Unformatted text preview: MS&E 120: Probabilistic Analysis
Fall 200708, Prof. Shachter HOMEWORK #1 SOLUTIONS
CHAPTER #1
PROBLEMS
3) 20 workers assigned to 20 jobs.
Order the jobs in a particular list. Each ordering of workers assigned to this list is a
different assignment. There are 20! ways of arranging the workers in a list.
Therefore, # of possible assignments = 20!.
7) Ordering of boys and girls in a row.
(a) There are 6 people in total (3 boys and 3 girls).
# of ways of arranging them in a row = 6! = 720.
(b) The boys and girls must each sit together in a group. Either the boys can be placed
first or the girls can be placed first. For each of these 2 configurations, the boys can be
arranged themselves in 3! ways (and similarly for the girls).
# of ways = 2*3!*3! = 72.
(c) The boys must sit together, so treat the boys as one individual person. Now these 4
people (the 3 girls and the group of boys) can be arranged in 4! ways. Once this order has
been assigned (i.e. the position of the girls and the group of boys has been determined),
the boys can be arranged in 3! ways among themselves.
# of ways = 4!*3! = 144.
(d) No two people of the same sex can sit together. The configuration can be of the type
BGBGBG or GBGBGB, where B represents a boy and G a girl. The argument is similar
to that in part (b). There are 2 basic configurations, and the boys and girls can be
arranged separately in 3! ways each.
# of ways = 2*3!*3! = 72.
8) Letter arrangements. MS&E 120: Probabilistic Analysis
Fall 200708, Prof. Shachter
This is a case of the number of permutations for n objects, given n1 similar objects, etc.
See Example 3d in the textbook.
(a) FLUKE
# of ways = 5! = 120.
(b) PROPOSE
# of ways = 7!/(2!*2!) = 1260.
(c) MISSISSIPPI
# of ways = 11!/(4!*4!*2!) = 34650.
(d) ARRANGE
# of ways = 7!/(2!*2!) = 1260.
12) 5 awards presented to 30 students.
(a) Each of the 5 awards can go to any of the 30 students. This is equivalent to placing
n=5 distinct balls into r=30 distinct urns.
# of ways = 305.
(b) Each student can get a maximum of one award, and the awards are distinct. The first
award can go to any of the 30 students, the second to any of the remaining 29, and so
on…
# of ways = (30)*(29)*(28)*(27)*(26).
13) The number of handshakes = number of pairs out of 20, since everyone shakes hands
with each other. Note that the ordering of the pair is not important, i.e. Debarun shaking
hands with Caleb is the same as Caleb shaking hands with Debarun!
æ 20 ö
# of ways = ç ÷ = 190.
è2ø
14) Poker hand.
There are 52 cards in a deck, and we can choose any 5 of them for a 5card poker hand.
æ 52 ö
# of ways = ç ÷
è5ø MS&E 120: Probabilistic Analysis
Fall 200708, Prof. Shachter
15) We can first choose the 5 women out of 10 and 5 men out of 12. Now we can place
the women in a line, and then arrange the men in 5! ways to pair the men and women
together. We have assumed that the order of the pairs is unimportant (i.e. if there is a
particular group of pairings involving Tom and Nancy as a pair, then Tom and Nancy as
the first pair is the same as Tom and Nancy as the fourth pair).
æ 10 ö æ 12 ö
# of ways = ç ÷ * ç ÷ *5! = 23950080.
è5ø è5ø
19) Committee formation.
(a) Total # of ways of forming the committee (without any constraint) is through
choosing 3 women out of 8 and 3 men out of 6.
æ8ö æ 6 ö
# of ways = ç ÷ * ç ÷
è 3ø è 3 ø
æ8ö æ 4ö
# of ways in which both these men are selected = ç ÷ * ç ÷ as now we only need to
è 3ø è 1 ø
choose 1 more man out of the remaining 4.
Therefore, # of ways in which both these men are not selected together
æ8ö æ 6 ö æ8ö æ 4ö
= ç ÷ * ç ÷  ç ÷ * ç ÷ = 896.
è 3ø è 3 ø è 3ø è 1 ø
(b) Now 2 of the women refuse to serve together. The argument is similar to that in (a).
# of ways in which both these women are not selected together
æ8ö æ 6 ö æ 6 ö æ 6 ö
= ç ÷ * ç ÷  ç ÷ * ç ÷ = 1000.
