Sheldon Ross Chapters 1+2+3+4

An urn contains 6 white and 9 black balls. If 4 balls are to be randomly selected without replacement, what is the probability that the first 2 selected are white and the last 2 are black?

(6/15)(5/14)(9/13)(8/12)=6/91

From a group of 8 women and 6 men, a committee consisting of 3 men and 3 women is to be formed. How many different committees are possible if 1 man and 1 woman refuse to serve together?

(7 choose 3)(5 choose 3) for when neither feuding man/women participates+(7 choose 2)(5 choose 3) for when the feuding woman participates+(7 choose 3)(5 choose 2) for when the feuding man participates=910

An urn initially contains 5 white and 7 black balls. Each time a ball is selected, its color is noted and it is replaced in the urn along with 2 other balls of the same color. Compute the probability that of the first 4 balls selected, exactly 2 are black.

(7/12)(9/14)(5/16)(7/18)=35/768 There are 6 ways to draw 4 balls of which 2 are black. 6*(35/768)=.2734=35/128

A gambler has a fair coin and a two-headed coin in his pocket. He selects one of the coins at random; when he flips it, it shows heads. What is the probability that it is the fair coin Suppose that he flips the same coin a second time and it shows heads. Now what is the probability that it is the fair coin?

P(FC given h)=(P(h given FC)P(FC))/P(h)=(.5*.5)/(.5*.5+.5)=1/3 P(FC given hh)=(P(hh given FC)P(FC))/(P(hh))=(.5*.25)/(.5*.25+.5)=1/5

Ninety-eight percent of all babies survive delivery. However, 15 percent of all births involve Cesarean (C) sections, and when a C section is performed, the baby survives 96 percent of the time. If a randomly chosen pregnant woman does not have a C section, what is the probability that her baby survives?

P(S given Cc)=P(S and Cc)/P(Cc) P(S)=P(S and Cc)+P(S and Cc) P(S and Cc)=P(S)-P(S given C)P(C) P(S and Cc)=.98-.96*.15=.836 P(S given Cc)=.836/.85=.9835

Two dice are thrown n times in succession. Compute the probability that double 6 appears at least once. How large need n be to make this probability at least 1/2?

a)1-(35/36)^n b)n=25

(N choose R) when R>N or when R<0

(N choose R)=0

Expand (x1+2x2+3x3)^4

Look it up

Stores A, B, and C have 50, 75, and 100 employees, respectively, and 50, 60, and 70 percent of them respectively are women. Resignations are equally likely among all employees, regardless of sex. One woman employee resigns. What is the probability that she works in store C?

P(C given W)=(P(W given C)P(C))/(P(W))=(.7*(100/225))/(140/225)=.5

What is the probability that someone has an accident in the second year given that he or she had no accidents in the first year? Given: Accident prone people have an accident within a one year period at a probability of .4 Nonaccident prone people have an accident within a one year period at a probability of .2 30% of the population is accident prone

P(accident given no accident)=P(no accident and accident)/P(no accident)=(.3*.6*.4+.7*.8*.2)/(.3*.6+.7*.8)=46/185

Five people, designated as A, B, C, D, E, are arranged in linear order. Assuming that each possible order is equally likely, what is the probability that a) there is exactly one person between A and B? b) there are exactly two people between A and B? c) there are three people between A and B?

a)2*18/120=3/10 b)2*12/120=1/5 c)2*6/120=1/10

If 8 new teachers are to be divided among 4 schools, how many divisions are possible? What if each school must receive 2 teachers?

A)4^8=65,536 B)8!/(2!*2!*2!*2!)=2,520

Suppose that n balls are randomly distributed into N compartments. Find the probability that m balls will fall into the first compartment. Assume that all N^n arrangements are equally likely.

Let us take 1 example to understand this problem... Suppose we have: 3 departments, 6 balls of which 2 will got into department 1 The 6 balls have the option to go into each of the three departments. Hence, the denominator is 6^3 To determine how many ways in which 2 balls can be put into department 1, (6 choose 2). Then to determine how many different distributions of the ways in which the remaining 4 balls can be distributed can be determined by 2^4. Hence, overall ((6 choose 2)(2^4))/(6^3). Or, in more general terms, ((n choose m)((N-1)^(n-m)))/(n^N)

Consider an experiment that consists of determining the type of job--either blue collar or white collar-- and the political affiliation--Republican, Democratic, or Independent--of the 15 members of an adult soccer team. How many outcomes are... a)In the sample space? b)in the event that at least one of the team members is a blue-collar worker? c)in the event that none of the team members considers himself or herself an Independent?

a)6^15 b)6^15-3^15 c)4^15

Fifty-two percent of the students at a certain college are females. Five percent of the students in this college are majoring in computer science. Two percent of the students are women majoring in computer science. If a student is selected at random, find the conditional probability that a) the student is female given that the student is majoring in computer science; b) the student is majoring in computer science given that the student is female

a)P(F given C)=P(F and C)/P(C)=.02/.05=.4 b)P(C given F)=P(C and F)/P(F)=.02/.52=.038

How many people have to be in a room in order that the probability that at least two of them celebrate their birthday in the same month is at least 1/2?

1 person has a unique personality 12/12 times. For the 2nd person to have a unique birthday, their odds are 11/12. For the 3rd person to have a unique birthday while the others birthdays are also unique, their odds are 10/12. For the 4th person to have a unique birthday while others have unique birthdays, their odds are 9/12. For the 5th person to have a unique birthday, their odds are 8/12. (12*11*10*9)/(12^4)=.572 (12*11*10*9*8)/(12^5)=.389

If a die is rolled 4 times, what is the probability that 6 comes up at least once?

1-(5/6)^4=671/1296

John, Jim, Jay, and Jack have formed a band consisting of 4 instruments. If each of the boys can play all 4 instruments, how many different arrangements are possible> What is John and Jim can play all 4 instruments, but Jack and Jay can each play only piano and drums?

1. 4!=24 2. 2*1*2*1=4 if you start w/ John and Jim

Ten weight lifers are competing in a team weight-lifting contest. Of the lifters, 3 are from the United States, 4 are from Russia, 2 are from China, and 1 is from Canada. If scoring takes account of the countries that lifters represent, but not their individual identities, how many different outcomes are possible in which the United states has 1 competitor in the top 3 and 2 in the bottom 3?

3(3 choose 2)(7!/(4!2!)=945

Two cards are randomly selected from an ordinary playing deck. What is the probability that they form a blackjack? That is, what is the probability that one of the cards is an ace and the other one is either a ten , a jack, a queen, or a king?

4 ways to pick an ace, 16 ways to pick a ten, jack, queen, or king. 4*16=64/(52 choose 2)=.0483

Ms. Jones has 10 books that she is going to put on her bookshelf. Of these, 4 are mathematics books, 3 are chemistry books, 2 are history books, and 1 is a language book. Ms. Jones want to arrange her books so that all the books dealing with the same subject are together on the shelf. How many different arrangements are possible?

4!*4!*3!*2!*1!=6,912 The first 4! is for ordering the subjects, the rest are for ordering the books within each set of books.

In the game of bridge, the entire deck of 52 cards is dealt out to 4 players. What is the probability that one of the players receives all 13 spades?Each player receives an ace?

4/(52 choose 13)

The game of bridge is played by 4 players, each of whom is dealt 13 cards. How many bridge deals are possible?

52!/(13!13!13!13!)=5.36*10^28

Delegates from 10 countries, including Russia, France, England, and the United States, are to be seated in a row. How many different seating arrangements are possible if the French and English delegates are to be seated next to each other and the Russian and U.S. delegates are not to be next to each other?

