AP Stats: Chapter 6: Probability

If I pick 2 cards from a standard deck without replacement, what is the probability that I select two spades?

(13/52)*(12/51) = 1/17 They are not independent.

Event and its complement should = ?

1

4 Fundamental Properties of Probability

1. Legitimate values 2. Sample space 3. Addition 4. Complement

Simulation steps:

1. Use random mechanisms to represent an observation (record the process) 2. Create an observation using method in step 1 and determine if outcome of interest has occurred. 3. Repeat step 2 a large number of times. 4. Calculate estimated probability by dividing the number of observations of outcome of interest by total number of observations generated.

Mutually exclusive (disjoint)

2 events have no outcomes in common (never happen simultaneously)

How many types of face cards?

3 face cards (1. Jacks, 2. Queens, 3. Kings) 4 cards per face

How many suits in a deck of cards?

4 suits (1. Spades, 2. Diamonds, 3. Hearts, 4. Clubs) 13 cards per suit

Suppose that the probabilities that an answer can be found on Google is .95, on Answers.com is .92, and on both websites is .874. Are the possibilities of finding the answer on the two websites independence? A) Yes, because (.95)(.92) = .874 B) No, because (.95)(.92) = .874 C) Yes, because .95 > .92 > .874. D) No, because .5(.95 + .92) = .874 E) There is insufficient information to answer this question.

A) If PandF = P(E)P(F) then E and F are independent

Three-fourths of college students change their major at least once. The reasons for changing and what they change to as following: 1. Reason Change in career interests = .55 Advisor suggestion = .1 Professor suggestion = .2 Other Reason = .15 2. Change to Related field = .4 Liberal arts = .25 Pre-professional = .35 Assuming reasons and what students change to are independent, what is the probability that a college student decides to change a major, based on an advisor suggestion, to pre-professional? A) .0263 B) .035 C) .0467 D) .3 E) .45

A) When events are independent, the probability of their intersection is the product of their probabilities. In this case (3/4)(.1)(.35) = .0265

Suppose P(X) = .25 and P(Y) = .40. If (X|Y) = .2, what is P(Y|X)? A) .1 B) .125 C) .32 D) .45 E) .5

C) P(XnY) = P(X|Y)P(Y) = .2*.4 = .08 Then P(Y|X) = P(XnY)/P(X) = .08/.25 = .32

Suppose that for a certain coastal city, in any given year the probability of a major hurricane hitting is .4, the probability of flooding is .3, and the probability of both a hurricane and flooding will occur is .2. What is the probability of flooding given that a major hurricane hits? A) .2 B) .286 C) .5 D) .667 E) .75

C) P(flooding|hurricane) = [P(flooding n hurricane)/P(hurricane)] = .2/.4 = .5

The sampling distribution of the sample mean is close to the normal distribution A) only if the parent population is unimodal, not badly skewed, and does not have outliers. B) no matter what the distribution of the parent population or what the value of n. C) if n is large, no matter what the distribution of the parent population. D) if the standard deviation of the parent population is known. E) only if both n is large and the parent population has a normal distribution.

C) This follows from the Central Limit Theorem.

Sample choice (s)

Collection of all possible outcomes of a chance experiment

A travel agent books passages on three different tours, with half her customers choosing tour T1, one-third choosing T2, and the rest choosing T3. The agent has noted that three-quarters of those who take tour T1, return to book passage again, two-thirds of those who take T2 return, and one-half of those who take T3 return. If a customer does return, what is the probability that the person first went on tour T2? A) 1/3 B) 2/3 C) 2/9 D) 16/49 E) 49/72

D) 16/49 P(T1 return) = (1/2)(3/4) = 3/8 P(T2 return) = (1/3)(2/3) = 2/9 P(T3 return) = (1/6)(1/2) = 1/12 P(T3) = 1 - (.5 + .333) = 1/6 P(return) = (3/8) + (2/9) + (1/12) = 49/72 P(T2/return) = (2/9)/(49.72) = 16/49

The following is from a particular region's mortality table. Age = Number Surviving 0 = 10,000 20 = 9,700 40 = 9,240 60 = 7,800 80 = 4,300 What is the probability that a 20 year old will survive to be 60? A) .1959 B) .44119 C) .7800 D) .8041 E) .9700 *****************************************

