and 1 P(A brown-eyed, and a blue-eyed parent have a blue-eyed child) = 2
γ γ + γ + α
.
(Communicati Communication on through a noisy channel) A transmitter wishes to send one of two alternative messages, “a” or “b”. Because of the way these messages originate, it is known that “b” messages are twice as likely as “a” messages. These messages are encoded into binary messages (codewords), in order to be transmitted over a digital channel. channel. Assume Assume that the two two codewords are 0110 and 1000, respectively respectively.. However, the channel is noisy and each bit transmitted may be received incorrectly, according to the probabilities shown below. For example, if a zero is sent, there is probability ability 0.2 that a 1 is received. Assume Assume that errors during the transmission transmission of different different bits are statistic statistically ally independent. independent. Given Given that the receiver receiver received received the sequence 0001, find the probability that message “a” was transmitted. Problem 25.
Homework 3.2
We are given given three coins. The first coin is a fair coin painted blue on the head side and white on the tail side. The other two coins are biased so that the probability of a head is p. They are painted blue on the tail side and red on the head side. side. Two Two of the three three coins are to be select selected ed at random random and tossed. tossed. Descri Describe be the outcomes outcomes in the sample space. It was experimenta experimentally lly determined determined that the probabilit probability y that the sides that land face up are of the same color is 29/ 29/96. What are the possible values of p? Problem 26.
There are two stages to the experiment: experiment: the selection selection of the two coins and the flipping of the coins. There are three different ways that two coins can be selected: the 1st and 2nd, the 1st and 3rd, and the 2nd and 3rd. Each of these pairs are equally likely likely to be select selected. ed. For each pair, pair, the flips have have four four possibl possiblee outcom outcomes: es: (heads (heads,, heads), heads), (heads, (heads, tails), (tails, heads), heads), (tails, (tails, tails). The sample space can be described described as follows: Solution.
The probability that the sides that land face up are the same is then:
P(same) = P (blue,blue) + P (red,red)
1 1 1 (1 − p) p) + (1 − p) p) + p + p2 + (1 − p) p)2 3 2 2 1 = (2 p2 − 3 p + 2). 2). 3 =
Setting this equal to p = 7/8.
29 96
and solving the quadratic equation for p yields p = 5/8 or
A certain test for a disease is only 60% accurate. accurate. In order to arrive at a somewhat trustworthy result, a blood sample is repeatedly tested until there are 4 test outcomes that indicate indicate the same result. Given Given that a particular particular blood sample came from a patient carrying the disease, find the probability that in exactly i tests, for i = 4, 5, 6, 7, the patient’s blood will be found to have the disease.
Problem 27.
14
heads, heads
blue, red
heads, tails
blue, blue
tails, heads
white, red
tails, tails
white, blue
(1/2)p
heads, heads
blue, red
(1/2)(1-p)
heads, tails
blue, blue
tails, heads
white, red
tails, tails
white, blue
heads, heads
red, red
heads, tails
red, blue
tails, heads
blue, red
tails, tails
blue, blue
(1/2)p (1/2)(1-p) 1&2
*
(1/2)p (1/2)(1-p)
1/3
1/3
1&3
*
(1/2)p (1/2)(1-p)
1/3
pp p(1-p) 2&3
*
(1-p)p (1-p)(1-p)
Figure 0.7:
15
*
Solution.We
use the binomial probabilities:
P(correct diagnosis in 4 tests) = P(correct diagnosis in 5 tests) = P(correct diagnosis in 6 tests) = P(correct diagnosis in 7 tests) =
4 4
· (0.6)4 = 0.1296,
5 4
· (0.6)4 · (0.4)1 = 0.2592,
6 4
· (0.6)4 · (0.4)2 = 0.31104,
7 4
· (0.6)4 · (0.4)3 = 0.29.
A new robot is equipped with the brains to solve any kind of problem set! Due to bugs in the design, however, and independent of the difficulty of problems, this robot gets problems right with probability 0.65, independently for each problem. This robot is very expensive, and you have to decide whether to keep it or not. You decide that if at any point the number of wrong problems exceeds the number of correct problems by more than 15, you will return the robot. If, however, at any point the number of correct problems exceeds the number of incorrect problems by 15, then you will keep the robot. What are the chances that you will return the robot? Hint : Use the solution to the gambler’s ruin problem. Problem 28.