9!*2!-8!*2!*2!=564,480

The dice game Craps is played as follows. The player throws two dice, and if the sum is 7 or 11, then he wins. If the sum is 2, 3 or 12, then he loses. If the sum is anything else, then he continues throwing until he either throws that number again (in which case he wins) or he throws a 7 (in which case he loses). Calculate the probability that the player wins.

= 2/9 + 2[(1/12)(1/3) + (1/9)(2/5) + (5/36)(5/11)] = 244/495 = 0.493

Basic Principle of Counting

If one experiment can result in any of M possible outcomes and if another experiment can result in any of N possible outcomes, then there are MN possible outcomes of the two experiments.

A poker hand consists of 5 cards. If the cards have distinct consecutive values and are not all of the same suit, we say that the hand is a straight. What is the probability that one is dealt a straight?

(10(4^5-4))/(52 choose 5)=.0039

(n choose r)=?

((n-1) choose (r-1))+((n-1) choose r) where 1<=r=n

If it is assumed that all (52 choose 5) poker hands are equally likely, what is the probability of being dealt c)two pairs(a,a,b,b,c where a, b, and c are unique denominations)

((13 choose 2)(4 choose 2)(4 choose 2)(44 choose 1))/(52 choose 5)

Shuffling n cards, whenever you get a tail, the card goes to the back of the deck. If you get a head, the card stays where it is. One the card has been flipped n times, a round has been completed. What is the probability that the ordering after one round is the same as the initial ordering?

(n+1)/(2^n)

An urn contains n white and m black balls, where n and m are positive numbers. b)If a ball is randomly withdrawn and then replaced before the second is drawn, what is the probability that the withdrawn balls are the same color?

(n/(n+m))^2+(m/(n+m))^2

What is the binomial theorem?

(x+y)^n=sum from k=0 to k=n of (n choose k)*(x^k)(y^(n-k))

How many different 7-place license plates are possible if the first three places are to be occupied by letters and the final four by numbers? A)With replacement? B)Without replacement?

A)26*26*26*10*10*10*10=175,760,000 B)26*25*24*10*9*8*7=78,624,000

In successive rolls of a pair of fair dice, what is the probability of getting 2 sevens before 6 even numbers?

All numbers that are not sevens or are are not even are not relevant. The probability of getting a seven among all numbers that are seven or are even is (1/6)/(1/6+1/2)=1/4. If we want 2 sevens before 6 even numbers, we can find this by finding its complement. 1-(zero sevens and 7 even numbers)-(one seven and 6 even numbers)=1-(3/4)^7-7(3/4)^6(1/4)=4547/8192

The Generalized Basic Principle of Counting

If R experiments that are to be performed are such that the first one may result in any of N1 possible outcomes; and if, for each of these N1 possible outcomes, there are N2 possible outcomes of the second experiment...then there is a total of N1*N2*...NR possible outcomes of the R experiments.

In a class, there are 4 first-year boys, 6 first-year girls, and 6 sophomore boys. How many sophomore girls must be present if sex and class are to be independent when a student is selected at random?

P(B given S)=6/(16+x) P(B)=10/(16+x) P(S)=6+x/(16+x) For independence, P(B given S)=P(B)P(S) 6/(16+x)=(60+10x)/((16+x)^2) 96+6x=60+10x 4x=36 x=9

A deck of cards is shuffled and then divided into two halves of 26 cards each. A card is drawn from one of the halves; it turns out to be an ace. The ace is then placed in the second half-deck. The half is then shuffled, and a card is drawn from it. Compute the probability that this drawn card is an ace.

Let A=The event that the card drawn from the second deck is an ace Let C=The event that the card drawn from the second deck is the same ace from the first deck A=AC or ACc P(A)=P(AC)+P(ACc) P(A)=P(A given C)P(C)+P(A given Cc)P(Cc) P(A)=1*(1/27)+(3/51)(26/27) P(A)=43/459

In how many ways can 8 people be seated in a row if there are 5 men and they must sit next to each other assuming the remaining 3 people are women?

Men are block w/ 3 other women. 4! permutations w/ man block and 3 women. Then, 5! for the permutations for the people within the block.5!*4!=2880

Permutations when some of the values are equivalent to each other(I.E. PEPPER, P's and E's are equivalent).

N!/(N1!*N2!*...NR!) when there are N permutations of all the objects, yet there are N1 that are alike, N2 are alike,...NR that are alike.

Given 20 people, what is the probability that among the 12 months in the year, there are 4 months containing exactly 2 birthdays and 4 containing exactly thee birthdays?

Overall, since each of the twenty people can be born in any of the 12 months, the probability of each possibility of their birthdays are 12^20. To choose which 4 months will be associated with having two birthdays, (12 choose 4). Then, to choose the other 4 months that have 3 birthdays within them, (12 choose 4). Then, to divide (3,3,3,3,2,2,2,2) among the 20 people, the multinomial theorem must be used. ((20!/((3!)^4(2!)^4))(12 choose 4)(8 choose 4))/(12^20)=.001

P(E U F U G)=?

P(E)+P(F)+P(G)-P(EF)-P(FG)-P(EG)+P(EFG)

A group of 6 men and 6 women is randomly divided into 2 groups of size 6 each. What is the probability that both groups will have the same number of men?

Since, you just need to determine how 3 men and 3 women can be put in any one group, then the other group will naturally have 3 men and 3 women. To pick 6 people from 12 people is (12 choose 6). To pick 3 men or 3 women is (6 choose 3). Hence ((6 choose 3)(6 choose 3))/(12 choose 6)=.4329

Suppose that you continually collect coupons and that there are m different types. Suppose also that each time a new coupon is obtained, it is a type i coupon with probability pi, i=1,..., m. Suppose that you have just collected your nth coupon, What is the probability that it is a new type?

Sum from i to m of P(new given i)pi=Sum from i to m of (1-pi)^(n-1)pi

From a group of 5 women and 7 men, how many different committees consisting of 2 women and 3 men can be formed if 2 of the men are feuding and refuse to serve on the committee together?

There are (2 choose 2)(5 choose 1)=5 out of the (7 choose 3)=35 possible groups of 3 men contain both of the feuding men. Hence 35-5=30 groups of men*(5 choose 2)=10 groups of women=300 possible committees.

How do permutations of N objects work?

There are N(N-1)(N-2)...1=N! different permutations. Note that in this case, order does matter. ABC is not the same as BAC.

A system is composed of 5 components, each of which is either working or failed. Consider an experiment that consists of observing the status of each component, and let the outcome of the experiment be given by the vector (x1,x2,x3,x4,x5), where xi is equal to 1 if the component i is working and is equal to 0 if the component i is failed. i)Suppose that the system will work if components 1 and 2 are both working, or if components 3 and 4 are both working, or if components 1,3, and 5 are working. Let W be the event that the system will work. Specify all the outcomes in W.

There should be 15, look them up at http://ozelmatematik.net/dokumanlar/First%20Course%20in%20Probability%207E%20-%20Ross.pdf

There are 15 tennis balls in a box, of which 9 have not previously been used. Three of the balls are randomly chosen, played with, and then returned to the box. Later, another 3 balls are randomly chosen from the box. Find the probability that none of these balls has ever been used.

