Solutions Manual For Introduction to Probability Models 12th Edition By Sheldon M. Ross
Chapter 1 1. S = {(R, R), (R, G), (R, B), (G, R), (G, G), (G, B), (B, R), (B, G), (B, B)} The probability of each point in S is 1/9. 2. S = {(R, G), (R, B), (G, R), (G, B), (B, R), (B, G)} 3. S = {(e1 , e2 , . . . , en ), n ≥ 2} where ei ∈(heads, tails}. In addition, en = en−1 = heads and for i = 1, . . . , n − 2 if ei = heads, then ei+1 = tails. P{4 tosses} = P{(t, t, h, h)} + P{(h, t, h, h)} 4 1 1 =2 = 2 8 4. (a) F(E ∪ G)c = F E c G c (b) E F G c (c) E ∪ F ∪ G (d) E F ∪ E G ∪ F G (e) E F G (f) (E ∪ F ∪ G)c = E c F c G c (g) (E F)c (E G)c (F G)c (h) (E F G)c 5. 43 . If he wins, he only wins $1, while if he loses, he loses $3. 6. If E(F ∪ G) occurs, then E occurs and either F or G occur; therefore, either E F or E G occurs and so E(F ∪ G) ⊂ E F ∪ E G
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Introduction to Probability Models
Similarly, if E F ∪ E G occurs, then either E F or E G occurs. Thus, E occurs and either F or G occurs; and so E(F ∪ G) occurs. Hence, E F ∪ E G ⊂ E(F ∪ G) which together with the reverse inequality proves the result. 7. If (E ∪ F)c occurs, then E ∪ F does not occur, and so E does not occur (and so E c does); F does not occur (and so F c does) and thus E c and F c both occur. Hence, (E ∪ F)c ⊂ E c F c If E c F c occurs, then E c occurs (and so E does not), and F c occurs (and so F does not). Hence, neither E or F occurs and thus (E ∪ F)c does. Thus, E c F c ⊂ (E ∪ F)c and the result follows. 8. 1 ≥ P(E ∪ F) = P(E) + P(F) − P(E F) 9. F = E ∪ F E c , implying since E and F E c are disjoint that P(F) = P(E) + P(F E)c . 10. Either by induction or use n
c ∪ E i = E 1 ∪ E 1c E 2 ∪ E 1c E 2c E 3 ∪ · · · ∪ E 1c · · · E n−1 En 1
and as each of the terms on the right side are mutually exclusive: P( ∪ E i ) = P(E 1 ) + P(E 1c E 2 ) + P(E 1c E 2c E 3 ) + · · · i
11. P{sum is i} =
c + P(E 1c · · · E n−1 En ) ≤ P(E 1 ) + P(E 2 ) + · · · + P(E n ) (why?) i−1 36 , 13−i 36 ,
i = 2, . . . , 7
i = 8, . . . , 12 12. Either use hint or condition on initial outcome as: P{E before F} = P{E before F|initial outcome is E}P(E) + P{E before F|initial outcome is F}P(F) + P{E before F|initial outcome neither E or F}[1 − P(E) − P(F)] = 1 · P(E) + 0 · P(F) + P{E before F} = [1 − P(E) − P(F)] P(E) Therefore, P{E before F} = P(E)+P(F) 13. Condition an initial toss
P{win} =
12 i=2
P{win|throw i}P{throw i}
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Now, P{win|throw i} = P{i before 7} ⎧ 0 i = 2, 12 ⎪ ⎪ ⎪ ⎪ i −1 ⎪ ⎨ i = 3, . . . , 6 5+1 = 1 i = 7, 11 ⎪ ⎪ ⎪ ⎪ 13 − i ⎪ ⎩ i = 8, . . . , 10 19 − 1 where above is obtained by using Problems 11 and 12. P{win} ≈ .49. 14. P{ A wins} = =
∞ n=0 ∞
P{ A wins on (2n + 1)st toss} (1 − P)2n P
n=0 ∞
=P
[(1 − P)2 ]n
n=0
1 1 − (1 − P)2 P = 2P − P 2 1 = 2− P P{B wins} = 1 − P{ A wins} 1− P = 2− P =P
16. P(E ∪ F) = P(E ∪ F E c ) = P(E) + P(F E c ) since E and F E c are disjoint. Also, P(E) = P(F E ∪ F E c ) = P(F E) + P(F E c ) by disjointness Hence, P(E ∪ F) = P(E) + P(F) − P(E F) 17. Prob{end} = 1 − Prob{continue} = 1 − P({H, H, H } ∪ {T, T, T }) = 1 − [Prob(H, H, H ) + Prob(T, T, T )].