è 3ø è 3 ø è 1 ø è 3 ø
(c) Now 1 man and 1 woman refuse to serve together. Similar to (a) and (b) again,
# of ways in which the man and woman are not selected together
æ8ö æ 6 ö æ 7 ö æ 5ö
= ç ÷ * ç ÷  ç ÷ * ç ÷ = 910.
è 3ø è 3 ø è 2 ø è 2 ø
21) Number of paths (going upwards and towards the right). MS&E 120: Probabilistic Analysis
Fall 200708, Prof. Shachter
We can choose to go right or up at any stage. A pathway can be represented as, say,
something like RRURRUU. Every possible arrangement of ‘U’s and ‘R’s is a separate
pathway from A to B. Thus the number of paths is the same as the number of ways of
choosing 3 out of the 7 ‘U’s (equivalent to choosing 4 out of the 7 ‘R’s).
æ7ö
# of paths = ç ÷ = 35.
è 3ø
OR
We can solve a dynamic program to find the number of paths. Let p(i,j) be the # of paths
with ‘i’ steps to go to the right and ‘j’ steps to go upwards. From a particular point we can
go right or we can go up. The number of paths from position (i,j) is the sum of paths from
the point to its right and the point above it.
Thus, p(i,j) = p(i1,j) + p(i,j1).
The boundary condition is that:
p(0,0) = 0; and,
p(i,0) = p(0,j) = 1 (There is only 1 path if you are on the top edge or the right edge of the
grid). Using the formulae, we can recursively find that,
p(4,3) = 35.
Note the similarity of this problem to the Pascal Triangle.
22) Let’s call the point in the middle, point C. We can solve the two problems of going
from A to C and from C to B separately, and multiply the number of paths in the two
problems (because a path from A to B comprises of both legs of the journey, A to C and
then C to B). From A to C, we need to take 2 steps up and 2 steps to the right; from C to
B, we need to take 1 step up and 2 steps to the right. Using the argument described in
problem #21,
æ 4ö æ 3ö
# of paths = ç ÷ * ç ÷ = 18.
è 2ø è 2ø
32) 8 people getting off an elevator with 6 floors.
(a) This is equivalent to finding the number of integral solutions to: MS&E 120: Probabilistic Analysis
Fall 200708, Prof. Shachter
x1 + x2 +…+ x6 = 8; x1, x2,…x6 ≥ 0
æ 8 + 6  1ö æ 13 ö
# of ways = ç
÷ = ç ÷ = 1287.
è 6 1 ø è 5 ø
(b) Now treat the men and women separately. Solve the two individual problems as
above, and multiply the two results.
æ 5 + 6  1ö æ 3 + 6  1ö æ 10 ö æ 8 ö
# of ways = ç
÷* ç
÷ = ç ÷ * ç ÷ = 14112.
è 6 1 ø è 6 1 ø è 5 ø è 5 ø
33) Investing $20K in some projects.
(a) 20 must be allocated to 4 projects, with minimum investments of 2,2,3,4 respectively.
Allocating the minimum investment in each, there is [20(2+2+3+4)] or 9 left for
investment.
This is equivalent to finding the number of integral solutions to:
x1 + x2 + x3 + x4 = 9; x1, x2,…x4 ≥ 0
æ 9 + 4  1ö æ 12 ö
# of ways = ç
÷ = ç ÷ = 220.
è 4 1 ø è 3 ø
(b) We must break the problem up into different cases.
There are 5 cases: Invest in all 4 projects; Invest in Projects 1,2,3; Invest in Projects
1,2,4; Invest in Projects 1,3,4; Invest in Projects 2,3,4.
We need to ADD the possibilities for all 5 cases. The first case has already been solved in
part (a). Let us solve for the second case, i.e. Invest in Projects 1,2,3.
20 must be allocated to 3 projects, with minimum investments of 2,2,3 respectively.
Allocating the minimum investment in each, there is [20(2+2+3)] or 13 left for
investment.
This is equivalent to finding the number of integral solutions to:
x1 + x2 + x3 = 13; x1, x2, x3 ≥ 0
æ 13 + 3  1ö æ 15 ö
# of ways = ç
÷=ç ÷
è 3 1 ø è 2 ø
A similar procedure can be adopted for the other 3 cases.
Adding the possibilities, MS&E 120: Probabilistic Analysis
Fall 200708, Prof. Shachter
æ 12 ö æ 15 ö æ 13 ö æ 14 ö æ 13 ö
# of ways = ç ÷ + ç ÷ + ç ÷ + ç ÷ + ç ÷ = 572.
è3ø è2ø è2ø è2ø è2ø
THEORETICAL EXERCISES
2) We can form a tree to count the number of possibilities for the 2 experiments: the
number of leaves of the tree give the number of possibilities. The first outcome of the
first experiment can result in n1 possibilities, the 2nd n2, and so on….