This problem can be best done by breaking it up into 4 parts: When you pick 3 of the 6 balls that have already been used, then you pick 3 of the 9 unused balls in the second try: ((6 choose 3)(9 choose 3))/((15 choose 3)(15 choose 3))=48/5915 When you pick 2 of the 6 balls that have already been used and 1 of the 9 balls that have yet to have been used, then you pick 3 balls of the 8 that have not been used yet: ((6 choose 2)(9 choose 1)(8 choose 3))/((15 choose 3)(15 choose 3))=216/5915 When you pick 1 ball from the 6 that have already been used and 2 from the 9 balls that have yet to have been used, then you pick 3 balls from the 7 balls that have not been used yet: ((6 choose 1)(9 choose 2)(7 choose 3))/((15 choose 3)(15 choose 3))=216/5915 When you pick 0 balls from the 6 that have been already used and 3 from the 9 balls that have yet to have been used, then you pick 3 balls from the 6 balls that have not been used yet: ((6 choose 0)(9 choose 3)(6 choose 3))/((15 choose 3)(15 choose 3))=48/5915 Add all values together=528/5915=.0893

Suppose that you are playing blackjack against a dealer. In a freshly shuffled deck, what is the probability that neither you nor the dealer is dealt a blackjack?

What you want to do is fine the probability of both winning or one winning, then subtract this value from 1. P(both winning)=(4*16*3*15)=2,880 options P(one winning)=2(which player wins)*(4*16)((3 choose 2)+(3 choose 1)(32 choose 1)+(47 choose 2))=151,040 options P(either you or the dealer)=(2,880+151,040)/((52 choose 2)(50 choose 2))=.0947 1-.0947=.9052

Each of 2 balls is painted either black or gold and then placed in an urn. Suppose that each ball is colored black with probability 1/2 and that these events are independent. a)Suppose that you obtain information that the gold paint has been used (and thus at least one of the balls is painted gold). Compute the conditional probability that both balls are painted gold.

a)P(GG given that 1 G)=P(GG and 1+ G)/(P(1+G) P(GG and 1+G)=1/4 P(1+G)=3/4 P(GG given that 1 G)=(1/4)/(3/4)=1/3

Consider an urn containing 12 balls of which 8 are white. A sample of size 4 is to be drawn with replacement(without replacement). What is the conditional probability(in each case) that the first and third balls drawn will be white given that the sample drawn contains exactly 3 white balls?

With replacement: There are only 2 ways to select a ball in which the 1st and 3rd balls are white, when 3 out of the 4 balls are white. 2/((8 choose 3)(4 choose 1))=1/112 Without replacement: There are 4 ways to select 3 white out of 4 balls. Only 2 of those options have the 1st and 3rd being white balls.

Expand (x1+x2+x3)^2

X1^2+x2^2+x3^2+2x1x2+2x1x3+2x2x3

Suppose that each child born to a couple is equally likely to be a boy or a girl, independently of the sex distribution of the other children in the family. For a couple having 5 children, compute the probabilities of the following events: a) All children are of the same sex. b) The 3 eldest are boys and the others girls. c)Exactly 3 are boys. d) The 2 oldest are girls. e) There is at least 1 girl.

a) (1/2)^5+(1/2)^5=1/16 b) (1/4)(1/8)=1/32 c) (5 choose 3)(1/8)(1/4)=5/16 d)1*(1/2)(1/2)=1/4 e)1-(1/2)^5=31/32

A closet contains 10 pairs of shoes. If 8 shoes are randomly selected, what is the probability that there will be a)no complete pair b)exactly 1 complete pair

a) First shoe will be moving towards being a pair 20/20 of the time. The second shoe has an 18/19 chance of not being paired with the first shoe...etc (20/20)(18/19)(16/18)(14/17)(12/16)(10/15)(8/14)(6/13)=.0045 b)First, you need to pick which of the 10 pairs will be chosen-->(10 choose 1). Then, we need to pick 6 colors for the remaining 9 pairs of shoes. This can be done in (9 choose 6) ways. Then, among those 6 colors, you have to options to decide which shoe from the pair to pick. 2^6. In total, you have (20 choose 8) shoes to pick. ((10 choose 1)(9 choose 6)2^6)/(20 choose 8)=.4267

How many functions defined on N points are possible if each functional value is either 0 or 1?

2^N

An urn contains 5 red, 6 blue, and 8 green balls. If a set of 3 balls is randomly selected, what is the probability that each of the balls will be the same color with replacement? What is the probability that they will be of different colors with replacement?

A)((5^3)+(6^3)+(8^3))/(19^3)=.124 B)(6*5*8*6)/(19^3)

Expand (x+y)^3

y^3+3xy^2+3x^2y+x^3

Expand (3x^2+y)^5

y^5+15x^2y^4+90x^4y^3+270x^6y^4+405x^8y^5+243x^10

A, B, C take turns flipping coins. The first one to get a head wins. The sample space of this experiment can be defined by...S=1,01,001,0001,...,0000... i)Define the following events in terms of S:(AUB)^c assuming A flips first,then B,then C, then A, and so on

{0000..., 001,000001,...}

Seven balls are randomly withdrawn from an urn that contains 12 red, 16 blue, and 18 green balls. Find the probability that either exactly 3 red balls or exactly 3 blue balls are withdrawn.

((12 choose 3)(34 choose 4)+(16 choose 3)(30 choose 4)-(12 choose 3)(16 choose 3)(18 choose 1))/(46 choose 7)

Compute in a hand of bridge the conditional probability that East has 3 spades given that North and South have a combined total of 8 spades.

((13 choose 8)(39 choose 18)(5 choose 3)(21 choose 10))/((52 choose 26)(26 choose 13))=.0548=P(E has 3 and N and S have 8 combined) ((13 choose 8)(39 choose 18))/((52 choose 26)=.1618=P(N and S have 8 combined) .0548/.1618=.339 .

In order to play a game of basketball, 10 children at a playground divide themselves up into two teams of 5 each. How many different divisions are possible?

(10!/(5!*5!))/2!

A dance class consists of 22 students, of which 10 are women and 12 are men. If 5 men and 5 women are to be chosen and then paired off, how many results are possible?

(12 choose 5) ways to choose the men. (10 choose 5) ways to choose the men. Then line up 5 men in a row, and there are 5! ways to line up the women next to each corresponding man. Hence, (12 choose 5)(10 choose 5)5!=23,950,080

If 3 balls are "randomly drawn" from a bowl containing 6 white and 5 black balls, what is the probability that one of the balls is white and the other two black?

(120+120+120)/990=4/11

Combination Formula/Terminology

(N choose R) where N>=R, (N choose R)= N!/((N-R)!R!). The number of possible combinations of N objects taken R at a time. When order is relevant, you can arrange, for example, 5 letters into groups of 3, 5*4*3=60 ways. Yet, for any hypothetical arrangement of ABC, CBA, CAB, etc, this arrangement is made 6 ways. Hence 60/6=10=(5 choose 3)

Let A and B play a series of games. Each games is independently won by A with probability p and by B with probability (1-p). They stop when the total number of wins of one of the players is two greater than that of the other player. The player with the greater number of total wins is declared the winner of the series. (a) Find the probability that a total of 4 games are plated. (b)Find the probability that A is the winner of the series.

