Combinatorial Probability
3.3 Applications of Counting Rules to Probability
3.56 Let X be the number of people (out of 20) who are cured. Note that the outcomes in the sample space Ω are equally likely. Then, according to the 1st method, (if N (E) denotes the number of outcomes in E)
P (X ≥ 15) = N ({X ≥ 15}) where the numerator is obtained by computing the number of ways to pick a set of k people, who are cured, from among 20 (which is%20
k
&
) and then summing the results over k = 15, · · · , 20.
The 2nd method gives:
P (X ≥ 15) =
20, 825 ≈ 0.000048, since there are 13 possibilities for the choice of the denom-ination and then there is just one way to pick 4 cards of that denomdenom-ination, while %52
4
&
is the number of ways to pick a set of four cards out of 52 and represents the total number of possible outcomes in the sample space.
b)
ways to pick four different denominations (out of a total of 13), and then there are 4 ways in which one can pick specific suits for each selected denomination. On the other hand, %52
4
&
represents the total number of ways to select a set of four different cards from the deck of 52.
3.58 1 −
45 ≈ 0.378, where we used the complementation rule and the fact that there are %8
2
&
ways to pick two winning tickets from among the eight that are not yours.
3.59 a) 1/13 ≈ 0.0769, since, by symmetry of the problem, the 7th card is equally likely to be of any one of the 13 denominations.
b) (48)6× 4
(52)7 ≈ 0.0524, since there are (48)6 ways to select the first six cards from the 48
“non-aces” (taking into account the order), and for each choice of those six cards there are 4 possibilities for the ace chosen next; on the other hand, (52)7 represents the total number of possible ordered selections of seven cards from the deck of 52, and is the number of outcomes in the sample space.
3.60
c) The required probabilities for N = 1, · · · , 70 are given in the following table:
N 1 2 3 4 5 6 7 8 9 10
Prob. 0 0.003 0.008 0.016 0.027 0.040 0.056 0.074 0.095 0.117
N 11 12 13 14 15 16 17 18 19 20
Prob. 0.141 0.167 0.194 0.223 0.253 0.284 0.315 0.347 0.379 0.411
N 21 22 23 24 25 26 27 28 29 30
Prob. 0.444 0.476 0.507 0.538 0.569 0.598 0.627 0.654 0.681 0.706
N 31 32 33 34 35 36 37 38 39 40
Prob. 0.730 0.753 0.775 0.795 0.814 0.832 0.849 0.864 0.878 0.891
N 41 42 43 44 45 46 47 48 49 50
Prob. 0.903 0.914 0.924 0.933 0.941 0.948 0.955 0.961 0.966 0.970
N 51 52 53 54 55 56 57 58 59 60
Prob. 0.974 0.978 0.981 0.984 0.986 0.988 0.990 0.992 0.993 0.994
N 61 62 63 64 65 66 67 68 69 70
Prob. 0.995 0.996 0.997 0.997 0.998 0.998 0.998 0.999 0.999 0.999 d) N = 23, by using the table in (c).
3.62 a)
%6
2
&
(6)4(4)3
(10)7 = 3
14 ≈ 0.2143; Note that a total of seven balls are drawn if and only if there are two red balls among the first six balls selected and the seventh ball is red. Let Ω consist of all ordered selections of seven balls chosen without replacement from a set of 10 (distinguishable) balls (where, say, ball #1 through ball #4 are red, while the rest are black).
Then N (Ω) = (10)7. On the other hand, there are %6
2
&
ways to decide which two of the first six balls selected are red. Moreover, for a given order in which red and black balls appear, there are (4)3 ways to pick the specific balls for the red ones, and there are (6)4 ways to pick the specific balls for the black balls.
b)
%6
2
&
6443
107 ≈ 0.1244; Here Ω is defined as in (a) but the selection is with replacement, resulting in N (Ω) = 107. As in (a), there are%6
2
&
ways to decide which two of the first six balls selected are red, but now, for a given order in which red and black balls are chosen, there are 43 ways to pick the specific red balls and 64 ways to pick the specific black ones.
3.63 3!4!3!5!
12! = 1
4620 ≈ 0.000216, since there are 3! ways of permuting the disciplines, 4! ways of permuting mathematicians among themselves, 3! ways of permuting chemists among themselves and 5! ways of permuting physicists among themselves.
3.64
%k
k
&%N−k
n−k
&
%N
n
& =
%N−k
n−k
&
%N
n
& .