We can reduce this problem to the gambler’s ruin problem with N = 30, k = 15, and p = 0.65. The event equivalent to returning the robot is that of G1 winning. The probability of this event is Solution.
P 15 =
Homework 3.3
1 − (0.35/0.65)15 ≈ 9.3 × 10−5 . 1 − (0.35/0.65)30
A particular jury consists of 7 jurors. Each juror has a 0.2 chance of making the wrong decision, independently of the others. If the jury reaches a decision by majority rule, what is the probability that it will reach a wrong decision? Problem 29.
Solution.The
number of jurors that make the “wrong” decision can be modeled as a binomial random variable with parameters n = 7 and p = .2. The jury as a whole will make the wrong decision if 4, 5, 6, or 7 jurors make the wrong decision. Denote these events by A, B, C , D, respectively. Since these events are mutually exclusive, the probability of their union is the sum of their probabilities, so P(Jury Error) = P(A) + P(B) + P(C ) + P(D)
=
7 (0.2)4 · (0.8)3 + 4
+
7 (0.2)6 · (0.8)1 + 6
= 0.033. 16
7 (0.2)5 · (0.8)2 5 7 (0.2)7 · (0.8)0 7
Problem 30.
Galton’s quincunx: Consider a mechanical device in which balls are
dropped through a triangular array of nails, starting at the top. The nails are placed in rows, with each nail having two nails symmetrically placed in the row underneath it. There is a total of n rows. Every time a ball hits a nail it has a probability 0.5 to fall to the left of the nail and a probability of 0.5 to fall to the right of the nail. The ball comes to rest at the nth row. What is the probability that the ball rests to the left of the kth nail of the nth row? Problem 31. Calculating the odds. Let A be an event such that 0 < P(A) < 1. The odds in favor of A are defined to be O(A) =
P(A) , P(Ac )
while the odds against A are defined to be the reciprocal of O(A). [To connect the term “odds” with its common usage, note for example that if the probability that a given horse wins a race at the track is 1/3, the odds against the horse winning are 2 to 1. A “fair” racetrack would then pay $2 for every $1 bet on the horse (plus the original $1 bet), if the horse wins; “fair” here means that the racetrack would break even on the average – this will become more precise in Chapter 2, when we will discuss the notion of expected value.] This problem deals with a formula for calculating “conditional odds,” that is, odds based on some partial information. If A and B are events with P(A) > 0 and P(B) > 0, the odds in favor of A given B are defined as O(A | B) =
P(A | B) . P(Ac | B)
Show that O(A | B) = L(B | A)O(A),
where L(B | A) is the so called likelihood ratio of B given A, defined as L(B | A) =
Solution.By
P(B | A) . P(B | Ac )
definition, O(A | B) is equal to
P(A | B) P(A ∩ B)/P(B) P(A ∩ B) P(B | A)P(A) = = = = L(B | A)O(A). P(Ac | B) P(Ac ∩ B)/P(B) P(Ac ∩ B) P(B | Ac )P(Ac )
Hypothesis testing. May B. Lucky is a compulsive gambler who is convinced that on any given day she is either “lucky,” in which case she wins each red/black bet she makes in the roulette with probability pL > 1/2, or she is “unlucky,” in which case she wins each red/black bet she makes in the roulette with probability pU < 1/2. May visits the casino every day, and believes that she knows the a priori probability that any one given visit is a “lucky” one (i.e., corresponds to pL rather than pU ). To improve her chances, May adopts a system whereby she estimates on-line whether she is lucky or unlucky on a given day, by keeping a running count of the numbers of bets that she wins and loses. In particular, she continues to play until the Problem 32.
17
conditional odds in favor of the event {lucky on the current day}, given the number of wins and losses so far, fall below a certain threshold (see the preceding problem). As soon as this happens, she stops playing. Provide a simple algorithm for updating May’s conditional odds with each play. Note : This example is typical of reasoning in sequential hypothesis testing systems, where the probability of correctness of a certain hypothesis, given some evidence, is calculated and sequentially updated. Solution.Let
A be the event that May is lucky on the current day, and let Bm,n be the event that m wins and n losses have occurred so far. We assume independence of the results of different spins/plays. Then we have, using the odds formula of the preceding problem and the binomial formula, O(A | Bm,n ) P(Bm,n | A) = L(Bm,n | A) = = O(A) P(Bm,n | Ac )
m+n m m+n m
n pm L (1 − pL )
n pm U (1 − pU )
=
m
pL pU
1 − pL 1 − pU
From this formula, a convenient recursive algorithm is obtained. After m + n plays, in which m wins and n losses occurred, we have pL , pU
O(A | Bm+1,n ) = O(A | Bm,n )
O(A | Bm,n+1 ) = O(A | Bm,n )
1 − pL , 1 − pU
if she wins in the next play,
if she loses in the next play.
