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Discrete Probability 离散概率
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A coin is flipped four times. For each of the events described below, express the event as a set in roster notation. Each outcome is written as a string of length 4 from {H, T}, such as HHTH. Assuming the coin is a fair coin, give the probability of each event.
The first and last flips come up heads.
E = {HHHH, HHTH, HTHH, HTTH}. p(E) = 4/16 = 1/4
A coin is flipped four times. For each of the events described below, express the event as a set in roster notation. Each outcome is written as a string of length 4 from {H, T}, such as HHTH. Assuming the coin is a fair coin, give the probability of each event.
There are at least two consecutive flips that come up heads.
E = {HHHH, HHHT, THHH, HHTH, HHTT, THHT, TTHH, HTHH}. p(E) = 8/16 = 1/2.
A coin is flipped four times. For each of the events described below, express the event as a set in roster notation. Each outcome is written as a string of length 4 from {H, T}, such as HHTH. Assuming the coin is a fair coin, give the probability of each event.
The first flip comes up tails and there are at least two consecutive flips that come up heads.
E = {THHH, THHT, TTHH}. p(E) = 3/16.
A class with n kids lines up for recess. The order in which the kids line up is random with each ordering being equally likely. There are two kids in the class named Celia and Felicity.
What is the probability that Celia is first in line?
The size of the sample space is n!. The number of ways to order the line in which Celia is first in line is (n-1)!. The probability that Celia is first in line is (n-1)!/n! = 1/n.
A class with n kids lines up for recess. The order in which the kids line up is random with each ordering being equally likely. There are two kids in the class named Celia and Felicity.
What is the probability that Celia is first in line and Felicity is last in line?
The size of the sample space is n!. The number of ways to order the line in which Celia is first in line and Felicity is last in line is (n-2)!. The probability that Celia is first in line and Felicity is last in line is (n-2)!/n! = 1/(n(n-1))
A class with n kids lines up for recess. The order in which the kids line up is random with each ordering being equally likely. There are two kids in the class named Celia and Felicity.
What is the probability that Celia and Felicity are next to each other in the line?
The size of the sample space is n!. The number of ways to order the line in which Celia and Felicity are next to each other is 2(n-1)!. (There are (n-1)! ways to order everyone except Celia. Once the order is fixed, Celia can be inserted either before or after Felicity). The probability that Celia and Felicity are next to each other in the line is 2(n-1)!/n! = 2/n.
There are 2n socks in a drawer, n white and n black. Two socks are selected at random from the drawer. Every way of selecting the two socks is equally likely.
How many ways are there to select the two socks?
C(2n, 2) = n(2n−1)
There are 2n socks in a drawer, n white and n black. Two socks are selected at random from the drawer. Every way of selecting the two socks is equally likely.
How many ways of selecting the socks result in two socks of the same color being chosen?
There are C(n, 2) ways of selecting a black pair and C(n, 2) ways of selecting a white pair. The total number of ways to select a pair of the same color is then 2*C(n, 2) = n(n−1).
There are 2n socks in a drawer, n white and n black. Two socks are selected at random from the drawer. Every way of selecting the two socks is equally likely.
What is the probability that a randomly chosen pair of socks are the same color? Simplify your final expression as much as possible so that it does not include any expressions of the form C(a, b).
n(n−1) / n(2n−1 ) = n−1 / 2n−1.
10 kids are randomly grouped into an A team with five kids and a B team with five kids. Each grouping is equally likely.
What is the size of the sample space?
Once the five kids for the A team are chosen, the B team is determined. There are C(10, 5) ways to pick the kids for the A team.
10 kids are randomly grouped into an A team with five kids and a B team with five kids. Each grouping is equally likely.
There are two kids in the group, Alex and his best friend Jose. What is the probability that Alex and Jose end up on the same team?
If Alex and Jose are both on the A team, then there are C(8, 3) ways to pick the remaining three kids for the A team. The number of ways for Alex and Jose to both end up on the B team is the same. Therefore the probability that Alex and Jose are together on either team is
2C(8, 3) / C(10, 5) = 4 / 9
What is the probability that the hand is a full house? A full house has three cards of the same rank and another pair of the same rank. For example, {4♠, 4♥, 4♦, J♠, J♣} is a full house.