3.65 a) (N )n
Nn = N (N − 1) · · · (N − (n − 1))
Nn .
b) For each n ∈ N ,
Thus, when a random sample is selected (with replacement) and the population size is large relative to the sample size, any member of the population is unlikely to be chosen more than once.
ways to select an un-ordered sample of size n (without replacement) containing the specified member, implying that there are n!%1
1
&%N−1
n−1
&
such ordered samples; on the other hand, the sample space consists of all ordered samples of size n taken (without replacement) from a population of size N , implying that N (Ω) = (N )n.
b) By symmetry of the problem, since all the members of the population are equally likely to be selected, the kth member selected is equally likely to be any one of the N possible mem-bers of the population. Thus, P (Ak) = 1/N . Since A1, A2, · · · , An are mutually exclusive and
3.68 a) By Example 3.21(c), for E =“At least one woman (among N ) gets her own key,” the required probability is given by:
P (E) =
c) The approximation error at N = 6 does not exceed 0.0002, and one can show that it decreases further quite rapidly as N grows larger. Thus,the approximation is extremely good for N ≥ 6.
3.69 a) 10!=3,628,800. b) 1/(10!) ≈ 2.756 × 10−7.
3.70 a) (N − 1)(N − 2) · · · (N − (n − 1)) · 1
N (N − 1) · · · (N − (n − 1)) = (N − 1)n−1 (N )n = 1
N.
b) Arrange the keys in a line in the order in which they will be tried. Then the correct key is equally likely to be in any one of the N positions, thus, the required probability is 1/N . c) (i) 1 − P (first n keys are incorrect) = 1 − (N − 1)n probability is given by:
P ( wrong keys in the first (n − 1) attempts, while Nn is the total number of ways to select n keys out of N with replacement, when the order of selection matters.
b) (i) 1 − P (first n keys are incorrect) = 1 − (N − 1)n
Then the required probability is given by:
P (
b) Let X be the number of questions that the student answers correctly. Then
P (X ≥ 9) =
3.73 a) Take the answers to Exercise 3.36 and divide by%52
5
&
= 2598960 to obtain the required probabilities. Namely,
P (straight flush) = 40
%52 P (straight) = 10200
%52
P (two pair) = 123552
%52
5
& ≈ 0.04754, P (one pair) = 1098240
%52
5
& ≈ 0.422557
b) No, the answers will be the same. The number of five-card poker studs, counted in each event, is equal to the number of respective five-card draw poker hands times 5!. On the other hand, the total number of possible five-card poker studs is equal to the 5! times the total number of possible five-card draw poker hands. The same factor 5!, that enters both the numerator and denominator (when we use equation (3.4) in the text), will cancel out.
3.74
184756 ≈ 0.0108, since one of the daughter cells consists of all normal subunits if and only if either daughter cell #1 consists of 10 normal subunits (and 0 mutant subunits) or it consists of 4 normal subunits and 6 mutant subunits.
3.75 n(n − 1)%n
nn . Exactly one box stays empty if and only if exactly one of the n boxes contains two balls, n − 2 boxes contain one ball each and one of the boxes contains 0 balls. Note that there are n ways to choose which box stays empty, then there are n − 1 choices of the box that contains two balls and there are %n
2
&
ways to fill it. Next there are (n − 2)!
ways to distribute the remaining (n − 2) balls between the remaining n − 2 boxes that must contain one ball each. On the other hand, the total number of outcomes in the sample space is equal to the number of ways to put n balls into n boxes (where some boxes could stay empty).
Since each of the n balls can be put into any of the n boxes, then N (Ω) = nnand the required conclusion follows. (Note that throughout this solution all the boxes and balls are taken to be distinguishable).
3.76 (n − k − 1) × 2 × ((n − 2)!)
n! = 2(n − k − 1)
n(n − 1) . Take the answer to Exercise 3.14 and divide it by n!.