D) 7800/9700 = .8041

The Air Force receives 40% of its parachutes from Company C1 and the rest from Company C2. The probability that a parachute will fail to open is .0025 or .002, depending on whether it is from company C1 or C2, respectively. If a randomly chosen parachute fails to open, what is the probability that is from Company C1? A) .0010 B) .0022 C) .4025 D) .4545 E) .5455

D) P(C2) = 1 - P(C1) = 1 - .4 = .6

Given that 49% of the US population are male, and 12.1% of the population are over 65 years old, can we conclude that (.490)(.121) = 5.93 percent of the population are men older than 65? A) Yes, by the multiplication rule B) Yes, by conditional probabilities C) Yes, by the Law of Large Numbers D) No, because the events are not independent E) No, because the events are mutually exclusive

D) P(EnF) = P(E)P(F) only if E and F are independent. In this case, men do not live as long as women, and so the events are not independent.

Which of the following is not a valid discrete probability distribution for the set {x1, x2, x3}? A) P(x1) = 1, P(x2) = 0, P(x3) = 0 B) P(x1) = 1/3, P(x2) = 1/3, P(x3) = 1/3 C) P(x1) = 1/2, P(x2) = 1/3, P(x3) = 1/6 D) P(x1) = 2/3, P(x2) = 2/3, P(x3) = -1/3 E) All of the above are valid probability distributions.

D) The probabilities must sum to 1, and no individual probability can be negative.

Which of the following is not a valid discrete probability distribution for the set {x1, x2, x3}? A) P(x1) = 1, P(x2) = 0, P(x3) = 0 B) P(x1) = 1/3, P(x2) = 1/3, P(x3) = 1/3 C) P(x1) = 1/2, P(x2) = 1/3, P(x3) = 1/6 D) P(x1) = 2/3, P(x2) = 2/3, P(x3) = -1/3 E) All the above are valid probability distributions. *****************************************

D) The probabilities must sum to 1, and no individual probability can be negative.

Addition

If two events E and F are disjoint, P(E or F) = P(E) + P(F).

In a set of eight boxes, three boxes each contain two red and two green marbles, while the remaining boxes each contain three red and two green marbles. A player randomly picks a box and then randomly picks a marble from that box. She wins if she ends up with a red marble. If she plays four times, what is the probability she wins exactly twice? A) .0606 B) .3164 C) .3221 D) .3634 E) .5625

D) The probability of winning is (3/8)(1/2) + (5/8)(3/5) = (9/16) and the probability of winning exactly twice in four games is 2(9/16)(9/16)(7/16)(7/16) = .3634.

General Rule for Addition

For any 2 events E and F: P(EuF) = P(E) + P(F) - P(EnF)

General Rule for Multiplication

For any 2 events E and F: P(EnF) = P(E|F)*P(F)

Complement

For any event E, P(E) + P(not E) = 1.

Probabilities of Equally Likely Outcomes

For any event E, P(E) = # of outcomes in E/total outcomes of E

Legitimate values

For any event all probabilities must be between 0 and 1 (decimal or fractions).

Sample space

If S is the sample space, P(S) = 1.

event

any collection of outcomes (subset) from the sample space of a chance experiment

Law of Large Numbers

as the number of repetitions of a chance experiment increase, the chance that the relative frequency of occurrence for an event will differ from the true probability by more than any small number approaches 0

Independent events _______ be disjoint. (can or cannot?)

cannot

If an event is independent, its _______ should also be independent.

complement

P(none) is ALWAYS the ________ of P(at least 1).

complement

Union

consists of all outcomes that are in at least one of the 2 events, that is, in A or B or in both

Intersection

consists of all outcomes that are in both of the events

Complementary

consists of all outcomes that are not in the event

If sample <5% of population treat as _________. (not independent or independent?)

independent

Independent

one event will occur (or has occurred) does not change the probability that the other occurs

The Subjective Approach

personal measure of the strength of belief that a particular outcome will occur

Relative Frequency Approach (Experimental)

probability of event E is the value approached by the relative frequency of occurrence of E in a long series of trials of a chance experiment P(E) = number of times E occurs/number of trials

Conditional Probability

probability that takes into account a given condition has occurred

Probability of an event E

ratio of number of outcomes favorable to E to the total number of outcomes in the sample space P(E) = # of favorable/total

Assume independence if it is with or without replacement?

with replacement

Two events E and F are independent, if and only if, P(EnF) =

P(E)*P(F)

If P(E|F) = P(E), it is also true that

P(F|E) = P(F)

P(B|A) equation is

P(AnB)/P(A) = overlap/prob. of 1st

If P(A) = .25 and P(B) = .34, what is P(AuB) if A and B are independent? A) .085 B) .505 C) .590 D) .675 E) There is insufficient information to answer this question.