(a) The only possible way 4 games to be played is when 3 games are won by A and 1 by B or 3 games by B and one by A. This can only happen if A and B split the first two games. Or, if p(1-p)+(1-p)p=2p(1-p) Then, either A wins the next two games: p^2 or B wins the next two games: (1-p)^2. Overall 2p(1-p)[p^2+(1-p)^2]=(2p-2p^2)[p^2+1-2p+p^2]=2p^3+2p-4p^2+2p^3-2p^4-2p^2+4p^3-2p^4=-4p^4+8p^3-6p^2+2p (b) P(A wins)=P(A wins given A wins first two)P(A wins first two)+P(A wins given B wins first two)P(B wins first two)+P(A wins given first two are split)P(first two are split)=1*p^2+0*(1-p)^2+P(A wins)*2p(1-p)=p^2+P(A wins)*(2p-p^3) Solving for P(A wins)=(p^2)/(1-2p+2p^2)

A certain organism possesses a pair of each of 5 different genes (which we will designate by the first 5 letters of the English alphabet). Each gene appears in 2 forms (which we designate by lowercase and capital letters). The capital letter will be assumed to be the dominant gene, in the sense that if an organism possesses the gene pair xX, then it will outwardly have the appearance of the X gene, For instance, if X stands for brown eyes and x for blue eyes, then an individual having either gene pair XX or xX will have brown eyes, whereas one having gene pair xx will have blue eyes. The characteristic appearance of an organism is called its phenotype, whereas its genetic constitution is called its genotype. (Thus 2 organisms with respective genotypes aA, bB, cc, dD, ee and AA, BB, cc, DD, ee would have different genotypes but the same phenotype,) IN a mating between 2 organisms, each one contributes at random, one of its gene pairs of each type. The 5 contributions of an organism (one of each of the 5 types) are assumed to be independent and are also independent of the contributions of the organism's mate. In a mating between organisms having genotypes aA, bB, cC, dD,eE and aa, bB, cc, Dd, ee what is the probability that the progeny will (i) phenotypically and (ii) genotypically resemble (a) the first parent? (b) the second parent? (c) either parent? (d) neither parent?

(i) (a) (1/2)(3/4)(1/2)(3/4)(1/2)=9/128 (b) (1/2)(3/4)(1/2)(3/4)(1/2)=9/128 (c) P(match 1st parent)+P(match 2nd parent)=P(either parent)=18/128=9/64 (d) (1-9/64)=55/64 (ii) (a)(1/2)^5=1/32 (b)=(1/2)^5=1/32 (c)=P(match first parent)+P(match 2nd parent)=P(match either)=2/32=1/16 (d)=1-P(match either)=15/16

Suppose that an insurance company classifies people into one of three classes: good risks, average risks, and bad risks. The company's records indicate that the probabilities that good-, average-, and bad-risk persons will be involved in accident over a 1-year span are, respectively, .05, .15, and .3. If 20 percent of the population is a good risk, 50 percent is an average risk, and 30 percent is a bad risk, what proportion of people have accidents in a fixed year? If policyholder A had no accidents in 2012, what is the probability that he or she is a good risk? Is an average risk?

1) P(A)=.05*.2+.15*.5+.3*.3=.175 2) P(G given NA)=(P(NA given G)P(G))/(P(NA))=(.2*.95)/(1-.175)=38/165 3)P(A given NA)=(P(NA given A)P(A))/(P(NA))=(.5*.85)/(1-.175)=17/33

For years, telephone area codes in the United States and Canada consisted of a sequence of three digits. The first digit was an integer between 2 and 9, the second digit was either 0 or 1, and the third digit was any integer from 1 to 9. How many area codes were possible? How many area codes starting with a 4 were possible?

1. 8*2*9=144 2. 1*2*9=18

Ten children are to be divided into an A team and a B team of 5 each. The A team will play in one league and the B team in another. How many different divisions are there?

10!/(5!*5!)

How many different 7 place license plates are possible if the first 2 places are for letters and the other 5 for numbers if no letter or number can be repeated in a single license plate?

19,656,000

An urn contains 3 red and 7 black balls. Players A and B withdraw balls from an urn consecutively (and in that order) until a red ball is selected. Find the probability that A selects the red ball.

1st attempt red: 3*9! 3rd attempt red: 7*6*3*7! 5th attempt red: 7*6*5*4*3*5! 7th attempt red: 7*6*5*4*3*2*3*3! Total number of permutations: 10! Ultimate result:7/12

How many subsets are there of a set consisting of n elements?

2^n. See http://www.mathsisfun.com/activity/subsets.html

A psychology laboratory conducting dream research contains 3 rooms, with 2 beds in each room. If 3 sets of identical twins are to be assigned to these 6 beds so that each set of twins sleeps in different beds in the same room, how many assignments are possible?

3! to decide which set of twins is in each room*2^3 for the permutations of which twin is sleeping in which bed=48 options

A person has 8 friends, of whom 5 will be invited to a party. A) How many choices are there if 2 of the friends are feuding and will not attend together? B)How many choices if 2 friends will only attend together?

A)(8 choose 5)-(6 choose 3)=36 options. Where (6 choose 3) is when the two feuding parties both attend. B)(6 choose 3) which is both parties+(6 choose 5) which is neither party=26 options

If N people, including A and B, are randomly arranged in a line, what is the probability that A and B are next to each other? What would the probability be if the people were randomly arranged in a circle?

A)Since A and B must be next to each other, they can be viewed as part of a block. There are thus (N-1)! possible permutations of all the people in the line if A and B must be next to each other. Yet, A and B can switch places in each of these permutations. Overall, there are N! permutations. Hence, 2(N-1)!/(N!) or 2/N B) This one is best done by example. Suppose there are 3 people. How many ways can three people be arranged around a circle? ABC=CAB=BCA is one option. ACB=BAC=CBA is the second option. 2 is the value in the denominator. For the numerator, lets say that A and B must sit next to each other. This happens in both situations. Let us try it with 4 people. How many ways can 4 people be arranged around a circle? ABCD=DABC=CDAB=BCDA is one option. ACBD=DACB=BDAC=CBDA is option two. ADBC=CADB=BCAD=DBCA is option 3. ABDC=CABD=DCAB=BDCA is option 4. ACDB=BACD=DBAC=CDBA is option 5. ADCB=BADC=CBAD=DCBA is option 6. Let us suppose that A and B must be next to each other. This happens in 4 out of 6 options or 2 out of 3. Overall, the probability must be 2/n-1 provided that n>=2.

Poker dice is played by simultaneously rolling 5 dice. A)What is the probability of 2 pair? B)What is the probability of 3 of a kind? C)Full House

A)Which 2 of the 6 numbers will be chosen. (6 choose 2) options Which of the 4 remaining numbers will be the final number=4 options Which of the 5 dice will be the lonely die=5 options Which of the 4 remaining dice will be the larger pair=(4 choose 2) options Which of the 2 remaining dice will be the smaller pair=(2 choose 2) options Overall... (6 choose 2)(4 choose 2)(2 choose 2)*5*4=1800/6^5=.2315 B)Which three dice will be chosen?: (5 choose 3) Which number will be represented on these three dice: 6 Probability that the other two dice numbers are not the same as the original number: (5/6)(4/6) Overall... (5 choose 3)*6*5*4/(6^5)=.1543 C) Which three dice will be chosen to be the three of a kind part of the full house?: (5 choose 3) options Which one of the numbers will be the up-facing side of the dice?: 6 options Which will be the number up-facing for the remaining 2 die?: 5 options Which of the remaining two die will be the pair?: (2 choose 2) options Overall... (5 choose 3)(6 choose 1)(2 choose 2)(5 choose 1)/(6^5)=.0385

Consider 3 urns. Urn A contains 2 white and 4 red balls, urn B contains 8 white and 4 red balls, and urn C contains 1 white and 3 red balls. If 1 ball is selected from each urn, what is the probability that the ball chosen from urn A was white given that exactly 2 white balls were selected?

A=Ball from urn A is white B=2 whites selected P(A given B)=P(A and B)/P(B) P(A and B)=P(A=White,B=White,C=Red)+P(A=White, B=Red, C=White)=(1/3)(2/3)(3/4)+(1/3)(1/3)(1/4) P(B)=P(A=white, B=white, C=Red)+P(A=white, B=Red, C=White)+P(A=Red,B=White, C=White)=(1/3)(2/3)(3/4)+(1/3)(1/3)(1/4)+(1/3)(2/3)(1/4) Answer=7/11

Suppose that E and F are mutually exclusive events of an experiment, Show that if independent trials of this experiment are performed, then E will occur before F with probability P(E)/[P(E)+P(F)].