There are 13 ways to select the rank for the three of a kind and 12 remaining ways to select the rank for the pair. Once the rank has been selected for the three of a kind, there are C(4, 3) ways to select the 3 cards from the four cards with that rank. Once the rank has been selected for the pair, there are C(4, 2) ways to select the two cards from the 4 cards with that rank. The total number of 5-card hands that are a full house is 13⋅12⋅C(4, 3) * C(4, 2). The probability that a random 5-card hand is a full house is
13⋅12⋅C(4, 3)C(4, 2) / C(52, 5) = 6 / 4165
What is the probability that the hand is a three of a kind? A three of a kind has 3 cards of the same rank. The other two cards do not have the same rank as each other and do not have the same rank as the three with the same rank. For example, {4♠, 4♦, 4♣, J♠, 8♥} is a three of a kind.
There are 13 ways to select the rank for the three of a kind. Once the rank has been selected for the three of a kind, there are C(4, 3) ways to select the three cards from the four cards with that rank. There are 12 ranks remaining for the other two cards. There are C(12, 2) ways to pick the ranks for the two other cards. Once the ranks have been chosen for those cards, there are four ways to pick the suit for the first card and four ways to pick the suit for the second card. The total number of 5-card hands that are a three of a kind is 13
C(4, 3)
C(12, 2)⋅4⋅4.The probability that a random 5-card hand is a three of a kind is
13
C(4, 3)
C(12, 2)⋅4⋅4 / C(52, 5) = 88 / 4165
What is the probability that all 5 cards have the same suit?
There are four possible ways to select the suit for the 5 cards. Once the suit has been selected, there are C(13, 5) ways to select the 5 cards from that suit. The total number of 5-card hands that all have the same suit is 4⋅C(13, 5).The probability that a random 5-card hand has 5 cards, all of the same suit is
4⋅C(13, 5) * C(52, 5) = 33 / 16660
What is the probability that the hand is a two of a kind? A two of a kind has two cards of the same rank (called the pair). Among the remaining three cards, not in the pair, no two have the same rank and none of them have the same rank as the pair. For example, {4♠, 4♦, J♠, K♣, 8♥} is a two of a kind.
There are 13 ways to select the rank for the pair. Once the rank has been selected for the pair, there are C(4, 2) ways to select the two cards from the four cards with that rank. There are 12 ranks remaining for the other three cards. There are C(12, 3) ways to pick the ranks for the three other cards. Once the ranks have been chosen for those cards, there are four ways to pick the suit for each of the three cards (resulting in a factor of 4 for each card). The total number of 5-card hands that are a two of a kind is 13
C(4, 2)
C(12, 3)⋅4⋅4⋅4. The probability that a random 5-card hand is a two of a kind is
13
C(4, 2)
C(12, 3)⋅4⋅4⋅4 / C(52, 5) = 352 / 833
A fair coin is flipped n times. Give an expression for each of the probabilities below as a function of n. Simplify your final expression as much as possible.
At least n - 1 flips come up heads.
The event that at least n - 1 flips come up heads is equivalent to the event that at most 1 flip comes up tails. Let T0 be the event that there are no tails and T1 be the event that there is exactly one tail. T0 and T1 are mutually exclusive, so the probability that at most one flip comes up tails is p(T0) + p(T1). | T0 | = 1 because the only way for there to be no tails is for the sequence to be all heads. |T1| = n because the flip that comes up tails can be any of the n flips in the sequence. The size of the sample space is 2n, so the solution is
p(T0) + p(T1) = 1/2^n + n/2^n = (n + 1)/2^n.
A fair coin is flipped n times. Give an expression for each of the probabilities below as a function of n. Simplify your final expression as much as possible.
There are at least two consecutive flips that are the same.
Call the event A. The event A is that the flips alternate between heads and tails. There are only two outcomes that alternate between heads and tails: one that starts with a heads and one that starts with a tails. Therefore, p(A) = 2/2^n = 1/2^n-1. p(A) = 1 - p(A) = 1 - (1/2^n-1).
A fair coin is flipped n times. Give an expression for each of the probabilities below as a function of n. Simplify your final expression as much as possible.
The number of heads is different from the number of tails. (Assume that n is even, otherwise the probability is one.)
Call the event A. The event A is that there are the same number of heads as tails (i.e., there are exactly n/2 heads and exactly n/2 tails). The number of sequences in which the number of heads and tails are the same is (nn/2). Therefore
p(A)=1−p(~A) = 1−[C(n, n/2) / 2^n]
A class with n kids lines up for recess. The order in which the kids line up is random with each ordering being equally likely. There are three kids in the class named Hubert, Celia and Felicity. The use of the word "or" in the description of the events, should be interpreted as the inclusive or. That is "A or B" means that A is true, B is true or both A and B are true.
What is the probability that Celia is first in line or Felicity is first in line?
