosen has blood type O) has absolutely no effect on the probability that the other person chosen will have blood type O (the probability that event B will occur). Therefore, the events are independent.
why can we only use the addition rule for disjoint events
the answer will be over 1
In 2012, researchers working with a very large population of health records found that 9.3% of all Americans had diabetes (source: National Diabetes Statistics Report, 2014). Suppose a medical researcher randomly selects two individuals from a large population. Let A represent the event "the first individual has diabetes." Let B represent the event "the second individual has diabetes." True or false? A and B are independent events.
true
when A and B are not necessarily disjoint, we use which rule
"General Addition Rule".
The complement to "getting at least one question right" is
"getting none of the questions right."
In a certain liberal arts college with about 10,000 students, 40% are males. If two students from this college are selected at random, what is the probability that they are of the same gender? a. .16 b. 0.96 c. .48 d. 0.52
.52 disjoint events P(both male)= .4(.4) + 0.6(0.6)
There are two traffic lights on Fred's route from home to work. Over the years Fred has tracked when he has to stop at these lights. He determined that 45% of the time he has to stop at the first light; 30% of the time he has to stop at the second light, and 20% of the time he has to stop at both lights. Let A be the event that Fred has to stop at the first light. Let B be the event that Fred has to stop at the second light. What is the probability that Fred has to stop at one or more lights on his way to work?
.55 P(A) = .45 P(B) = .30 P(A and B) = .20 P(A) + P(B) - P(A and B) .45 + .30 - (.20) .75 - .20
T or F in venn diagrams, disjoint events overlap
FALSE do not share any possible outcomes
1st basic probability rule
For any event A, 0 ≤ P(A) ≤ 1. -basically the likelihood of an event can range anywhere from 0 to 1
5th basic probability rule (multiplication rule for independent events)
If A and B are two independent events, then P(A and B) = P(A) * P(B).
what does it mean when two events are dependent
If whether or not one event occurs does affect the probability that the other event will occur
P(A and B) =
P(event A occurs and event B occurs)
4th basic rule of probability (restricted to disjoint events)
The Addition Rule for Disjoint Events: If A and B are disjoint events, then P(A or B) = P(A) + P(B).
An engineering school reports that 55% of its students were male (M), 40% of its students were between the ages of 18 and 20 (A), and that 25% were both male and between the ages of 18 and 20. What is the probability of a random student being female between the ages of 18 and 20? Assume P(F) = P(not M). a. 0.15 b. 0.16 c. 0.30 d. 0.40
a. 0.15 P(A) = .40 P(A and M) = .25 P(A) − P(M and A) or build a table
the word "or" will always be associated with the operation of
addition
the information in a probability two-way table is for
an entire population
random experiment
an experiment that produces an outcome that cannot be predicted in advance (hence the uncertainty).
In a recent school election at a large school we know that 45% of the students supported candidate X and the other 55% supported candidate Y. Assume everyone has a strong opinion about one candidate or the other. If we select 2 students at random, what is the probability that they both support candidate X? a. 0 b. 0.2025 c.0.69
b. 0.2025 make a two way table
According to the most updated data gathered by the American Association of Suicidology, 80% of suicides in the U.S. are committed by men. Two suicide cases are selected at random. What is the probability that both suicides were committed by a person of the same gender? a. 0.64 b. 0.68 c. 1.6 d. Cannot be determined since the outcomes are not independent cannot be determined since the events are not disjoint.
b. 0.68 P(both suicides same gender) = P(both males) + P(both females) = (0.8)(0.8) + (0.2)(0.2) = 0.68
For a criminal trial, 8 active and 4 alternate jurors are selected. Two of the alternate jurors are male and two are female. During the trial, two of the active jurors are dismissed. The judge decides to randomly select two replacement jurors from the 4 available alternates. What is the probability that both jurors selected are female? a. 1/4 b. 1/12 c. 1/6 d. 1/2
c. 1/6 The order that the jurors are chosen does not matter, so F1F2 and F2F1 are the same outcome. Thus, the sample space contains six equally likely outcomes. S = {F1F2, F1M1, F1M2, F2M1, F2M2, M1M2}. Only one of the six outcomes is two females.
Four students attempt to register online at the same time for an Introductory Statistics class that is full. Two are freshmen and two are sophomores. They are put on a wait list. Prior to the start of the semester, two enrolled students drop the course, so the professor decides to randomly select two of the four wait list students and gives them a seat in the class. What is the probability that both students selected are freshmen? a. 1/2 b. 1/4 c.1/6
c.1/6 The order that the students are chosen does not matter, so F1F2 and F2F1 are the same outcome. Thus, the sample space contains six equally likely outcomes. S = {F1F2, F1S1, F1S2, F2S1, F2S2, S1S2}.