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Introduction to Probability Models
Fair coin: Prob{end} = 1 − =
3 4
Biased coin:P{end} = 1 −
1 1 1 1 1 1 · · + · · 2 2 2 2 2 2
1 1 1 3 3 3 · · + · · 4 4 4 4 4 4
9 16 18. Let B = event both are girls; E = event oldest is girl; L = event at least one is a girl. =
P(B) 1/4 1 P(B E) = = = P(E) P(E) 1/2 2 1 3 (b) P(L) = 1 − P(no girls) = 1 − = , 4 4 P(B L) P(B) 1/4 1 P(B|L) = = = = P(L) P(L) 3/4 3 (a) P(B|E) =
19. E = event at least 1 six P(E) number of ways to get E 11 = = number of samples pts 36 D = event two faces are different P(D) = 1 − Prob(two faces the same) 5 P(E D) 10/36 1 6 = P(E|D) = = = =1− 36 6 P(D) 5/6 3 20. Let E = event same number on exactly two of the dice; S = event all three numbers are the same; D = event all three numbers are different. These three events are mutually exclusive and define the whole sample space. Thus, 1 = P(D) + P(S) + P(E), P(S) = 6/216 = 1/36; for D have six possible values for first die, five for second, and four for third. ∴ Number of ways to get D = 6 · 5 · 4 = 120. P(D) = 120/216 = 20/36 ∴ P(E) = 1 − P(D) − P(S) 1 5 20 − = =1− 36 36 12 21. Let C = event person is color blind. P(Male|C) =
P(C|Male)P(Male) P(C|MaleP(Male) + P(C|Female)P(Female)
.05 × .5 .05 × .5 + .0025 × .5 2500 20 = = 2625 21
=
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22. Let trial 1 consist of the first two points; trial 2 the next two points, and so on. The probability that each player wins one point in a trial is 2 p(1 − p). Now a total of 2n points are played if the first (a − 1) trials all result in each player winning one of the points in that trial and the nth trial results in one of the players winning both points. By independence, we obtain P{2n points are needed} = (2 p(1 − p))n−1 ( p 2 + (1 − p)2 ), n ≥ 1 The probability that A wins on trial n is (2 p(1 − p))n−1 p 2 and so P{ A wins} = p
2
∞
(2 p(1 − p))n−1
n=1
=
p2 1 − 2 p(1 − p)
23. P(E 1 )P(E 2 |E 1 )P(E 3 |E 1 E 2 ) . . . P(E n |E 1 . . . E n−1 ) P(E 1 E 2 ) P(E 1 E 2 E 3 ) P(E 1 . . . E n ) = P(E 1 ) ... P(E 1 ) P(E 1 E 2 ) P(E 1 . . . E n−1 ) = P(E 1 . . . E n ) 24. Let a signify a vote for A and b one for B. (a) P2,1 = P{a, a, b} = 1/3 (b) P3,1 = P{a, a} = (3/4)(2/3) = 1/2 (c) P3,2 = P{a, a, a} + P{a, a, b, a} = (3/5)(2/4)[1/3 + (2/3)(1/2)] = 1/5 (d) P4,1 = P{a, a} = (4/5)(3/4) = 3/5 (e) P4,2 = P{a, a, a} + P{a, a, b, a} = (4/6)(3/5)[2/4 + (2/4)(2/3)] = 1/3 (f) P4,3 = P{always ahead|a, a}(4/7)(3/6) = (2/7)[1 − P{a, a, a, b, b, b|a, a} − P{a, a, b, b|a, a} − P{a, a, b, a, b, b|a, a}] = (2/7)[1 − (2/5)(3/4)(2/3)(1/2) − (3/5)(2/4) − (3/5)(2/4)(2/3)(1/2)] = 1/7 (g) P5,1 = P{a, a} = (5/6)(4/5) = 2/3 (h) P5,2 = P{a, a, a} + P{a, a, b, a} = (5/7)(4/6)[(3/5) + (2/5)(3/4)] = 3/7 By the same reasoning we have
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Introduction to Probability Models
(i) P5,3 = 1/4 (j) P5,4 = 1/9 (k) In all the cases above, Pn,m = n−n n+n 25. (a) P{ pair } = P{second card is same denomination as first} = 3/51 P{pair|different suits} (b) P{pair, different suits} = P{different suits} = P{pair}/P{different suits} 3/51 = = 1/13 39/51 39.38.37 52 = 13 51.50.49 26.25 3 36 39 P(E 2 |E 1 ) = = 1 12 13 38.37 2 24 26 = 13/25 P(E 3 |E 1 E 2 ) = 1 12 13 P(E 4 |E 1 E 2 E 3 ) = 1 39.26.13 P(E 1 E 2 E 3 E 4 ) = 51.50.49 27. P(E 1 ) = 1 P(E 2 |E 1 ) = 39/51, since 12 cards are in the ace of spades pile and 39 are not. P(E 3 |E 1 E 2 ) = 26/50, since 24 cards are in the piles of the two aces and 26 are in the other two piles.