The initial condition is O(A | B0,0 ) is equal to the initial (unconditional) odds O(A) (which May knows by assumption). Homework 3.4
Problem 33. *
Let A and B be events such that A ⊂ B. Can A and B be
independent? Solution.The
events A and B are independent if and only if P(A)P(B) = P(A ∩ B) = P(A), where the last equality follows from the fact that A ⊂ B. This can be the case if and only if P(A) = 0 or P(B) = 1. Problem 34. * Alice starts by flipping a coin until she obtains tails for the first time.
After that, Bob starts flipping until he obtains tails for the first time, and they keep alternating similarly. Assume that coin flips are independent and that the probability of heads at each flip is p. The game ends when either Alice has accumulated m heads, in which case she wins, or when Bob has accumulated n heads, in which case he wins, whichever comes first. Show that the probability qm,n that Alice wins satisfies qm,n = pqm−1,n + (1 − p)(1 − qn,m ). Solution.
SECTION 1.6. Counting A parking lot contains 100 cars that all look quite nice from the outside. However, K of these cars happen to be lemons. The number K is known to lie in the range {0, 1, . . . , 9}, with all values equally likely. Problem 35.
18
n
(a) We testdrive 20 distinct cars chosen at random, and to our pleasant surprise, none of them turns out to be a lemon. Given this knowledge, what is the probability that K = 0? (b) Repeat part (a) when the 20 cars are chosen with replacement; that is, at each testdrive, each car is equally likely to be selected, including those that were selected earlier. Solution.(a)
Let A be the event that all 20 cars tested are good. We are asked to find P(K = 0 | A). Using Bayes’ rule, we have P(K = 0 | A) =
P(K = 0)P(A | K = 0) 9
i=0
P(K = i)P(A | K = i)
.
It is given that P(K = i) = 1/10 for all i. To compute P(A | K = i), we condition on the event of exactly i lemons, and reason as follows. The first selected car has probability (100 − i)/100 of being good. Having succeeded in the first selection, we are left with 99 cars out of which i are lemons; thus, the second selected car has probability (99 − i)/99 of being good. Continuing similarly, and using the multiplication rule, we obtain (100 − i)(99 − i) · · · (81 − i) P(A | K = i) = , 100 · 99 · 81 from which we can then obtain P(K = 0 | A). (b) We use the exact same argument as in part (a), except that we need to recalculate P(A | K = i). Since the cars are chosen with replacement, we are dealing with 20 independent Bernoulli trials. The probability of finding a good car in any one trial is (100 − i)/100. The probability of finding good cars in all 20 trials is P(A | k = i) =
100 − i 100
20
,
from which we can then obtain P(K = 0 | A). A certain department offers 8 lower level courses: {L1 , L2 , . . . , L8 } and 10 higher level courses: {H 1 , H 2 , . . . , H10 }. A valid curriculum consists of 4 lower level courses, and 3 higher level courses. Problem 36.
(a) How many different curricula are possible? (b) Suppose that {H 1 , . . . , H5 } have L1 as a prerequisite and {H 6 , . . . H10 } have L2 and L3 as prerequisites, i.e., any curricula which involve, say, one of {H 1 , . . . , H5 } must also include L1 . How many different curricula are there?
There are 84 ways to pick 4 lower level classes, and 3 higher level classes, so there are Solution.(a)
10 3
ways to choose
8 4
10 3
valid curricula. (b) This part is more involved. We need to consider several different cases 19
(i) Suppose we do not choose L1 . Then both L2 and L3 must be chosen; otherwise no higher level courses would be allowed. Thus, we need to choose 2 more lower level classes out of the remaining 5, and 3 higher level classes from the available 5. We then obtain 52 53 valid curricula.
(ii) If we choose L1 but choose neither L2 nor L3 , we have
5 3
5 3
choices.