Call C the event that Celia is first in line and F the event that Felicity is first in line. C and F are mutually exclusive, so p(C ∪ F) = p(C) + p(F). The size of the sample space is n!. The number of ways to order the kids in which Celia is first is (n-1)! because there are (n-1)! ways to order the rest of the kids besides Celia. Therefore p(C) = (n-1)!/n! = 1/n. The reasoning for F is the same, so p(F) = 1/n. Therefore, p(C ∪ F) = p(C) + p(F) = 1/n + 1/n = 2/n.
A class with n kids lines up for recess. The order in which the kids line up is random with each ordering being equally likely. There are three kids in the class named Hubert, Celia and Felicity. The use of the word "or" in the description of the events, should be interpreted as the inclusive or. That is "A or B" means that A is true, B is true or both A and B are true.
What is the probability that Celia is first in line or Felicity is last in line?
Let C be the event that Celia is first in line and F' be the event that Felicity is last in line. Using the inclusion-exclusion principle, p(C ∪ F') = p(C) + p(F') - p(C ∩ F'). Using the reasoning from the previous problem, p(C) = p(F') = 1/n. The number of ways to order the kids in which Celia is first and Felicity is last is (n-2)! because there are (n-2)! ways to order the rest of the kids, besides Celia and Felicity. Therefore p(C ∩ F') = (n-2)!/n! = 1/n(n-1).
p(C ∪ F') = p(C) + p(F') - p(C ∩ F') = 1/n + 1/n - 1/n(n-1) = (2n - 3)/n(n-1).
A class with n kids lines up for recess. The order in which the kids line up is random with each ordering being equally likely. There are three kids in the class named Hubert, Celia and Felicity. The use of the word "or" in the description of the events, should be interpreted as the inclusive or. That is "A or B" means that A is true, B is true or both A and B are true.
What is the probability that Celia is not next to Felicity in line?
Let N be the event that Celia is next to Felicity. The question asks for p(N) = 1 - p(N). There are 2(n-1)! ways to order the kids so that Celia is next to Felicity (first order all (n-1) kids besides Celia and then put Celia into the line either before or after Felicity). Therefore p(N) = 2(n-1)!/n! = 2/n. p(N) = 1 - 2/n.
A class with n kids lines up for recess. The order in which the kids line up is random with each ordering being equally likely. There are three kids in the class named Hubert, Celia and Felicity. The use of the word "or" in the description of the events, should be interpreted as the inclusive or. That is "A or B" means that A is true, B is true or both A and B are true.
What is the probability that Hubert is next to Celia or Felicity in line?
Let C be the event that Hubert is next to Celia. Let F be the event that Hubert is next to Felicity. Using the previous argument, p(C) = p(F) = 2/n.
The event (C ∩ F) is the event that Hubert is next to Celia and Felicity. The number of ways to order the kids so that Hubert is next to Celia and Felicity is 2(n-2)!. (Order all n-2 kids besides Celia and Felicity. Then put in Celia before Hubert and Felicity after Hubert or put in Celia after Hubert and Felicity before Hubert). Using the inclusion-exclusion principle:
p(C∪F) = p(C) + p(F)− p(C∩F) = 2 / n + 2/n − 2(n−2)!/n! = 2(2n−3) / n(n−1)
A class with n kids lines up for recess. The order in which the kids line up is random with each ordering being equally likely. There are three kids in the class named Hubert, Celia and Felicity. The use of the word "or" in the description of the events, should be interpreted as the inclusive or. That is "A or B" means that A is true, B is true or both A and B are true.
What is the probability that Hubert is not next to either Celia or Felicity in line?
The event that Hubert is not next to either Celia or Felicity is the complement of the event from the previous problem. Therefore the probability that Hubert is not next to Celia or Felicity is
1−[2(2n−3) / n(n−1)]
10 kids are randomly grouped into an A team with five kids and a B team with five kids. Each grouping is equally likely. There are three kids in the group, Alex and his two best friends Jose and Carl. What is the probability that Alex ends up on the same team with at least one of his two best friends?
Let J be the event that Alex ends up on the same team as Jose and C be the event that Alex ends up on the same team as Carl. The question asks for p(C ∪ J).
Alex and Carl can be either on the A team together or the B team together. If they are on the A team together, then there are (83) ways to select the other three members of the A team. Therefore p(C) = p(J) = 2 * C(8, 3) / C(10, 5)
The event C ∩ J is the event that Alex, Carl and Jose all end up on the same team together. The three could be on the A or B team (2 choices). Once the team is determined, there are C(7, 2) ways to select the other two members of their team. Therefore
p(C∪J) = p(C)+p(J)−p(C∩J) = 2⋅[2
C(8, 3) / (105)] − [2
] − [2 * C(7, 2) / C(10, 5)] = 13 / 18
The hand has at least one club.