26.
P(E 1 ) =
4 1
48 12
P(E 4 |E 1 E 2 E 3 ) = 13/49 So P{each pile has an ace} = (39/51)(26/50)(13/49) 28. Yes. P(A|B) > P(A) is equivalent to P(AB) > P(A)P(B), which is equivalent to P(B| A) > P(B). 29. (a) P(E|F) = 0 (b) P(E|F) = P(E F)/P(F) = P(E)/P(F) ≥ P(E) = .6 (c) P(E|F) = P(E F)/P(F) = P(F)/P(F) = 1 P{George, not Bill} 30. (a) P{George|exactly 1 hit} = P{exactly 1} P{G, not B} = P{G, not B} + P{B, not G)} (.4)(.3) = (.4)(.3) + (.7)(.6) = 2/9
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(b) P{G|hit} = P{G, hit}/P{hit} = P{G}/P{hit} = .4/[1 − (.3)(.6)] = 20/41 31. Let S = event sum of dice is 7; F = event first die is 6. 1 P(F|S) 1 P(F S) = P(F|S) = 6 36 P(S) 1 1/36 = = 1/6 6
P(S) =
32. Let E i = event person i selects own hat. P (no one selects own hat) = 1 − P(E 1 ∪ E 2 ∪ · · · ∪ E n ) P(Ei 1 ) − P(Ei 1 Ei 2 ) + · · · =1− i 1 <i 2
i1
P(E 1 E 2 E n ) + (−1) =1− P(Ei 1 ) − P(Ei 1 Ei 2 ) n+1
i 1 <i 2
i1
−
P(Ei 1 Ei 2 Ei 3 ) + · · ·
i 1 <i 2 <i 3 n
+ (−1) P(E 1 E 2 E n ) Let k ∈ {1, 2, . . . , n}. P(Ei 1 E I2 Ei k ) = number of ways k specific men can select own hats ÷ total number of ways hats can be arranged = (n − k)!/n!. Number of terms in summation i1 <i2 <···<ik = number of ways to choose k variables out of n n variables = = n!/k!(n − k)!. k Thus, P(Ei 1 Ei 2 · · · Ei k ) i 1 <···<i k
=
i 1 <···<i k
(n − k)! n!
1 n (n − k)! = = k n! k! ∴ P(no one selects own hat) 1 1 1 1 + − + · · · + (−1)n 1! 2! 3! n! 1 1 1 − + · · · + (−1)n = 2! 3! n!