Theory Exercises
3.77 Note that the number of members (in the entire population), who have the specified at-tribute, is equal to N p. Therefore, we have that:
a)
ways to pick k members of the sample from among the N p members of the population who have the specified attribute, and there are %N−N p
n−k
&
=%N(1−p)
n−k
&
ways to pick the remaining n − k members of the sample from the N − N p members of the population who lack the specified attribute, while %N
n
&
represents the total number of ways to pick an (unordered) sample of size n without replacement from the population of size N . b)
pk(1 − p)n−k. If we consider ordered selection with replace-ment, then there are Nn different samples possible. On the other hand, there are%n
k
&
possible ways to pick locations of the k members (who have the specified attribute) in the sample of size n, and, given any one such choice, there are (N p)k(N − N p)n−k ways to pick the specific members.
c) The proof is the same as in the solution to Exercise 5.70.
d) If the population size N is large relative to the sample size n, then, when sampling is done without replacement, the probability of the event, that exactly k members with the specified attribute are selected, can be well approximated by the corresponding probability for sampling with replacement.
-. Note that exactly n of the N women get their own keys if and only if n women get their own keys while the other N − n women get the wrong keys. There are %N
n
&
ways to fix the set of those n women who get their own keys. Once that set is fixed, let us consider the remaining N − n women and their N − n keys. Applying the result of Example 3.21(c), we obtain that
P (none of the remaining (N − n) women gets her own key) = 1 −
N−n*
k=1
(−1)k+1 k! .
Note also that the above probability, that none of the remaining (N − n) women gets her own key, is also equal to
# of ways to arrange (N − n) keys among (N − n) women so that none has her own
(N − n)! ,
thus, the number of ways to arrange (N − n) remaining keys among the (N − n) remaining women in such a way that none has her own key is equal to
(N − n)!
Therefore, the total number of ways, in which exactly n of the N women get their own keys, is equal to
Finally, note that there are N ! possible ways to arrange all N keys among the N women.
Therefore, the required probability is equal to 'N
3.79 Since the number of the U.S. adults is very large relative to the sample size n = 10, then, by Exercise 3.65, when sampling is done with replacement, any member of the population is unlikely to be selected more than once. Therefore, in the current problem we can approximate the desired answers by considering sampling with replacement instead of the sampling without replacement. Therefore,
a) '10
c) The probabilities obtained in (a) and (b) are only approximately correct since we replaced sampling without replacement with sampling with replacement.
d) For even N ,
P (exactly five are Democrats) =
%0.5N
(If N is not even, then 0.5N has to be replaced with the appropriate closest larger and smaller integers).
3.80 a) 1/52. We need to calculate how many of the 52! possible orderings of the cards have the ace of spades immediately following the first ace. Note that each ordering of the 52 cards can be obtained by first ordering the 51 cards different from the ace of spades and then inserting the ace of space into that ordering. There are 51! possible orderings of the 51 cards different from the ace of spades. Moreover, for each such ordering, there is only one way in which the ace of spades can be inserted immediately after the first ace. Therefore,
P (the ace of spades follows the first ace) = 51!
52! = 1 52. b) 1/52, by the same argument as in (a).
c) 4/52. Let A1 be the event that the ace of spades follows right after the first ace, A2 be the event that the ace of diamonds follows right after the first ace, A3 be the event that the ace of hearts follows right after the first ace, and A4 be the event that the ace of clubs follows right after the first ace. Then, clearly, events A1, A2, A3, A4 are mutually exclusive and d) 4/52, by an argument similar to the one given in (c).
3.81 a) Infinitely many, since at noon the box contains all the N -numbered balls except for those numbered 10k, k = 1, 2, . . . .
b) 0, since for every k ∈ N , the ball #k is removed from the box at 2k−11 minute to noon.
c) The number of basketballs left in the box at noon is equal to 0 with probability 1. Indeed, let En be the event that ball number 1 is still in the box after the first n withdrawals. Then
P (En) = 9 × 18 × 27 × · · · × (9n) 10 × 19 × 28 × · · · × (9n + 1). The event that ball number 1 is still in the box at noon is equal to.∞
n=1En. Note that events En are decreasing (i.e. En⊃ En+1 for each n ∈ N ), thus, by the continuity property of probability
measure,
P (ball #1 is still in the box at noon) = P ( /∞ n=1
En) = lim
n→∞P (En) = 0∞ n=1
9n 9n + 1.
Let us show that 0∞ n=1
9n
9n + 1 = 0 or, equivalently, that 1∞
0
n=1
9n 9n + 1
2−1
= 0∞ n=1
9n + 1
9n = ∞.