G=E^cF^c P(E occurs before F)=P(E occurs before F given that trial 1 is E)P(E)+P(E occurs before F given that trial 1 is F)P(F)+P(E occurs before F given that trial 1 is G)P(G)=P(E)+P(E before F)P(G)=P(E before F)=P(before F)(1-G)=P(E)=P(E)/(1-P(G))=P(E)/(P(E)+P(F)

How many different letter arrangements can be formed from the letters PEPPER?

If you distinguish each of the letters, there are 6! permutations of the letters. Yet, if you alternate the P letters or the E letters, you can still get the same arrangement. I.E. PEPPER. Hence, to account for this, note that PEPPER can be created in 12 different ways by alternating the P's and the E's. In any possible arrangement of these letters, 12 permutations are made to create the exact same phrase. Hence, 6!/(12)=6,912

Ms. Aquina has just had a biopsy on a possible cancerous tumor. Not wanting to spoil a weekend family event, she does not want to hear any bad news in the next few days. But if she tells the doctor to call only if the news is good, then if the doctor does not call, Ms. Aquina can conclude that the news is bad. So, being a student of probability, Ms. Aquina instructs the doctor to flip a coin. If it comes up heads, the doctor is to call if the news is good and not call if the news is bad. If the coin comes up tails, the doctor is not to call. In this way, even if the doctor doesn't call, the news is not necessarily bad. Let P be the probability that the tumor is cancerous; let Y be the conditional probability that the tumor is cancerous given that the doctor does not call. Find Y in terms of P to prove whether Y or P is larger.

Let C=cancerous, let H=the doctors coin flip is H We want to know what is the conditional probability that the tumor is cancerous given that the doctor does not call. The doctor will not call if CH or if Hc. Hence P(C given (CH or Hc))=P(C and (CH or Hc))/(P(CH or Hc))=(P(CH or CHc))/(P(CH)+P(Hc))=(P(CH)+P(CHc))/(P(CH)+P(Hc))=(C/2+C/2)/(C/2+1/2)=C/(C+1)/2=2C/(C+1)>P for all values of C less than 1.

The following method was proposed to estimate the number of people over the age of 50 who reside in a town of known population 100,000: "As you walk along the streets, keep a running count of the percentage of people you encounter whoa re over 50. Do this for a few days; then multiply the percentage you obtain by 100,000 to obtain the estimate." Prove what needs to be true to make this estimate effective.

Let p=your estimate P(O)=P(people in a town over the age of 50) let S=you spotting someone on the street let Y=Amount of time a person under the age of 50 spends in the street let O=Amount of time a person over the age of 50 spends in the street. Ultimately what you are trying to find is P(O given S)=(P(S given O)P(O))/((P(S given O)(P(O)+P(S given Y)P(Y) P(O)=p P(S given O)=O P(Y)=1-p P(S given Y)=Y Overall pO/(pO+Y(1-p)) When O=Y, pY/(pY+Y-pY)=p. Hence, when people who are old and young spend the same amount of time on the streets as off the streets, your estimates will be most accurate

Two cards are randomly chosen without replacement from an ordinary deck of 52 cards. Let B be the event that both cards are aces, let As be the event that the ace of spades is chosen, and let A be the event that at least one ace is chosen. Find a)P(B given As) b)P(B given A)

P(A)=((4 choose 1)(48 choose 1)+(4 choose 2))/(52 choose 2)=33/221 P(B)=(4 choose 2)/(52 choose 2)=1/221 P(A given B)=1 P(As)=(51 choose 1)/(52 choose 2)=1/26 P(As given B)=(3 choose 1)/(4 choose 2)=1/2 a)P(B given As)=(P(As given B)P(B))/P(As)=((1/2)(1/121))/(1/26)=1/17 b)P(B given A)=(P(A given B)P(B))/(P(A)=(1/221)/(33/221)=1/33

On the morning of September 30, 1982, the won-lost records of the three leading baseball teams in the Western Division of the National League were as follows: Team Won Lost Atlanta Braves 87 72 San Francisco Giants 86 73 Los Angeles Dodgers 86 73 Each team had 3 games remaining. All 3 of the Giants' games were with the Dodgers, and the 3 remaining games of the Braves were against the San Diego Padres. Suppose that the outcomes of all remaining games are independent and each game is equally likely to be won by either participant. For each team, what is the probability that it will win the division title? If two teams tie for first place, they have a playoff game, which each team has an equal chance of winning.

P(Braves winning)=P(B given B win 3)P(B win 3)+P(B given B win 2)P(B win 2)+P(B win given B win 1)P(B win 1)+P(B win given B win 0)P(B win 0)= (1/8)+(3/8)((2/8)(1/2)+(6/8))+(3/8)((6/8)(1/2))=38/64=19/32 P(Dodgers or Giants winning)=(1-(19/32))/2=13/64

Suppose that an ordinary deck of 52 cards (which contains 4 aces) is randomly divided into 4 hands of 13 cards each. We are interested in determining p, the probability that each hand has an ace. Let Ei be the event that the ith hand has exactly one ace. Determine p=P(E1E2E3E4) by using the multiplication rule.

P(E1E2E3E4)=P(E1)*P(E2 given E1)*P(E3 given E1E2)*P(E4 given E1E2E3) P(E1)=((4 choose 1)(48 choose 12))/(52 choose 13) P(E2 given E1)=((3 choose 1)(36 choose 12))/(39 choose 13) P(E3 given E1E2)=((2 choose 1)(24 choose 12))/(26 choose 13) P(E4 given E1E2E3)=((1 choose 1)(12 choose 12))/(13 choose 13) P(E1E2E3E4)=((4 choose 1)(48 choose 12))/(52 choose 13)*((3 choose 1)(36 choose 12))/(39 choose 13)*((2 choose 1)(24 choose 12))/(26 choose 13)*((1 choose 1)(12 choose 12))/(13 choose 13)=.1054

Suppose we have 10 coins such that if the ith coin is flipped, heads will appear with probability i/10, i= 1,2,...,10. When one of the coins is randomly selected and flipped, it shows heads. What is the conditional probability that it was the coin?

P(F given H)=(P(H given F)P(F))/(P(H given F)P(F)+P(H given Fc)P(Fc))=(.5*.1)/(.55)=1/11

A parallel system functions whenever at least one of its components works. Consider a parallel system of n components, and suppose that each component works independently with probability 1/2. Find the conditional probability that component 1 works given that the system is functioning.

P(O given F)=(P(F given O)(P(O))/P(F)=(1*1/2)/(1-P(Fc))=(1/2)/(1-(1/2)^n)

Each of 2 cabinets identical in appearance has 2 drawers. Cabinet A contains a silver coin in each drawer, and cabinet B contains a silver coin in one of its drawers and a gold coin in the other. A cabinet is randomly selected, one of its drawers is opened, and a silver coin is found. What is the probability that there is a silver coin in the other drawer?

S=silver is the first one found A=you are looking in cabinet A P(A given S)=(P(S given A)P(A))/(P(S))=.5/(.5(1+.5))=2/3

(x1+x2+...xr)^n=

Sum of (n1,...,nr):n1+n2+...nr=n of (n choose (n1,n2,...nr))x1^n1x2^n2...xr^nr. Note, the sum is over all nonnegative integer-valued vectors (n1,n2,...nr) such that n1+n2+...nr=n. Where (n choose n1,n2,...nr) are known as multinomial coefficients

Poisson Formula

Suppose there are and N is the average number of successes over a time frame, U is the actual number of successes over a time frame. P(X,N)=((e^(-N))(N^U))/(U!)

If 8 rooks(castles) are randomly placed on a chessboard, compute the probability that none of the rooks can capture any of the others.