=1−
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Introduction to Probability Models
33. Note that each set played is independently golden with probability 2(1/2)24 = (1/2)23 . Condition on whether the match has 2 or 3 sets. Using that there will be a total of 2 sets with probability 1/2 yields the solution: (1 − (1 − (1/2)23 )2 )(1/2) + (1 − (1 − (1/2)23 )3 )(1/2). It should be noted that the preceding uses that whether a set is golden is independent of who won the set. That is, the conditional probability that a set is golden given that it is won by player 1 is (1/2)23 . To see why, it is obvious that who wins is independent of the event that the set was golden (since the winner given that it was golden is still equally likely to be either 1 or 2). But independence is symmetric, thus the event that a set is golden is independent of who won the set. 34. Letting A and B be respectively the events that the computer is fixed by A and that it is fixed by B. As A and B are mutually exclusive P (A ∪ B) = .4 + .6(.2) = .52 35. (a) 1/16 (b) 1/16 (c) 15/16, since the only way in which the pattern H, H, H, H can appear before the pattern T, H, H, H is if the first four flips all land heads. 36. Let B = event marble is black; Bi = event that box i is chosen. Now B = B B1 ∪ B B2 P(B) = P(B B1 ) + P(B B2 ) = P(B|B1 )P(B1 ) + P(B|B2 )P(B2 ) 1 1 2 1 7 = · + · = 2 2 3 2 12 37. Let W = event marble is white. P(B1 |W ) =
P(W |B1 )P(B1 ) P(W |B1 )P(B1 ) + P(W |B2 )P(B2 ) 1 1 1 3 2 · 2 4 = = 1 1 1 5 5 2 · 2 + 3 · 2 12
= 1
38. Let TW = event transfer is white; TB = event transfer is black; W = event white ball is drawn from urn 2. P(TW |W ) =
P(W |TW )P(TW ) P(W |TW )P(TW ) + P(W |TB )P(TB ) 2 2 4 4 7 · 3 = 21 = 2 1 1 5 5 · + · 7 3 7 3 21
= 2
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39. Let W = event woman resigns; A, B, C are events the person resigning works in store A, B, C, respectively. P(C|W ) = = =
P(W |C)P(C) P(W |C)P(C) + P(W |B)P(B) + P(W | A)P(A) .70 × 100 225
75 50 .70 × 100 225 + .60 × 225 + .50 × 225
70 225
1 140 = 225 2
40. (a) F = event fair coin flipped; U = event two-headed coin flipped. P(H |F)P(F) P(H |F)P(F) + P(H |U )P(U ) 1 1 1 · 1 = 1 12 2 1 = 43 = 3 · + 1 · 2 2 2 4
P(F|H ) =
(b)
(c)
P(H H |F)P(F) P(H H |F)P(F) + P(H H |U )P(U ) 1 1 1 · 1 = 1 14 2 1 = 85 = 5 · + 1 · 4 2 2 8
P(F|H H ) =
P(H H T |F)P(F) P(H H T |F)P(F) + P(H H T |U )P(U ) P(H H T |F)P(F) = =1 P(H H T |F)P(F) + 0
P(F|H H T ) =
since the fair coin is the only one that can show tails. 41. Note first that since the rat has black parents and a brown sibling, we know that both its parents are hybrids with one black and one brown gene (for if either were a pure black then all their offspring would be black). Hence, both of their offspring’s genes are equally likely to be either black or brown. P(2 black genes) (a) P(2 black genes|at least one black gene) = P(at least one black gene) 1/4 = = 1/3 3/4 (b) Using the result from part (a) yields the following: P(2 black genes) P(5 black offspring) 1/3 = 1(1/3) + (1/2)5 (2/3) = 16/17
P(2 black genes|5 black offspring) =
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Introduction to Probability Models
where P(5 black offspring) was computed by conditioning on whether the rat had 2 black genes. 42. Let B = event biased coin was flipped; F and U (same as above). P(H |U )P(U ) P(H |U )P(U ) + P(H |B)P(B) + P(H |F)P(F) 1 1 · 13 4 3 = = = 1 3 1 1 1 9 9 1· 3 + 4 · 3 + 2 · 3 12
P(U |H ) =
43. Let B be the event that Flo has a blue eyed gene. Using that Jo and Joe both have one blue-eyed gene yields, upon letting X be the number of blue-eyed genes possessed by a daughter of theirs, that P(B) = P(X = 1|X < 2) =