Let C be the event that the hand has at least one club. p(C) = 1 - p(C). The event C is that there are no clubs in the hand. The number of 5-card hands with no clubs is (395) because there are 39 non-clubs from which to pick the five cards. The size of the sample space is (525), so
p(C) = 1−p(~C) = 1−C(39, 5) / C(52, 5)
The hand has at least two cards with the same rank.
Let S be the event that the hand has at least two cards with the same rank. p(S) = 1 - p(S). The event S is that every card has a different rank. There are C(13, 5) ways to pick the 5 different ranks of the cards in the hand. For each rank, there are four ways to select the suit for the card of that rank. Therefore
p(S) = 1−p(~S) = 1− [C(13, 5)⋅4^5 / C(52,5)]
The hand has exactly one club or exactly one spade.
Let C be the event that the hand has exactly one club and S be the event that the hand has exactly one spade. If the hand has exactly one club, then there are 13 ways to pick the club and (394) ways to select the other 4 cards from the non-clubs. Therefore
p(C) = p(S) = 13⋅C(39, 4) / C(52, 5)
If the hand has exactly one club and exactly one spade, there are 13 ways to select the club, 13 ways to select the spade, and C(26, 3) ways to select the other three cards from the cards in the deck that are not clubs or spades. Therefore p(C∪S) = 2⋅13⋅C(39, 4) − 13⋅13⋅C(26, 3) / C(52,5)
The hand has at least one club or at least one spade.
Let O be the event that the hand has at least one club or at least one spade. p(O) = 1 - p(~O). The event O is the event that the hand does not have any clubs or spades. There are C(26, 5) ways to select the five cards from the cards that are neither clubs nor spades. Therefore
p(O)=1−p(~O) = 1− C(26, 5) / C(52, 5)
A red and a blue die are thrown. Both dice are fair. The events A, B, and C are defined as follows:
A: The sum on the two dice is even
B: The sum on the two dice is at least 10
C: The red die comes up 5
Calculate the probability of each individual event. That is, calculate p(A), p(B), and p(C).
For every outcome of the red die, there are three outcomes of the blue die that cause the sum to be even. (If the red die is even, the blue die must also be even. If the red die is odd, the blue die must also be odd.) Therefore |A| = 18. The size of the sample space is 36, and p(A) = 18/36 = 1/2.
B = {(4, 6), (5, 5), (5, 6), (6, 4), (6, 5), (6, 6)}. Therefore |B| = 6, and p(B) = 6/36 = 1/6.
There are six possible outcomes for the blue die, so |C| = 6, and p(C) = 1/6.
A red and a blue die are thrown. Both dice are fair. The events A, B, and C are defined as follows:
A: The sum on the two dice is even
B: The sum on the two dice is at least 10
C: The red die comes up 5
What is p(A|C)?
A ∩ C = {(5, 1), (5, 3), (5, 5)}. Therefore, p(A|C) = |A ∩ C|/|C| = 3/6 = 1/2.
A red and a blue die are thrown. Both dice are fair. The events A, B, and C are defined as follows:
A: The sum on the two dice is even
B: The sum on the two dice is at least 10
C: The red die comes up 5
What is p(B|C)?
B ∩ C = {(5, 5), (5, 6)}. Therefore p(B|C) = |B ∩ C|/|C| = 2/6 = 1/3.
A red and a blue die are thrown. Both dice are fair. The events A, B, and C are defined as follows:
A: The sum on the two dice is even
B: The sum on the two dice is at least 10
C: The red die comes up 5
What is p(A|B)?
A ∩ B = {(4, 6), (5, 5), (6, 4), (6, 6)}. Therefore p(A|B) = |A ∩ B|/|B| = 4/6 = 2/3.
A red and a blue die are thrown. Both dice are fair. The events A, B, and C are defined as follows:
A: The sum on the two dice is even
B: The sum on the two dice is at least 10
C: The red die comes up 5
Which pairs of events among A, B, and C are independent?
A and C are independent. p(A|C) = p(A) = 1/2.
B and C are not independent. p(B|C) = 1/3 ≠ 1/6 = p(B).
A and B are not independent. p(A|B) = 2/3 ≠ 1/2 = p(A).
The letters {a, b, c, d, e, f, g} are put in a random order. Each permutation is equally likely. Define the following events:
A: The letter b falls in the middle (with three before it and three after it)
B: The letter c appears to the right of b, although c is not necessarily immediately to the right of b. For example, "agbdcef" would be an outcome in this event.