Suppose you have put down i rooks, you have (8-i) possibilities for which row to put the next rook in and (8-i) possibilities for the next column to put the next rook in. Hence, over all for the next rook, you have (8-i)^2 possibilities for the next rook. Thus... (64)(49)(36)(25)(16)(9)(4)(1)/(64)(63)(62)(61)(60)(59)(58)(57)=9.10*10^-6

In a hand of bridge, find the probability that you have 5 spades and your partner has the remaining 8.

When you pick your spades, you can pick 5 spades in (13 choose 5) ways. However, for the remaining 39 cards, they must be divided so that you get 8, your partner gets 5, and the remaining two people each get 13. Overall, the four sets of 13 have to be divided among 3 people. ((13 choose 5)(39!/(8!5!13!13!)))/(52!/(13!13!13!13!))=2.608*10^-6

Suppose that an ordinary deck of 52 cards is shuffled and the cards are then turned over one at a time until the first ace appears. Given that the first ace is the 20th card to appear, what is the conditional probability that the card following it is the a)ace of spades? b)two of clubs?

a) Let A= probability that the 20th card to appear is an ace Let B=probability that the 21st card is the ace of spades P(B given A)=P(B and A)/P(A) P(B and A)=(3/33)(1/32)=1/352 P(A)=4/33 P(B given A)=3/128 b)Let A=probability that the 20th card to appear is an ace Let B=probability that the 21st card is the two of clubs P(B given A)=P(B and A)/P(A) P(B and A)=(47 choose 19)/(48 choose 19)=.604 Next is ace=1 Next is two of clubs=1/32 .604*1/32=.018

The probability of the closing of the ith relay in the circuits shown is given by pi, for i=1,...,5. If all relays function independently, what is the probability that a current flows between A and B for the respective circuits in figure 1 and 2 Figure 1: A<1 or 3<2 or 4, 2 to 5, 4 to 5-->B Figure 2: A to 1 to 4 to B. A from 1 to 3 to 5 to B. A from 2 to 5 to B. A from 2 to 3 to 4 to B.

a) Let E1 represent 1, 2, and 5 being closed. Let E2 represent 3,4, and 5 being closed. Let E3 represent 1,2,3,4, and 5 being closed. P(A to B=P(E1)+P(E2)-P(E3)=p1p2p5+p3p4p5-p1p2p3p4p5 b)Let E1 represent 1 and 4 closed. Let E2 represent 2 and 5 closed. Let E3 represent 1,3, and 5 closed. Let E4 represent 2,3, and 4 closed. Using the inclusion-exclusion principle, we want to sum P(Ei) for i=1 to 4)-P(Ei and Ej) from i and j 1 to 4+P(Ei and Ej and Ek) from i,j, and k from 1 to 4-P(Ei+Ej+Ek+El) for i,j,k, and l from 1 to 4. =p1p4+p2p5+p1p3p5+p2p3p4-p1p2p4p5-p1p3p4p5-p1p2p3p4-p1p2p3p5-p2p3p4p5-p1p2p3p4p5+4p1p2p3p4p5-p1p2p3p4p5

A woman has n keys, of which one will open her door. a) If she tries the keys at random, discarding those that do not work, what is the probability that she will open the door on her kth try? b) What if she does not discard previously tried keys?

a) Let us say that she succeeds on her second try. And she has 5 keys. The probability that she fails to pick the right key on her first attempt is (4/5). Then, to succeed, her odds of success are (1/4). If these two numbers are multiplied together, the result is 1/5 or 1/n. b)Let us say that she succeeds on her second try. And she has 5 keys. The probability that she fails to pick the right key on her first attempt is (4/5)/ Then, to succeed, her odds are 1/5. Lets say that she succeeds on her 4rd try and she has 6 keys. To fail on her first, second, and third tries, her odds are (5/6)^3. To succeed on her 4th try, her odds are 1/6. In summary, ((n-1)/n)^(k-1)(1/n)

In a certain village, it is traditional for the eldest son (or the older son in a two-son family) and his wife to be responsible for taking care of his parents as they age. In recent years, however, the women of this village, not wanting that responsibility, have not looked favorably upon marrying an eldest son. a)If every family in the village has two children, what proportion of all sons are older sons? b)If every family in the village has three children, what proportion of all sons are eldest sons? Assume that each child is, independently, equally likely to be either a boy or a girl.

a) Out of every four families... BB=.25 BG=.25 GB=.25 GG=.25 Hence, out of every four families, there is an older son 1/2 the time. Then add in the fact that in the family that has two boys, 1 out of every two boys is the older one. Hence, (3/4) or excluding the younger boy in BB families, of boys are older the older boy. b)Out of every eight families... BBB=1/8 BBG=1/8 BGG=1/8 GGG=1/8 GGB=1/8 GBB=1/8 BGB=1/8 GBG=1/8 Hence, in families with boys, we see a total of 12 boys. Out of these 12 boys, 7 are the oldest of boys. Hence 7 out of 12 boys are oldest boys.

Given that an engineering system consisting of n components is said to be a k-out-of-n system (k is less than or equal to n) if the system functions if and only if at least k of the n components function. Suppose that all the components function independently of one another. If system 1 is A to 1 to 2 to 5 to B or A from 3 to 4 to 5 to B, and if the ith component functions with probability Pi, what is the conditional probability that relays 1 and 2 are both closed given that a current flows from A to B?

a) P(1 and 2 Closed given Flows)=(P(flows given 1 and 2 are closed)P(1 and 2 are closed))/(P(flows)=(P5P1P2)/(P1P2P5+P3P4P5-P1P2P3P4P5)

A simplified model for the movement of the price of a stock supposes that on each day the stock's price either moves up 1 unit with probability p or moves down 1 unit with probability (1-p). The changes on different days are assumed to be independent. a) What is the probability that after 2 days the stock will be at its original price? b) What is the probability that after 3 days the stock's price will have increased by 1 unit? c) Given that after 3 days the stock's price has increased by 1 unit, what is the probability that it wen up on the first day?

a) UD=0 UU=+2 DU=0 DD=-2 p(1-p)+(1-p)p=2(p-p^2) b) UDU=+1 DUD=-1 UUU=+3 UUD=+1 UDD=-1 DDD=-3 DDU=-1 DUU=+1 p(1-p)p+pp(1-p)+(1-p)pp=3(p^2-p^3) c) See the instances above in which 2/3 of the instances in which after 3 days the stock price goes up 1, the price also goes up on the first day.

A total of 46 percent of the voters in a certain city classify themselves as Independents, whereas 30 percent classify themselves as Liberals and 24 percent say that they are Conservatives. In a recent local election, 35 percent of the Independents, 62 percent of the Liberals, and 58 percent of the Conservatives voted. A voter is chosen at random. Given that this person voted in the local election, what is the probability that he or she is a)an Independent? b) a Liberal? c) a Conservative? d)What percent of voters participated in the local election?

a)(.46*.35)/(.46*.35+.3*.62+.58*.24)=.331 b)(.3*.62)/(.46*.35+.3*.62+.58*.24)=.383 c)(.58*.24)/(.46*.35+.3*.62+.58*.24)=.286 d)(.46*.35+.3*.62+.58*.24)=.4862 or 48.62 percent of voters voted in the local election

A recent college graduate is planning to take the first three actuarial examinations in the coming summer. She will take the first actuarial exam in June. If she passes that exam, then she will take the second exam in July, and if she also passes that one, then she will take the third exam in September. If she fails an exam, then she is not allowed to take any others. The probability that she passes the first exam is .9. If she passes the first exam, then the conditional probability that she passes the second one is .8, and if she passes both the first and the second exams, then the conditional probability that she passes the third exam is .7. a)What is the probability that she passes all three exams? b)Given that she did not pass all three exams, what is the conditional probability that she failed the second exam?