1/2 = 2/3 3/4
Hence, with C being the event that Flo’s daughter is blue eyed, we obtain P(C) = P(C B) = P(B)P(C|B) = 1/3 44. Let W = event white ball selected. P(W |T )P(T ) P(W |T )P(T ) + P(W |H )P(H ) 1 1 · 12 = 1 1 5 25 1 = 37 5 · 2 + 12 · 2
P(T |W ) =
45. Let Bi = event ith ball is black; Ri = event ith ball is red. P(B1 |R2 ) = =
P(R2 |B1 )P(B1 ) P(R2 |B1 )P(B1 ) + P(R2 |R1 )P(R1 ) r b b+r +c · b+r r b r +c r b+r +c · b+r + b+r +c · b+r
rb r b + (r + c)r b = b+r +c =
46. Let X (=B or =C) denote the jailer’s answer to prisoner A. Now for instance, P{ A to be executed|X = B} P{ A to be executed, X = B} = P{X = B} P{ A to be executed}P{X = B|A to be executed} = P{X = B} (1/3)P{X = B| A to be executed} = . 1/2
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Now it is reasonable to suppose that if A is to be executed, then the jailer is equally likely to answer either B or C. That is, P{X = B| A to be executed} =
1 2
P{ A to be executed|X = B} =
1 3
and so,
Similarly, P{ A to be executed|X = C} =
1 3
and thus the jailer’s reasoning is invalid. (It is true that if the jailer were to answer B, then A knows that the condemned is either himself or C, but it is twice as likely to be C.) 47. 1. 0 ≤ P(A|B) ≤ 1 P(B) P(S B) = =1 2. P(S|B) = P(B) P(B) 3. For disjoint events A and D P((A ∪ D)B) P(B) P( AB ∪ D B) = P(B) P( AB) + P(D B) = P(B) = P( A|B) + P(D|B)
P(A ∪ D|B) =
Direct verification is as follows: P(A|BC)P(C|B) + P(A|BC c )P(C c |B) P(ABC) P(BC) P(ABC c ) P(BC c ) = + P(BC) P(B) P(BC c ) P(B) c P( ABC ) P(ABC) + = P(B) P(B) P(AB) = P(B) = P( A|B) 49. Apply Proposition 1.1 to the increasing events Acn , n ≥ 1. 50. (a) It is always the case that liminf An ⊂ limsup An ⊂ ∪n An
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Introduction to Probability Models
Now, if An is increasing then a ∈ ∪n An then a ∈ Am for some m. But if An is increasing then this implies that a ∈ An for all n ≥ m, and thus that a ∈ liminf An . Hence, in this case ∪n An ⊂ liminf An , thus proving the result. 51. Using that P (T ) = P (MT ) + P (M c T ), yields that P (M c T ) = .3 − .2 = .1. Hence, P (T |M c ) =
P (M c T ) = 1/6. P (M c )
Chapter 2 1. P{X = 0} =
7 2
14 10 = 2 30
2. −n, −n + 2, −n + 4, . . . , n − 2, n 1 3. P{X = −2} = = P{X = 2} 4 1 P{X = 0} = 2 4. (a) 1, 2, 3, 4, 5, 6 (b) 1, 2, 3, 4, 5, 6 (c) 2, 3, . . . , 11, 12 (d) −5, −4, . . . , 4, 5 11 = P{min = 1} 36 1 P{max = 5} = = P{min = 2} 4 7 P{max = 4} = = P{min = 3} 36 5 P{max = 3} = = P{min = 4} 36 1 P{max = 2} = = P{min = 5} 12 1 P{max = 1} = = P{min = 6} 36 6. (H, H, H, H, H ), p 5 if p = P{heads} 7. p(0) = (.3)3 = .027 p(1) = 3(.3)2 (.7) = .189
5. P{max = 6} =
p(2) = 3(.3)(.7)2 = .441 p(3) = (.7)3 = .343 8. p(0) =
1 , 2
p(1) =
1 2
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1 1 1 , p(1) = , p(2) = , 2 10 5 1 1 p(3) = , p(3.5) = 10 10 2 3 200 5 3 1 3 1 = − 10. 1 − 3 6 2 6 6 216 9. p(0) =
3 8 4 5 10 + 1 11 2 5 1 5 1 = + = 12. 5 3 4 3 3 243 243
11.
10 10 1 10 13. i 2 i=7
6 1 1 = 2 64 6 1 6 P{X = 1} = P{X = 5} = 6 = 2 64 6 15 6 1 P{X = 2} = P{X = 4} = = 2 2 64 6 20 6 1 P{X = 3} = = 3 2 64
14. P{X = 0} = P{X = 6} =
15.
n! k n−k P{X = k} (n−k)!k! p (1 − p) = n! k−1 (1 − p)n−k+1 P{X = k − 1} (n−k+1)!(k−1)! p
=
n−k+1 p k 1− p
Hence, P{X = k} ≥ 1 ↔ (n − k + 1) p > k(1 − p) P{X = k − 1} ↔ (n + 1) p ≥ k The result follows. 16. 1 − (.95)52 − 52(.95)51 (.05) n! 17. Follows since there are x1 !···x permutations of n objects of which x1 are alike, x2 r! are alike,…, xr are alike.