C: The letters "def" occur together in that order (e.g. "gdefbca")
Calculate the probability of each individual event. That is, calculate p(A), p(B), and p(C).
|A| = 6! because once the b is placed in the middle, there are 6! ways to permute the rest of the letters. The size of the sample space is 7!, so p(A) = 6!/7! = 1/7.
For every way to permute the letters with c to the right of b, there is exactly one way to permute the letters with c to the left of b (obtained by swapping the letters b and c). Therefore, in exactly half of the permutations, c appears to the right of b. |B| = 7!/2, p(B) = 7!/2·7! = 1/2.
Permute the 4 letters {a, b, c, g}. There are five possible places to put the def in the ordering of {a, b, c, g}. Therefore |C| = 5·4! = 5!. p(C) = 5!/7! = 1/42.
The letters {a, b, c, d, e, f, g} are put in a random order. Each permutation is equally likely. Define the following events:
A: The letter b falls in the middle (with three before it and three after it)
B: The letter c appears to the right of b, although c is not necessarily immediately to the right of b. For example, "agbdcef" would be an outcome in this event.
C: The letters "def" occur together in that order (e.g. "gdefbca")
What is p(A|C)?
The event A ∩ C is the set of orderings in which b is in the middle and def appear together. If the def is to the left of the b, then there are 3! ways to permute {a, c, g} to the right of the b. If the def is to the right of the b, then there are 3! ways to permute {a, c, g} to the left of the b. Therefore, |A ∩ C| = 2·3! = 12. p(A|C) = |A ∩ C|/|C| = 12/5! = 1/10.
The letters {a, b, c, d, e, f, g} are put in a random order. Each permutation is equally likely. Define the following events:
A: The letter b falls in the middle (with three before it and three after it)
B: The letter c appears to the right of b, although c is not necessarily immediately to the right of b. For example, "agbdcef" would be an outcome in this event.
C: The letters "def" occur together in that order (e.g. "gdefbca")
What is p(B|C)?
For every way to permute the letters with def occurring together and c to the right of b, there is exactly one way to permute the letters with def occurring together and c to the left of b (obtained by swapping b and c). Therefore, in exactly half of the permutations in which def occur together, c appears to the right of b. |B ∩ C| = 5!/2, p(B|C) = 5!/(2·5!) = 1/2.
The letters {a, b, c, d, e, f, g} are put in a random order. Each permutation is equally likely. Define the following events:
A: The letter b falls in the middle (with three before it and three after it)
B: The letter c appears to the right of b, although c is not necessarily immediately to the right of b. For example, "agbdcef" would be an outcome in this event.
C: The letters "def" occur together in that order (e.g. "gdefbca")
What is p(A|B)?
A ∩ B is the event that b is in the middle and c appears to the right of b. If b is in the middle, there are 3 possible locations for the c to the right of b. Once c is placed in one of those locations, there are 5! ways to permute the rest of the letters. Therefore |A ∩ B| = 3·5!. p(A|B) = |A ∩ B|/|B| = (3·5!)/(7!/2) = 3·2·5!/7! = 1/7.
The letters {a, b, c, d, e, f, g} are put in a random order. Each permutation is equally likely. Define the following events:
A: The letter b falls in the middle (with three before it and three after it)
B: The letter c appears to the right of b, although c is not necessarily immediately to the right of b. For example, "agbdcef" would be an outcome in this event.
C: The letters "def" occur together in that order (e.g. "gdefbca")
Which pairs of events among A, B, and C are independent?
A and C are not independent. p(A|C) = 1/10 ≠ 1/7 = p(A).
B and C are independent. p(B|C) = 1/2 = p(B).
A and B are independent. p(A|B) = 1/7 = p(A).
A wedding party of eight people is lined up in a random order. Every way of lining up the people in the wedding party is equally likely.
What is the probability that the bride is next to the groom?
The number of ways for the people to be ordered such that the bride is next to the groom is 2·7!. (There are 7! ways to order everyone except the groom, then there are two ways to place the groom - either to the left or the right of the bride). The size of the sample space is 8!. Therefore the probability that the bride is next to the groom is 2·7!/8! = 1/4.
A wedding party of eight people is lined up in a random order. Every way of lining up the people in the wedding party is equally likely.
What is the probability that the maid of honor is in the leftmost position?
Once the maid of honor is placed in the leftmost position, there are 7! ways to order the rest of the people, so the probability that the maid of honor is in the leftmost position is 7!/8! = 1/8.
A wedding party of eight people is lined up in a random order. Every way of lining up the people in the wedding party is equally likely.
Determine whether the two events are independent. Prove your answer by showing that one of the conditions for independence is either true or false.