a).9*.8*.7=63/125 b)(.9*.2)/(.1+.9*.2+.9*.8*.3)=45/124

An engineering system consisting of n components is said to be a k-out-of-n system (k is less than or equal to n) if the system functions if and only if at least k of the n components function. Suppose that all components function independently of one another. a) If the ith component functions with probability pi, i=1,2,3,4, compute the probability that a 2-out-of-4 system functions. b)Repeat part (a) for a 3-out-of-5 system. c)Repeat for a k-out-of-n system when all the Pi equal p(that is, Pi=p, i=1,2,...n).

a)P(1st,2nd, no 3rd, no 4th)+P(1st, no 2nd, 3rd, no 4th)+P(1st, no 2nd, no 3rd, 4th)+P(no 1st, 2nd, 3rd, no 4th)+P(no 1st, 2nd, no 3rd, 4th)+P(no 1st, no 2nd, 3rd, 4th)+P(1st, 2nd, 3rd, no 4th)+P(1st, 2nd, no 3rd, 4th)+P(no 1st, 2nd, 3rd, 4th)+P(1st, no 2nd, 3rd, 4th)+P(1st,2nd,3rd,4th) where no P3, for example, would be (1-p3). So, for the 1st list of the string P1P2(1-p3)(1-p4). b)P(1st, 2nd, 3rd, no 4th, no 5th)+P(1st, 2nd, no 3rd, 4th, no 5th))+P(1st, no 2nd, 3rd, 4th, no 5th)+P(no 1st, 2nd, 3rd, 4th, no 5th)+P(1st, 2nd, no 3rd, no 4th, 5th)+P(1st, no 2nd, 3rd, no 4th, 5th)+P(1st, no 2nd, no 3rd, 4th, 5th)+P(no 1st, no 2nd, 3rd, 4th, 5th)+P(no 1st, 2nd, no 3rd, 4th, 5th)+P(no 1st, 2nd, 3rd, no 4th, 5th)+P(1st, 2nd, 3rd, 4th, no 5th)+P(1st, 2nd, 3rd, no 4th, 5th)+P(no 1st, 2nd, 3rd, 4th, 5th)+P(1st, 2nd, no 3rd, 4th, 5th)+P(1st, no 2nd, 3rd, 4th, 5th)+P(1st, 2nd, 3rd, 4th, 5th) c)Sum from i=k to n of (n choose i)pi(1-p)^(n-i)

A high school student is anxiously waiting to receive mail telling her whether she has been accepted to a certain college. She estimates that the conditional probabilities of receiving notification on each day of next week, given that she is accepted and that she is rejected, are as follows: P(mail given accepted) P(mail given rejected) Monday .15 .05 Tuesday .2 .1 Wednesday .25 .1 Thursday .15 .15 Friday .1 .2 She estimates that her probability of being accepted is .6. a) If there is no mail through Wednesday, what is the conditional probability that she will be accepted? b) What is the conditional probability that she will be accepted if no mail arrives that week?

a)P(A given McTcWc)=(P(McTcWc given A)P(A))/(P(McTcWc)=((1-.15-.2-.25)(0.6))/((1-.15-.2-.25)0.6+(1-.05-.1-.1)0.4)=12/27 b) P(A given no mail)=(P(no mail given A)P(A))/(P(no mail)=(.15*.6)/(.15*.6+.4*.4)=9/25

A and B are involved in a duel. The rules of the duel are that they are to pick up their guns and shoot at each other simultaneously. If one or both are hit, then the duel is over. If both shots miss, then they repeat the process. Suppose that the results of the shots are independent and that each shot of A will hit B with probability pa, and that each shot of B will hit A with probability pb. What is a) the probability that A is not hit? b) the probability that both duelists are hit? c) the probability that the duel ends after the nth round of shots? d) the conditional probability that the duel ends after the nth round of shots given that A is not hit? e) the conditional probability that the duel ends after the nth round of shots given that both duelists are hit?

a)P(A not given One hit)=(pa(1-pb))/(pa+pb-papb) b)P(Both given One hit)=(papb)/(pa+pb-papb) c)P(End n given one hit)=((pa+pb-papb)(1-pa)^(n-1)(1-pb)^(n-1)). Essentially, no one is hit for the first n-1 attempts, then someone is hit on the nth attempt. d)P(End n given a not hit)=(P(a not hit given n end)p(n end))/(p(n end)=(Pa(1-pb)(1-pa)^(n-1)(1-pb)^(n-1))/((papb)/(pa+pb-papb))=((1-pa)^(n-1)(1-pb)^(n-1))/(pa+pb-papb) e)P(nth given both)=P(both and nth)/p(both)=(papb(1-pa)^(n-1)(1-pb)^(n-1))/(papb/(pa+pb-papb))=((1-pa)^(n-1)(1-pb)^(n-1))/(pa+pb-papb)

Barbara and Dianne go target shooting, Suppose that each of Barbara's hots hits a wooden duck target with probability p1, while each shot of Dianne's hits it with probability p2. Suppose that they shoot simultaneously at the same target. If the wooden duck is knocked over (indicating that it was hit), what is the probability that a)both shots hit the duck? b)Barbara's shot hit the duck?

a)P(Both given One or more)=(P(One or more given Both)P(Both))/(P(One or more))=(1*p1p2)/(1-(1-p1)(1-p2))=(p1p2)/(1-(1-p1-p2+p1p2))=(p1p2)/(p1+p2-p1p2) b)P(B hit given One or more)=(P(One or more given B hit)P(B hit))/(P(One or more))=p1/(p1+p2-p1p2)

Prostrate cancer is the most common type of cancer found in males. As an indicator of whether a male has prostrate cancer, doctors often perform a test that measures the level of the prostrate-specific antigen (PSA) that is produced only by the prostrate gland. Although PSA levels are indicative of cancer, the test is notoriously unreliable. Indeed, the probability that a noncancerous man will have an elevated PSA level is approximately .135, increasing to approximately .268 if the man does have cancer. If, on the basis of other factors, a physician is 70 percent certain a male has prostrate cancer, what is the conditional probability that he has the cancer given that a) the test indicated an elevated PSA level b) the test did not indicate an elevated PSA level

a)P(C given E)=(P(E given C)P(C))/(P(E given C)P(C)+P(E given Cc)P(Cc))=(.268*.7)/(.268*.7+.135*.3)=.8224 b)P(C given L)=(P(L given C)P(C))/(P(L given C)P(C)+P(L given Cc)P(Cc))=(.732*.7)/(.732*.7+.865*.3)=.6638

There us a 50-50 chance that the queen carries the gene for hemophilia. If she is a carrier, then each prince has a 50-50 chance of having hemophilia. If the queen has three princes, without the disease, what is the probability that the queen is a carrier? If there is a fourth prince, what is the probability that he will have hemophilia? -Assuming that if the prince does not hemophilia, then the prince cannot get the disease

a)P(C given no p1,p2,p3)=(P(no p1,p2,p3 given C)P(C))/(P(no p1,p2,p3 given C)P(C)+P(no p1,p2,p3 given no C)P(C))=((1/8)(1/2))/((1/8)(1/2)+1*1/2)=1/9 b)P(4th have given 1,2,3 do not)=(P(1,2,3 not given 4th has)P(4th has))/(P(1,2,3 do not)=((1/2)^5+0*1/2)/((1/2)^4+1/2)=1/18

On rainy days, Joe is late to work with probability .3; on nonrainy days, he is late with probability .1. With probability .7, it will rain tomorrow. a)Find the probability that Joe is early tomorrow. b)Given that Joe was early, what is the conditional probability that it rained?

a)P(E)=P(E given R)P(R)+P(E given Rc)P(Rc)=.7*.7+.3*.9=.76 b)P(R given E)=(P(E given R)P(R))/P(E)=(.7*.7)/(.76)=49/76