(iii) If we choose L1 and choose one of L2 or L3 , we have 2 · 52 53 choices. This is because there are two ways of choosing between L2 and L3 , 52 ways of choosing 2 lower level classes from L3 , . . . , L8 , and 53 ways of choosing 3 higher level classes from H 1 , . . . , H5 .
(iv) Finally, if we choose L1 , L2 , and L3 , we have
5 1
10 3
choices.
Note that we are not double counting, because there is no overlap in the cases we are considering, and furthermore we have considered every possible choice. The total is obtained by adding the counts for the above four cases. How many 6-word sentences can be made using each of the 26 letters of the alphabet exactly once? A word is defined as a nonempty (possibly jibberish) sequence of letters. Problem 37.
Solution.Think
of a 6-word sequence using the 26 letters as a sequence of length 31: there are 26 letters, plus five blanks to separate one word from the next. Note that the blanks cannot be at the beginning or the end of the sequence, which only leaves 29 available positions. Thus, there are 29 possible choices for the locations of the blanks. 5 Then, the 26 letters can be placed in arbitrary order (permutation) in the remaining 26 locations (26! choices). Thus, the number of possible sentences is 26! 29 . 5
A candy factory has an endless supply of red, orange, yellow, green, blue, and violet jelly beans. The factory packages the jelly beans into jars of 100 jelly beans each. One possible color distribution, for example, is a jar of 56 red, 22 yellow, and 20 green jelly beans. As a marketing gimmick, the factory guarantees that no two jars have the same color distribution. What is the maximum number of jars the factory can produce? Problem 38.
Solution.Think
of lining up the jelly beans, by first placing the red ones, then the orange ones, etc. We also place 5 dividers to indicate where one color ends and another starts. (Note that two dividers can be adjacent if there are no jelly beans of some color.) By considering both jelly beans and dividers, we see that there is a total of 105 positions. Choosing the number of jelly beans of each color is the same as choosing the positions of the dividers. Thus, there are 105 possibilities, and this is the number of possible 5 jars.
We have m married couples (2m individuals). After a number of years, each person has died, independently, with probability p. Let N be the number of surviving individuals. Let C be the number of couples in which both individuals are alive. Find the conditional probability P(C = c | N = n). Problem 39.
Solution.Use
Bayes rule. Find P(N = n | C = c) first. The c couples give 2c alive individuals. Then find the number of ways that we can have n − 2c alive ones in the remaining couples without having any alive couples.NEEDS TO BE WORKED OUT - TRICKY
20
Homework 3.5
Problem 40.
Consider three independent rolls of a fair six-sided die.
(a) What is the probability that the sum of the three rolls is 11? (b) What is the probability that the sum of the three rolls is 12? (c) In the seventeenth century, Galileo explained the experimental observation that a sum of 10 is more frequent than a sum of 9, even though both 10 and 9 can be obtained in six distinct ways. Can you retrace Galileo’s thinking? (a) The number of outcomes that leads to a sum of 11 is the number of outcomes of the first two rolls that lead to a sum greater than or equal to 11 − 6 = 5 and less than or equal to 11 − 1 = 10. The number of outcomes of the first two rolls that leads to sum less than 5 or greater than 10 is 9. So the desired probability is Solution.
62 − 9 . 63 (b) The number of outcomes that leads to a sum of 12 is the number of outcomes of the first two rolls that lead to a sum greater or equal to 12 − 6 = 6 and less than or equal to 12 − 1 = 11. The number of outcomes of the first two rolls that leads to sum less than 6 or greater than 11 is 11. So the desired probability is 62 − 11 . 63 (c) Each of the sums 9 and 10 can be obtained in six distinct ways: 9 = 1 + 2 + 6 = 1 + 3 + 5 = 1 + 4 + 4 = 2 + 2 + 5 = 2 + 3 + 4 = 3 + 3 + 3, 10 = 1 + 3 + 6 = 1 + 4 + 5 = 2 + 2 + 6 = 2 + 3 + 5 = 2 + 4 + 4 = 3 + 3 + 4. However, the number of outcomes that sum to 9 is 25, while the number of outcomes that sum to 10 is 27. Thus, a sum of 10 has probability 27/63 and is more frequent than a sum of 9, which has probability 25/63 . The weather on any given day can be sunny, cloudy, rainy, or snowy. Assume that a snowy day can happen only during the winter, that a rainy day cannot happen in the summer, and that each season has 90 days. What is the number of all possible distinct 360-day weather sequences? Problem 41.