The number of ways to order the people so that the maid of honor is in the leftmost position and the bride is next to the groom is 2·6!. (Place the maid of honor on the left, then permute the remaining 6 people besides the groom and then place the groom to the left or right of the bride). The probability that the maid of honor is in the leftmost position and the bride is next to the groom is 2·6!/8! = 1/28. Since 1/28 ≠ (1/4)(1/8), the two events are not independent.
10 kids are randomly grouped into an A team with five kids and a B team with five kids. Each grouping is equally likely. There are three kids in the group, Alex and his two best friends, Jose and Carl. Define the events J and C as:
J: Alex ends up on the same team as Jose
C: Alex ends up on the same team as Carl
Are the events J and C independent? Prove your answer by showing that one of the conditions for independence is either true or false.
The size of the sample space is C(10, 5). The number of ways to select the teams so that Alex and Jose are both on the A team is C(8, 3) because once Alex and Jose are placed on the A team, there are C(8, 3) ways to select the remaining three kids for the A team. A similar argument holds for Alex and Jose to be placed on the B team together. Therefore |J| = 2
C(8, 3) and p(J), the probability that Jose and Alex are on either team together, is 2
C(8, 3)/C(10, 5) = 4/9.
p(J|C) = |J ∩ C|/|C|. We can use the same reasoning above to conclude that |C| = 2
C(8, 3). J ∩ C is the event that Alex is on the same team with both Carl and Jose. The number of ways for Alex, Jose and Carl to be all on the same team is 2
C(7, 2). (There are two possible teams. Then once the three kids are placed on one of the two teams, there are C(7, 2) ways to select the two other kids for that team.) p(J|C) = |J ∩ C|/|C| which is 2
C(7, 2) / 2
/ 2 * C(8, 3) = 3/8.
p(J) = 4/9 ≠ 3/8 = p(J|C). The events J and C are not independent.
A 5-card hand is dealt from a perfectly shuffled deck. Define the events:
A: the hand is a four of a kind (all four cards of one rank plus a 5th card).
B: at least one of the cards in the hand is an ace
Are the events A and B independent? Prove your answer by showing that one of the conditions for independence is either true or false.
Events A and B are independent if and only if
|A||S|⋅|B||S|=|A∩B||S|
The equality holds if and only if |A|·|B| = |A ∩ B|·|S|.
|S| = C(52, 5).
|A| = 13·48. There are 13 ways to pick the rank for the four cards of the same rank. Then there are 48 cards left in the deck from which to select the 5th card.
|B| = C(52,5) − C(48,5). The number of 5-card hands that have no aces is C(48,5).
A ∩ B is the set of 5-card hands that have at least one ace and also have four cards of the same rank. There are two disjoint possibilities: the four of a kind are aces or the 5th card which is not part of the four of the same rank is an ace. There are 48 possibilities for the first case because there are 48 ways to select the 5th non-ace card. In the second case, there are 4 ways to select the ace and 12 ways to select the rank of the four cards that have the same rank, for a total of 4·12 = 48 hands for the second case. Since the two cases are disjoint, |A ∩ B| = 2·48.
Events A and B are not independent because
13⋅48(C(52,5)−C(48,5)) ≠ 2⋅48 * C(52,5)
A biased coin is flipped 10 times. In a single flip of the coin, the probability of heads is 1/3 and the probability of tails is 2/3. The outcomes of the coin flips are mutually independent. What is the probability of each event?
Every flip comes up heads.
(1/3)^10
A biased coin is flipped 10 times. In a single flip of the coin, the probability of heads is 1/3 and the probability of tails is 2/3. The outcomes of the coin flips are mutually independent. What is the probability of each event?
The first 5 rolls come up heads. The last 5 flips come up tails.
(1/3)^5*(2/3)^5
A biased coin is flipped 10 times. In a single flip of the coin, the probability of heads is 1/3 and the probability of tails is 2/3. The outcomes of the coin flips are mutually independent. What is the probability of each event?
The first roll comes up heads. The rest of the rolls come up tails.
(1/3)(2/3)^9
Sally has two coins. The first coin is a fair coin and the second coin is biased. The biased coin comes up heads with probability .75 and tails with probability .25. She selects a coin at random and flips the coin ten times. The results of the coin flips are mutually independent. The result of the 10 flips is: T,T,H,T,H,T,T,T,H,T. What is the probability that she selected the biased coin?
Let B be the event that Sally selects the biased coin. p(B) = p(B) = 1/2.