B) If A and B are independent P(AnB) = P(A)P(B) and thus P(AuB) = .25 + .34 - (.25)(.34) = .505

Given the probabilities P(A) = .3 and P(AuB) = .7, what is the probability P(B) if A and B are mutually exclusive? If A and B are independent? A) .4, .3 B) .4, 4/7 C) 4/7, .4 D) .7, 4/7 E) .7, .3

B) If A and B are mutually exclusive, P(AnB) = 0. Thus, .7 = .3 + P(B) - 0, and so P(B) = .4. If A and B are independent, then P(AnB) = P(A)P(B). Thus .7 = .3 + P(B) - .3 P(B), and so P(B) = 4/7.

A basketball player makes one out of his first two free throws. From that point on, the probability that he makes the next shot is equal to the proportion of shots made up to that point. If he takes two more shots, what is the probability he ends up making a total of two free throws? A) 1/4 B) 1/3 C) 1/2 D) 2/3 E) 3/4

B) P(2 shots made) = (1/2)(1/3) + (1/2)(1/3) = 1/3

Given two events, E and F, such that P(E) = .34, P(F) = .45, and P(EuF) = .637, then the two events are: A) independent and mutually exclusive B) independent, but not mutually exclusive C) mutually exclusive, but not independent D) neither independent nor mutually exclusive E) there is not enough information to answer this question

B) P(EuF) = P(E) + P(F) - P(EnF), so .637 = .340 + .45 - P(EnF) and P(EnF) = .153. Since P(EnF) cannot equal 0, E and F are not mutually exclusive. However, P(EnF) = .153 = (.34)(.45) = P(E)P(F), which implies that E and F are independent.

Two events, E and F, are independent if P(E|F) =

P(E)

Suppose a 6-sided die is rolled. The outcome that the die would land on an even number would be:

E = {2, 4, 6}

Suppose a 6-sided die is rolled. The event that the die would land on a prime number would be P = {2, 3, 5} and E = {2, 4, 6}. What would be the event E and P happening?

E and P = {2}

P(E|F) =

E happens after F

Suppose a 6-sided die is rolled. The event that the die would land on a prime number would be P = {2, 3, 5}. What would be the event E or P happening?

E or P = {2, 3, 4, 5, 6}

Suppose you toss a fair coin ten times and it comes up heads every time. Which of the following is a true statement? A) By the Law of Large Numbers, the next toss is more likely to be tails than another heads. B) By the properties of conditional probability, the next toss is more likely to be heads given that ten tosses in a row have been heads. C) Coins actually do have memories, and this what comes up on the next toss is influenced by the past tosses. D) The Law of Large Numbers tells how many tosses will be necessary before the percentages of heads and tails are again in balance. E) None of the above are true statements. *****************************************

E) Coins have no memory, and so the probability that the next toss will be heads is .5, and the probability that it will be tails is .5. The Law of Large Numbers says that as the number of tosses becomes larger, the percentage of heads tends to become closer to .5.

There are 5 outcomes to an experiment and a student calculates the respective probabilities of the outcomes to be .34, .5, .42, 0, and -.26. The proper conclusion is that A) the sum of the individual probabilities is 1 B) one of the outcomes will never occur C) one of the outcomes will occur 50 percent of the time D) all of the above are true E) the students made an error *****************************************

E) Probabilities are never less than 0

Suppose a 6-sided die is rolled. What would the event be that is the die NOT landing on an even number?

E^c = {1, 3, 5}

When A= {2} and B = {6}. On a single die roll, is it possible for A and B to happen at the same time?

No

Chance experiment

Uncertainty about which of 2 or more plausible outcomes will result