A couple decides to have children until they have one boy and one girl, but they will not have more than three children. Choose the correct sample space for this random experiment. a. S = {GGG, GGB, GBG, GBB, BBB, BBG, BGB, BGG} b. S = {GGG, GGB, BBB, BGG} c. S = {GB, BG, BBG, GGB} d. S = {GB, BG, BBG, GGB, BBB, GGG}
d. S = {GB, BG, BBG, GGB, BBB, GGG} Each outcome has one boy and one girl, except if the couple has three children of the same gender (in which case they stop having children because they originally decided to have no more than 3.)
Two events that cannot occur at the same time are called
disjoint or mutually exclusive
In a recent school election at a large school we know that 45% of the students supported candidate X and the other 55% supported candidate Y. Assume everyone has a strong opinion about one candidate or the other. If we select 2 students at random what is the probability that they both supported the same candidate? a. 0.2025 b. 0.2475 c. 0.3025 d. 0.4950 e. 0.5050
e. 0.5050 .45(.45) + (.55)(.55)
a statement about the nature of the outcome that we're actually going to get once the experiment is conducted
event
Consider the following two events: A—exactly one of the three children is a girl C—exactly one of the three children is a boy. S={BBB, BBG, BGB, GBB, GGB, GBG, BGG, GGG} answer the following i. What are the possible outcomes for each of these events? ii. Do the events share any of the outcomes? (i.e., is there an overlap between the two events?) iii. Based on ii, are the events disjoint or not?
i. A=BBG, BGB, GBB B=GGB, GBG, BGG ii. no, they have no outcome in common. iii. yes they are disjoint—they can never occur together.
A quiz consists of 10 multiple-choice questions, each with 4 possible answers, only one of which is correct. A student who does not attend lectures on a regular basis has no clue what the answers are, and therefore uses an independent random guess to answer each of the 10 questions. i. What is the probability that the student gets at least one question right (R) and wrong (W) ii. find P(not L) iii. find P(L)
i. P(R) = 1/4 = .25 P(W) = 3/4 = .75 ii. P(not L) = P(WWWWWWWWWW) = .75 * .75 * .75 * .75 * .75 * .75 * .75 * .75 * .75 * .75 = .0563. iii. P(L) = 1-P(not L) 1 - .0563 .9437
if it is a large population, we can assume the events are independent or dependent
independent
independent calls for which rule disjoint events call for which rule
independent=multiplication disjoint=addition
practical use of the 1st basic probability rule ( 0 ≤ P(A) ≤ 1.)
it can be used to identify any probability calculation that comes out to be more than 1 as wrong.
is P(A or B) larger or smaller than either one of the individual properties
larger more general
the word "and" will always be associated with the operation of
multiplication
what does it mean when two events are independent
when one event has occurred and does not affect the probability that the other event will occur.
A woman's pocket contains two quarters and two nickels. She randomly extracts one of the coins and, after looking at it, replaces it before picking a second coin. Let Q1 be the event that the first coin is a quarter and Q2 be the event that the second coin is a quarter. Are Q1 and Q2 independent events?
yes Since the first coin that was selected is replaced, whether or not Q1 occurred (i.e., whether the first coin was a quarter) has no effect on the probability that the second coin will be a quarter, P(Q2).
Consider the following two events: A—a randomly chosen person has blood type A, and B—a randomly chosen person has blood type B. are A and B disjoint
yes A and B can not occur at the same time, someone can not have both A and B blood
Two people are selected at random from all people in the United States. Let B1 be the event that one of the people has blue eyes and B2 be the event that the other person has blue eyes. are these independent events
yes. in this case, since they were chosen at random, whether one of them has blue eyes has no effect on the likelihood that the other one has blue eyes, and therefore B1 and B2 are independent. LARGE POPULATION
Three people are chosen at random. (Assume the choices are independent events). What is the probability that they all have the same blood type? O = .44 A = .42 B = 0.10 AB = 0.04
(.44 * .44 * .44) + (.42 * .42 * .42) + (.10 * .10 * .10) + (.04 * .04 * .04) = .160336
if A and B are disjoint, then P(A and B) =
0
Let A and B be two disjoint events such that P(A) = 0.20 and P(B) = 0.60. What is P(A and B)? a. 0 b. 0.12 c. 0.68 d. 0.80
0 If two events are disjoint, then by definition, P(A and B) = 0 (the two events cannot happen together).
On the "Information for the Patient" label of a certain antidepressant it is claimed that based on some clinical trials, when taking this medication - there is a 14% chance of experiencing sleeping problems, or insomnia (denote this event by I) What is the probability that a patient taking this drug will not experience insomnia? a. 0.14 b. 0.86 c. 0.74
b. 0.86 P(not I) = 1 - P(I) = 1 - 0.14 = 0.86.
T or F order does not always matter in random experiments
true