Note that for all m ≥ 1, 0∞
n=1
9n + 1
9n =
0∞ n=1
' 1 + 1
9n (
≥ 0m n=1
' 1 + 1
9n (
> 1 9 + 1
18 + 1
27+ · · · + 1 9m = 1
9
*m i=1
1 i. Taking m → ∞, and, in view of )∞
i=1 1
i = ∞, it follows that 3∞ n=19n+1
9n = ∞. Therefore, P (ball #1 is still in the box at noon) = 0.
Similarly, one shows that for arbitrary k ∈ N ,
P (ball #k is still in the box at noon) = 0.
Therefore, by the Boole’s inequality (Exercise 2.75),
P (box is not empty at noon) = P '∞
+
k=1
{ball #k is in the box at noon}
(
≤
*∞ k=1
P (ball #k is in the box at noon) =
*∞ k=1
0 = 0.
3.4 Review Exercises
Basic Exercises
3.82 a) The required tree diagram is of the form:
President Secretary
•
•a
• b ab
• c ac
• d ad
• e ae
•b
• a ba
• c bc
• d bd
• e be
•c
• a ca
• b cb
• d cd
• e ce
•d
• a da
• b db
• c dc
• e de
•e
• a ea
• b eb
• c ec
• d ed
b) The required tree diagram is the same as in (a), but one should switch the words “President”
and “Secretary”.
c) The required tree diagram is given by:
Size-2 subset President Secretary
•
{a,b}
• a • b
• b • a
{a,c}
• a • c
• c • a
{a,d}
• a • d
• d • a
{a,e}
• a • e
• e • a
{b,c}
• b • c
• c • b
{b,d}
• b • d
• d • b
{b,e}
• b • e
• e • b
{c,d}
• c • d
• d • c
{c,e}
• c • e
• e • c
{d,e}
• d • e
• e • d
d) The required tree diagram is given by:
President Secretary Treasurer
•
• a
• b • c
• c • b
• b
• a • c
• c • a
• c
• a • b
• b • a
e) The first required diagram has the form:
Two Democrats One Republican
•
{a,b}
• d
• e
{a,c}
• d
• e
{b,c}
• d
• e The second required diagram is given by:
One Republican Two Democrats
•
• d
•{a,b}
•{a,c}
•{b,c}
• e
•{a,b}
•{a,c}
•{b,c}
3.83 a) (10)6 = 10 × 9 × 8 × 7 × 6 × 5 = 151, 200.
e) 0.0511, since let A1 denote the event that a bridge hand is void in hearts, A2 be the event that a bridge hand is void in diamonds, A3 be the event that a bridge hand is void in spades, and let A4 be the event that a bridge hand is void in clubs. Then, by the inclusion-exclusion formula, where i, j, k are all distinct. Also, P
' 4
/
i=1
Ai
(
= 0. Therefore, upon applying the result of Exercise 3.38, we obtain that
P
/2 = 5/12 ≈ 0.4167, since the events “green die shows a larger number than the red die” and “red die shows a larger number than the green die” have the same probability (by symmetry), while the complement of their union, i.e. the event “both dice show the same number” has probability 6/36.
b)
3.99 Let x1 and x2 be the values on the first and second selected chips, respectively. Then a) 4/45. For sampling without replacement, the event of interest is given by
{(x1, x2) ∈ {1,···, 10}2 : x1+x2 = 10, x1 ̸= x2} = {(x1, x2) ∈ {1,···, 10}2 : x1+x2 = 10}\{(5, 5)}.
Then the number of elements in the event of interest is equal to the number of positive integer-valued solutions of the equation x1+ x2 = 10 minus one. Thus, upon using Exercise 3.51, we obtain that the number of elements in the event of interest equals to %10−1
2−1& − 1 = 8. Therefore, since 10 × 9 = 90 is the total number of elements in the sample space, the required probability
equals to 8/90 = 4/45 ≈ 0.0889
b) 0.09, since there are 102 elements in the sample space, while the required event, {(x1, x2) ∈ {1,···, 10}2 : x1+ x2 = 10},
by Exercise 3.51, contains 9 elements.
3.100 a)
606 ≈ 0.1389, since there are 6!/(2!3!) different color arrangements possible for the ordered selection of six balls, of which 2 are white and 3 are blue (and 1 is red); moreover, for every such arrangement of colors, there are 10 × 202× 303 possible choices of balls.
3.101
& , since if the kth item inspected is the last defective, then there are M − 1 defective items among the first k − 1 inspected, and, thus, there are%k−1
M−1
&
ways to pick which of the first k − 1 items are defective. On the other hand, the total number of ways to pick which of the N items are defective is equal to%N
M
&
, thus, the required answer follows.