Solution.
The number of possible sequences is (4 − 1)90490 (4 − 1)90 (4 − 2)90.
Consider a backgammon match with 25 games, each of which can have one of two outcomes: win (1 point), or loss (0 points). Find the number of all possible distinct score sequences under the following alternative assumptions. Problem 42.
(a) All 25 games are played. (b) The match is stopped when one player reaches 13 points. Solution.
(a) 225 .
(b) First note that under this rule, each match will be stopped after a number of games ranging from 13 to 25. If a match will be stopped at the k’th game with player 1 having 21
13 points, then the last game was a win and k − 13 of the previous games was a loss. 1 So, there are kk−−13 matches that ends at the k’th game with player 1 having a score of 13. Taking player 2 into consideration and summing over k, we obtain
25
2
k−1 k − 13
k=13
possible distinct score sequences. Alice and Bob each have a deck of playing cards. Each turns over a randomly selected card. Determine the following: Problem 43.
Homework 3.6
(a) The probability that the two cards are of the same suit. (b) The probability that at least one card is a spade. (c) The probability that neither card is a diamond. (d) The probability that both cards are hearts or spades. Solution.
(a) P(the two cards are of the same suit) = 4
13 52
2
1 4
=4
2
.
(b) P(at least one card is a spade) = 1 − P(neither card is a spade)
3 · 13 52
2
= 1− (c)
P(neither card is a diamond) =
3 · 13 52
2
3 4
(d) P(both cards are hearts or spades) =
Homework 3.7
3 4
2
= 1−
=
2 · 14 52
2
2
.
1 2
2
=
.
A parking lot contains 100 cars, k of which happen to be lemons. N of these cars are randomly selected and taken for a testdrive. Find the probability that n of these cars turn out to be lemons. Problem 44.
Solution.Clearly
if n > N or if n > k the probability must be zero. If n ≤ N and n ≤ k, then we can find the probability that the testdrive found n of the 100 cars defective by counting the total number of size N subsets, and then the number of size N subsets that contain n lemons. Clearly, there are 100 different subsets of size N . To count the N number of size N subsets with n lemons, we first choose the lemons from the k total lemons, and then choose the N − n good cars from the remaining 100 − k cars. Thus the number of ways to choose a subset of size N from 100 cars, and get n lemons, is
k n
·
100 − k N − n
22
and thus the desired probability is k n
100−k
·
N −n
100
N
Homework 3.8
Ninety students, including Joe and Jane, are to be split into three classes of equal size, and this is to b e done at random. What is the probability that Joe and Jane end up in the same class? Problem 45.
Suppose we label the classes A, B, and C . The probability that Joe and Jane will both be in class A is the number of possible combinations for class A that involve both Joe and Jane, divided by the total number of combinations for class A. Therefore the desired probability is Solution.
88 28 90 30
.
Since there are three classrooms, the probability that Joe and Jane end up in the same classroom is simply three times the answer we found above: 3·
88 28 90 30
.
Find the probability of winning in the following lottery. You choose 5 distinct integers in the range from 1 to 100. Then, the lottery operator chooses randomly 10 distinct integers in the same range (all outcomes being equally likely). You win if all of your 5 numbers are among those chosen by the operator. Problem 46.
A bank has a vault with a combination lock. The lock has a combination that consists of 10 integers in increasing order ranging from 1 to 90, but will open if any 8 of the numbers are entered. Find the probability that a burglar will open the vault on the first try. Problem 47.
Solution.To
find the probability, we will find the number of favorable outcomes, and divide by the total number of possible outcomes. There are 10 favorable outcomes, 8 i.e., successful combinations that will open the lock. There are 90 total number of 8 ways to choose 8 numbers out of 90, and therefore the probability that the burglar will open the vault on his first try is
10 8 90 8
Homework 3.9
.
A standard 52-card deck is distributed between 4 players (as in bridge). Find the probability that: Problem 48.
(a) Player 1 gets all 13 spades. (b) Some player gets all 13 spades. Solution.(a)
P(A) =
1 52
( )
= 1.57 × 10−12.
13
(b) P(B) = 4P(A) =
4 52
( )
.
13
23
