Sheldon Ross Chapters 1+2+3+4
Seven balls are randomly withdrawn from an urn that contains 12 red, 16 blue, and 18 green balls. Find the probability that either exactly 3 red balls or exactly 3 blue balls are withdrawn.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)
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)
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 .
(n choose r)=?
((n-1) choose (r-1))+((n-1) choose r) where 1<=r=n
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 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
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
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
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)
(N choose R) when R>N or when R<0
(N choose R)=0
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))
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
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
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 functions defined on N points are possible if each functional value is either 0 or 1?
2^N
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
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
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
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)
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
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
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
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
(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
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
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 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
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
P(E U F U G)=?
P(E)+P(F)+P(G)-P(EF)-P(FG)-P(EG)+P(EFG)
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 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
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)
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
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
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
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
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
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
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
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
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)
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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.
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.
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
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)
Expand (x1+2x2+3x3)^4
Look it up
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.
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
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
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.
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
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.
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
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.
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,...}