3.102 a)
d) 0.00000019N , by (a) and assuming that the N tickets are all different.
3.103 1 − 5/8 = 3/8 = 0.375, since, by the matching problem, i.e. Example 3.21(c), with N = 4, we have that
P (at least one man sits across from his wife) =
*4
ways to select the six floors where someone gets off, and, for every such choice, there are 6! ways to pick the order in which the six people leave the elevator.
& . (Same as Exercise 3.64).
3.106 a)
& . Note that a total of n members of the population are marked, and the two samples have exactly k members in common if in the sample of size m exactly k members are marked. Thus, one has to count the number of ways to choose k members out of n marked
members times the number of ways to choose m − k remaining members of the second sample out of N − n “unmarked” members, and then divide by the total number of ways to pick a sample of size m from the population of size N .
b)
& . Note that now a total of m members of the population will be marked, and the two samples have k members in common if k members of the first sample are selected from those m that are later marked, while the remaining n − k members of the first sample are selected from those N − m members that are left “unmarked”.
c) The equality holds since, by (a),(b), both expressions represent the probability that exactly k members of the population are selected both times.
d)
& . The answer follows by the same argument as in Exercise 3.106.
b) Assuming that the proportion of tagged deer in the sample of n recaptured animals is the same as in the entire population of N deer, one arrives at the equation:
M N = k
n,
thus, N = M nk = 10×83 ≈ 27.
3.110 a) Suppose 0! = x, then 1 = 1! = (0+1)! = (0+1)·(0!) = 0! = x, unless (n+1)! = (n+1)·n!
is allowed to be violated for n = 0.
b) Suppose that when you press the CLEAR key, the display reads x. Then, when you press “14, ENTER”, the display should show the product of x and 14, then, when you subsequently press
“3, ENTER”, the display should show the product of 14x and 3. Finally, when we subsequently press “6, ENTER”, the display should show the product of 14x × 3 and 6, i.e. the final result should equal to x × 14 × 3 × 6 = 252x. Since the display shows 252, then 252x = 252, implying that x = 1. Thus, multiplication of no numbers gives 1. Since 90 represents the multiplication of 9 zero times, then 90= 1.
c) Since the number of ways to choose three men and no women must be the same as the number of ways to choose three men, then%5
3
&%2
0
&
=%5
3
&
, implying that 10%2
0
&
= 10, thus%2
0
&
= 1.
Therefore, by the combinations rule, %2
0
&
= 0!2!2! = 0!1 = 1, which implies that 0! = 1.
3.111 a)%2N
N
&
, since every polygonal line is uniquely determined by the locations of N people who have $10 bills in a line of length 2N .
b) Each polygonal line should terminate at the point (2N, 0), since, overall, the line should travel from the origin N steps up and N steps down (regardless of the order in which those steps are made).
c) Event Ec =“Someone has to wait for change” occurs if and only if there are more people with $20 than those with $10 among the first k people in line for some k = 1, · · · , 2N . The latter is equivalent to the fact that the corresponding polygonal line travels more steps up than down at some point k in time, which is equivalent to the line going above the x-axis. Therefore, event E occurs if and only if the polygonal line does not go above the x-axis.
d) %2N
N+1
&
= (N +1)!(N −1)!(2N )! . Let us call our polygonal line good if it never goes above the x-axis, otherwise let us call it bad. In order to count the number of bad polygonal lines, note that for each bad polygonal line one can construct a polygonal path that starts at (0, 0) and ends at (2N, 2), by reflecting the portion of the original polygonal line to the right of the first bad point with respect to the horizontal line y = 1. Conversely, to any polygonal path that starts at (0, 0) and ends at (2N, 2) there corresponds a bad polygonal line of the original form. Thus, the total number of bad polygonal lines equals to the number of possible polygonal paths that start at (0, 0) and end at (2N, 2), and that number is equal to%2N
N+1
&
, since the path should climb N + 1 steps up (out of 2N ) while taking the remaining N − 1 steps down, in an arbitrary order.
e) By (a),(c),(d),
P (Ec) = P (random polygonal line is bad) =
%2N
N+1
&
%2N
N
& =
(2N )!
(N + 1)!(N − 1)!· N !N !
(2N )! = N N + 1. Thus,
P (E) = 1 − P (Ec) = 1 − N
N + 1 = 1 N + 1.