Consider an unending sequence of independent trials, where each trial is equally likely to result in any of the outcomes 1,2,or 3/ given that outcome 3 is the last of the three outcomes to occur, find the conditional probability that (a) the first trial results in outcome 1; (b) The first two trials both result in outcome 1.

a)P(FT=1 given 03=last)=(P(03=last given FT=1)P(FT=1))/(P(03=last))=(1/2)(1/3)/(1/3)=1/2 b)P(trials 1 and 2=1 given 03=last)=P(03=last given that trials 1 and 2=1)P(trials 1 and 2=1)/(P(03=last)=(1/2)(1/9)/(1/3)=1/6

The color of a person's eyes is determined by a single pair of genes. If they are both blue-eyed genes, then the person will have blue eyes; if they are both brown-eyed genes, then the person will have brown eyes; and if one of them is a blue-eyed gene and the other a brown-eyed gene, then the person will have brown eyes. (Because of the latter fact we say that the brown-eyed gene is dominant over the blue-eyed one.) A newborn child independently receives one eye gene from each of its parents, and the gene it receives from a parent is equally likely to be either of the two eye genes of that parent. Suppose that Smith and both of his parents have brown eyes, but Smith's sister has blue eyes. a) What is the probability that Smith possesses a blue-eyed gene? b) Suppose that Smith's wife has blue eyes. What is the probability that their first child will have blue eyes? c) If their first child has brown eyes, what is the probability that their next child will also have brown eyes?

a)P(Smith has blue-eyed gene)=P(X=1, X is less than or equal to 1)=(1/2)/(3/4)=2/3 b) P(child blue)=P(blue given blue gene)(2/3)+P(blue given no blue gene)(1/3)=(1/2)(2/3)=1/3 c)P(Smith blue given child brown)=(P(child brown given Smith blue)(2/3))/(2/3)=1/2 P(2nd child brown)=P(2nd child brown given given smith blue)(1/2)+P(2nd child brown given smith not blue)(1/2)=1/4+1/2=3/4

A true-false question is to be posed to a husband-and-wife team on a quiz show. Both the husband and the wife will independently give the correct answer with probability p. Which of the following is a better strategy for the couple? a) Choose one of them and let that person answer the question. b) Have them both consider the question, and then either give the common answer if they agree or, if they disagree, flip a coin to determine which answer to give.

a)Probability is p b)P(win)=P(win given both right)P(both right)+P(win given One agrees Right)P(One agrees Right)+P(win given neither right)P(neither right)=p^2+(1/2)(2p(1-p))+0(1-p)^2=p^2+p-p^2=p Either way, your odds of being correct are equal.

The chess clubs of two schools consist of, respectively, 8 and 9 players. Four members from each club are randomly chosen to participate in a contest between two schools. The chosen players from one team are then randomly paired with those from the other team, and each pairing plays a game of chess. Suppose that Rebecca and her sister Elise are on the chess clubs at different schools. What is the probability that... a) Rebecca and Elise will be paired? b) Rebecca and Elise will be chosen to represent their schools but will not play each other? c)either Rebecca or Elise will be chosen to represent her school?

a)Rebecca and Elise are chosen, then ((7 choose 3)(8 choose 3))3!/((8 choose 4)(9 choose 4))4!=1/18 to choose everyone else and to permeate who who will be seated at the other tables b)Rebecca and Elise are chosen, then ((7 choose 3)(8 choose 3))/((8 choose 4)(9 choose 4))-1/18=1/6 c)Rebecca is chosen, Elise is not: ((7 choose 3)(8 choose 4))/((8 choose 4)(9 choose 4)) Elise is chosen, Rebecca is not: ((7 choose 4)(8 choose 3))/((8 choose 4)(9 choose 4))=1/2

An elementary school is offering 3 language classes: one in Spanish, one in French, and one in German. The classes are open to any of the 100 students in the school. There are 28 students in the Spanish class, 26 in the French class, and 16 in the German class. There are 12 students who are in both Spanish and French, 4 are in both Spanish and German, and 6 who are in both French and German. In addition, there are 2 students taking all 3 classes. i)If 2 students are chosen randomly, what is the probability that at least 1 is taking a language class?

i)(50 choose 2)/(100 choose 2)=49/198 are taking no language classes, thus 1-((50 choose 2)/(100 choose 2)) include those in which at least 1 student is taking a language class=149/198

A set of n distinct items is to be grouped into r distinct groups of size n1, n2...nr. How many distinct divisions are possible?

n!/(n1!*n2!*...nr!) divisions

If n1+n2+...nr=n, we define (n choose (n1,n2,...nr)) by...

n!/(n1!*n2!*...nr!). (n choose (n1!*n2!*...nr!)) represents the # of possible divisions of n distinct objects into r distinct groups of respective sizes n1, n2,...nr.

Three prisoners are informed by their jailer that one of them has been chosen at random to be executed and the other two are to be freed. Prisoner A asks the jailer to tell him privately which of his fellow prisoners will be set free, claiming that there would be no harm in divulging this information because he already knows that at least one of the two will go free. The jailer refuses to answer the question pointing out that if A knew which of his fellow prisoners were to be set free, then his probability of being executed would rise from 1/3 to 1/2 because he would then be one of two prisoners. What do you think of the jailer's reasoning?

Let A be the prisoner to be executed and similar for B and C. Let b be the event that the jailer says "Prisoner B will not be executed" and similar for c. We need to show that P(A given b)=1/3 This assumes that P(c given A)=P(b given A)=1/2 P(A given b)=(P(b given A)P(A))/(P(b given A)P(A)+P(b given Ac)P(Ac))=(.5*1/3)/(.5*1/3+.5*2/3)=1/3

A pair of dice is rolled until a sum of either 5 or 7 appears. Find the probability that a 5 occurs first.

P(5 occurring when two dice are rolled)=1/9 P(7 occurring when two dice are rolled)=1/6 P(Neither on a given rolling of 2 dice)=((1/9)+(1/6))/36=13/18 Sum from 1 to infinity of (13/18)^(n-1)(1/9)=(1/9)(1/(1-13/18))=2/5 *Note that this can be simplified in this way because 13/18 or r is between -1 and 1. The formula for the sum of an infinite geometric series(what this is) is a/(1-r), where a is the first term.

A and B alternate rolling a pair of dice, stopping either when A rolls the sum 9 or when B rolls the sum 6. Assuming that A rolls first, find the probability that the final roll is made by A.

When A wins, the probability that he rolls a 9 is 1/9. Until then, then probability that he loses is (8/9)^n-1. Also, the probability that B loses before then is (31/36)^n-1. Hence, we want to find the sum, where j>or equal to 0 and are odd numbers of (1/9) sum of ((8/9)(31/36))^j=(1/9)/(1-(8/9)(31/36))=9/19=P(A winning)

Consider a sample of size 3 drawn in the following manner: We start with an urn containing 5 white and 7 red balls. At each stage, a ball is drawn and its color is noted. The ball is then returned to the urn, along with an additional ball of the same color. Find the probability that the sample will contain exactly a)1 white ball b) 2 white balls

a)3 ways to arrange 1 white ball: 3(7/12)(8/13)(5/14)=5/13 b)3 ways to arrange 2 white balls: 3(5/12)(6/13)(7/14)=5/32

If two fair dice are rolled, what is the conditional probability that the first one lands on 6 given that the sum of the dice is i? Compute for all values of i between 2 and 12.

i=2 p=0 i=3 p=0 i=4 p=0 i=6 p=0 i=7 p=1 i=8 p=1/5 i=9 p=1/4 i=10 p=1/3 i=11 p=1/2 i=12 p=1