Thus, the number of outcomes = n1+n2+…nm = m ån
i =1 i 3) The order of the objects is relevant, thus we need to form permutations. The first
object can be selected in n ways, the 2nd in n1 ways and so on…
# of ways = n*(n1)*(n2)…..*(nr+1). CHAPTER #2
PROBLEMS
1) Experiment with 3 marbles. Let S be the sample space.
With replacement: you can pick up the ball with any of the 3 colors from the box.
S = {RR, GG, BB, RG, GR, RB, BR, GB, BG}.
Without replacement: Colors are not allowed to repeat now.
S = {RG, GR, RB, BR, GB, BG}.
3) Two dice and events.
E = event that sum of dice is odd.
F = at least one dice shows 1.
G = event that the sum is 5.
EF = sum is odd and at least one dice shows 1 = {1,2; 2,1; 1,4; 4,1; 1,6; 6,1}. MS&E 120: Probabilistic Analysis
Fall 200708, Prof. Shachter
E È F = sum is odd or at least one dice shows one = {1,2; 2,1; 1,4; 4,1; 1,6; 6,1; 2,3; 3,2;
2,5; 5,2; 3,4; 4,3; 3,6; 6,3; 4,5; 5,4; 5,6; 6,5; 1,3; 3,1; 1,5; 5,1}.
FG = at least one dice is 1 and sum is 5 = {1,4; 4,1}.
EFc = sum is odd and no dice shows 1 = {2,3; 3,2; 2,5; 5,2; 3,4; 4,3; 3,6; 6,3; 4,5; 5,4;
5,6; 6,5}.
EFG = sum is five and sum is odd and at least one dice is 1 = FG = {1,4; 4,1}.
9) Let P(A) be the percentage of people carrying an American express card = 24%.
Let P(B) be the percentage of people carrying a visa card = 61%.
Percentage of people carrying both = P(AB) = 11%.
Percentage of people carrying either one = P(B È A) = P(A) + P(B) – P(AB) = 74%.
13) There are 3 newspapers in circulation in a certain town.
Total number of people = 100,000.
P(I) = proportion of people reading Newspaper I = 10%
Using similar notation below:
P(II) = 30%; P(III) = 5%
P(I ∩ II) = 8%
P(I ∩ III) = 2%
P(II ∩ III) = 4%
P(I ∩ II ∩ III) = 1%
The events I, II and III (and their unions and intersections) are shown in Figure 1. All the
answers can be explained with the help of the figure. MS&E 120: Probabilistic Analysis
Fall 200708, Prof. Shachter I II
1 19 7
1 1 3
68 0 III
Figure 1: Venn Diagram with percentage of people using Newspapers (Problem #13)
(a) # of people who read only 1 newspaper = (1+ 0+19)% = 20,000.
(b) # of people who read at least 2 newspapers = (7+1+3+1)% = 12,000.
(c) # of people reading at least 1 morning paper (I or III) AND the evening paper (II) =
(7+1+3)% = 11,000.
(d) # of people who don’t read newspapers = (100 – (1+1+1+7+3+19))% = 68,000.
(e) # of people reading only 1 morning paper AND an evening paper = (7+3)% = 10,000.
THEORETICAL EXERCISES
6) See Figures 2 and 3 for the Venn Diagrams for parts (a)(f) and (g)(j) respectively.
The shaded areas show the required regions (there are several different solutions to many
of the parts). Using the Venn diagrams in these figures, we get:
(a) E ∩ Fc ∩ Gc
(b) (E ∩ G) ∩ Fc
(c) E U F U G
(d) (E ∩ F) U (F ∩ G) U (E ∩ G)
(e) E ∩ F ∩ G
(f) (E U F U G)c
(g) [(E ∩ F) U (F ∩ G) U (E ∩ G)]c
(h) [E ∩ F ∩ G]c
(i) [(E ∩ F) U (F ∩ G) U (E ∩ G)] – (E ∩ F ∩ G) MS&E 120: Probabilistic Analysis
Fall 200708, Prof. Shachter
(j) S (entire space).
E F E F G G c)
E F G G
E F E F G Figure 2: Parts (a)(f) of Theoretical Exercise #6. E F G MS&E 120: Probabilistic Analysis
Fall 200708, Prof. Shachter F E F G F E G E F G Figure 3: Parts (g)(j) of Theoretical Exercise #6. E G ...
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