Let Y be the event that the outcome is T,T,H,T,H,T,T,T,H,T. Because the results of the flips are independent, p(Y|B) = (1/2)^10. Also p(Y|B) = (1/4)^7*(3/4)^3. Using Bayes' Theorem:
p(B|Y)=p(Y|B)p(B)p(Y|B)p(B)+p(Y|~B)p(~B) = (1/4)^7*(3/4)^3*(1/2) / (1/4)^7
(3/4)^3
(3/4)^3
Y|B)p(B)p(Y|B)p(B)+p(Y|~B)p(~B) = (1/4)^7*(3/4)^3*(1/2) / (1/4)^7*(3/4)^3*(1/2)+(1/2)^10*(1/2) = 27 / 27+2^10
Assume that you have two dice, one of which is fair, and the other is biased toward landing on six, so that 0.25 of the time it lands on six, and 0.15 of the time it lands on each of 1, 2, 3, 4 and 5. You choose a die at random, and roll it six times, getting the values 4, 3, 6, 6, 5, 5. What is the probability that the die you chose is the fair die? The outcomes of the rolls are mutually independent.
Let F be the event that the fair die is chosen. p(F) = p(F) = 1/2.
Let Y be the event that the outcome is 4, 3, 6, 6, 5, 5. Because the results of the flips are independent, p(Y|F) = (.15)^4*(.25)^2. Also p(Y|F) = (1/6)^6. Using Bayes' Theorem:
p(F|Y) = p(Y|F)p(F) / p(Y|F)p(F) + p(Y|~F)p(~F) = (1/6)^6
(1/2) / (1/6)^6(1/2) + (.15)^4
(.25)^2*(1/2)
The national flufferball association decides to implement a drug screening procedure to test its athletes for illegal performance enhancing drugs. 3% of the professional flufferball players actually use performance enhancing drugs. A test for the drugs has a false positive rate of 2% and a false negative rate of 4%. In other words, a person who does not take the drugs will test positive with probability 0.02. A person who does take the drugs will test negative with probability 0.04. A randomly selected player is tested and tests positive. What is the probability that she really does take performance enhancing drugs?
Let D be the event that the randomly chosen player takes drugs. Let T be the event that the test indicates that the player takes drugs. Because 3% of all flufferball players take drugs, p(D) = 0.03 and p(~D) = 0.97.
The false positive rate for the test is 2%, so p(T|~D) = 0.02. The false negative rate for the test is 4%, so p(~T|D) = 0.04. Using the fact that p(T|D) + p(~T|D) = 1, p(T|D) = 0.96.
p(D|T) = p(T|D)p(D) / p(T|D)p(D)+p(T|~D)p(~D) = (0.96)(0.03) / (0.96)(0.03) + (0.02)(0.97)
Assume one person out of 10,000 is infected with HIV, and there is a test in which 2.5% of all people test positive for the virus although they do not really have it. If you test negative on this test, then you definitely do not have HIV. What is the chance of having HIV, assuming you test positive for it?
Let D be the event that a person has HIV. Let T be the event that the person tests positive for HIV. Since one out of every 10,000 people are infected with HIV, a randomly selected person has the disease with probability p(D) = 0.0001. Therefore p(~D) = 0.9999.
The false positive rate p(T|~D) = 0.025.
The problem states that if a person tests negative, then that person definitely does not have HIV. Therefore, it is impossible for a person to test negative and have HIV, so p(~T ∩ D) = 0. Since p(~T|D) = p(~T ∩ D)/p(D), then p(~T|D) = 0. Since p(~T|D) + p(T|D) = 1, we know that p(T|D) = 1.
p(D|T) = p(T|D)p(D) / p(T|D)p(D) + p(T|~D)p(~D) = (0.0001) / (0.0001)+(0.025)(0.9999)
Consider an experiment in which a red die and a blue die are thrown. Let X be the random variable whose value is the product of the numbers on the red and blue dice.
What is the range of X?
{1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, 16, 18, 20, 24, 25, 30, 36}
Consider an experiment in which a red die and a blue die are thrown. Let X be the random variable whose value is the product of the numbers on the red and blue dice.
What is the probability that X = 6?
There are four outcomes for which X = 6: {(2, 3), (1, 6), (3, 2), (6, 1)}. Therefore, p(X = 6) = 4/36 = 1/9.
A hand of 5 cards is dealt from a perfectly shuffled deck of playing cards. Let the random variable A denote the number of aces in the hand.
What is the range of A?
{0, 1, 2, 3, 4}
A hand of 5 cards is dealt from a perfectly shuffled deck of playing cards. Let the random variable A denote the number of aces in the hand.
Give the distribution over the random variable A.
see 5.5.2
Two student council representatives are chosen at random from a group of 7 girls and 3 boys. Let G denote the distribution over the number of girls chosen.
What is the range of G?
{0, 1, 2}
Two student council representatives are chosen at random from a group of 7 girls and 3 boys. Let G denote the distribution over the number of girls chosen.
Give the distribution over the random variable G.
(0,1/15), (1,7/15), (2,7/15)
In a network of 40 computers, 5 hold a copy of a particular file. Suppose that 7 computers at random fail. Let F denote the number of computers that fail and have a copy of the file.
What is the range of F?
{0, 1, 2, 3, 4, 5}
In a network of 40 computers, 5 hold a copy of a particular file. Suppose that 7 computers at random fail. Let F denote the number of computers that fail and have a copy of the file.
What is the probability that F = 2?
C(5,2)*C(35, 5) / C(40,7)
A coin is tossed ten times. Let X be the random variable denoting the number of heads minus the number of tails.
{-10, -8, -6, -4, -2, 0, 2, 4, 6, 8, 10}
A coin is tossed ten times. Let X be the random variable denoting the number of heads minus the number of tails.
What is the distribution over X?
see 5.5.5
Two student council representatives are chosen at random from a group of 7 girls and 3 boys. Let G be the random variable denoting the number of girls chosen. What is E[G]?
The size of the sample space is C(10, 2) = 45. The number of ways to pick the two representatives in which both are girls is C(7, 2) = 21, so p(G = 2) = 21/45 = 7/15. The number of ways to pick the two representatives in which there is one girl and one boy is 7·3 = 21, so p(G = 1) = 21/45 = 7/15. The number of ways to pick the representatives so that no girls are chosen is C(3,2) = 3, so p(G = 0) = 3/45 = 1/15.
E[G] = 2·(7/15) + 1·(7/15) + 0·(1/15) = 21/15 = 7/5.
Consider a game in which a fair die is rolled. If the die comes up 1, the player wins $2. If the die comes up 2, the player wins $1. For all other outcomes, the player loses $1. What is the expected amount that the player wins or loses? Round to the nearest cent.
2·(1/6) + 1·(1/6) + (-1)·(4/6) ≈ -.167
The player loses 17 cents.
A lottery is run in which every ticket has six numbers in a particular order. Each of the six numbers is in the range from 1 through 50. The person purchasing a lottery ticket can select the numbers on their ticket. On each Friday of the week, the state lottery commission holds a drawing in which six random numbers are generated. Each number in the range from 1 through 50 is equally likely.
The order of the numbers matters, so if the random numbers selected are (25, 7, 12, 37, 2, 19) then the ticket (7, 25, 12, 37, 20, 19) matches in location 3, 4, and 6. If a ticket matches in all six locations, then the ticket holder wins $100,000,000. If the ticket matches in five of the six locations, then the ticket holder wins $1,000,000. The ticket holder does not win any money for any of the other outcomes. What are the expected winnings?
The size of the sample space is (50)6. For any particular ticket, there is only one way to draw the numbers so that the ticket is worth $100,000,000, so the probability that the ticket is worth $100,000,000 is 1/(50)^6.
The number of ways to pick the numbers so that a particular ticket is worth $1,000,000 is 6·49. First select the location that does not match (6 choices). There are 49 ways to select a number that does not match. The rest of the numbers are determined. The probability that the ticket is worth $1,000,000 is 6·49/(50)^6.
Let W be the random variable denoting the amount the ticket holder wins. The expected winnings are
E[W]=(100,000,000)⋅(1/50^6)+(1,000,000)⋅(294/50^6)+0⋅(1−1/50^6− 294/506)
A fair die is rolled once. Let X be the random variable that denotes the square of the number that shows up on the die. For example, if the die comes up 5, then X = 25. What is E[X]?
E[X] = 1·(1/6) + 4·(1/6) + 9·(1/6) + 16·(1/6) + 25·(1/6) + 36·(1/6) = 91/6.
A fair coin is tossed three times. Let Y be the random variable that denotes the square of the number of heads. For example, in the outcome HTH, there are two heads and Y = 4. What is E[Y]?
For {TTT}, Y = 0. For {HTT, THT, TTH}, Y = 1. For {HHT, HTH, THH}, Y = 4. For {HHH}, Y = 9.
E[Y] = 0·(1/8) + 1·(3/8) + 4·(3/8) + 9·(1/8) = 24/8 = 3.
don't forget to study bernoulli
yup, you aint done, biych
Consider a game in which a fair die is rolled. If the die comes up 1, the player wins $2. If the die comes up 2, the player wins $1. For all other outcomes, the player loses $1. What is the expected amount that the player wins or loses? Round to the nearest cent.
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