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Introduction to Probability Models
18. (a) P(X i = xi , i = 1, . . . , r − 1|X r = j) P(X i = xi , i = 1, . . . , r − 1, X r = j) = P(X r = j) =
xr−1 j x1 n! x1 !...xr −1 ! j! p1 . . . pr −1 pr j n! n− j j!(n− j)! pr (1 − pr ) r −1
(n − j)! pi = x1 ! . . . xr −1 ! 1 − pr
xi
i=1
(b) The conditional distribution of X 1 , . . . , X r −1 given that X r = j is multinomial with parameters n − j, 1−pipr , i = 1, . . . , r − 1. (c) The preceding is true because given that X r = j, each of the n − j trials that did not result in outcome r resulted in outcome i with probability 1−pipr , i = 1, . . . , r − 1. 19. P{X 1 + · · · + X k = m} n = ( p1 + · · · + pk )m ( pk+1 + · · · + pr )n−m m 20. (a) This is the multinomial probability (2!)i (1!)n−2in!(0!)365−n+i (1/365)n = 2n!i (1/365)n . (365)! , (b) i!(n−2i)!(365−n+i)! [n/2] (365)! n! (c) 1 − i=0 i!(n−2i)!(365−n+i)! (1/365)n , where this uses that all possibilities 2i for having no days with 3 or more birthdays are mutually exclusive.
21. (a) P (X1 = 0 or X2 = 0) = q1n1 + q2n2 − q1n1 q2n2 , where qi = 1 − pi . (b) P (X1 = 1, X2 = 0) + P (X1 = 0, X2 = 1) = n1 p1 q1n1 −1 q2n2 + q1n1 n2 p2 q2n2 −1. (c) P (X1 = X2 = 1) + P (X1 = 2, X2 = 0) + P (X1 = 0, X2 = 2) = n1 2 n1 −2 n2 n1 −1 n2 −1 n1 n2 + p 1 q 1 q 2 + q1 p22 q2n2 −2 . n 1 p1 q 1 n 2 p2 q 2 2 2 1 32 23. In order for X to equal n, the first n − 1 flips must have r − 1 heads, and then the nth flip must land heads. By independence the desired probability is thus n − 1 r −1 p (1 − p)n−r x p r −1
22.
24. It is the number of tails before heads appears for the r th time. 25. A total of 7 games will be played if the first 6 result in 3 wins and 3 losses. Thus, P{7 games} =
6 3 p (1 − p 3 ) 3
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Differentiation yields d P{7} = 20 3 p 2 (1 − p)3 − p 3 3(1 − p)2 dp = 60 p 2 (1 − p)2 [1 − 2 p] Thus, the derivative is zero when p = 1/2. Taking the second derivative shows that the maximum is attained at this value. 26. Let X denote the number of games played. (a) P{X = 2} = p 2 + (1 − p)2 P{X = 3} = 2 p(1 − p) E[X ] = 2 p 2 + (1 − p)2 + 6 p(1 − p) = 2 + 2 p(1 − p) Since p(1 − p) is maximized when p = 1/2, we see that E[X ] is maximized at that value of p. (b) P{X = 3} = p 3 + (1 − p)3 P{X = 4} = P{X = 4, I has 2 wins in first 3 games} + P{X = 4, II has 2 wins in first 3 games} = 3 p 2 (1 − p) p + 3 p(1 − p)2 (1 − p) P{X = 5} = P{each player has 2 wins in the first 4 games} = 6 p 2 (1 − p)2 E[X ] = 3 p 3 + (1 − p)3 + 12 p(1 − p) p 2 + (1 − p)2 + 30 p 2 (1 − p)2 Differentiating and setting equal to 0 shows that the maximum is attained when p = 1/2. P{ A = i, B = i} 27. P{same number of heads} = i
k n−k (1/2)k (1/2)n−k = i i i
k = i i
=
i
=
n−k (1/2)n i
k k −i
n (1/2)n k
n−k (1/2)n i
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Introduction to Probability Models
Another argument is as follows: P{# heads of A = # heads of B} = P{# tails of A = # heads of B} since coin is fair = P{k − # heads of A = # heads of B} = P{k = total # heads} 28. (a) Consider the first time that the two coins give different results. Then P{X = 0} = P{(t, h)|(t, h) or (h, t)} p(1 − p) 1 = = 2 p(1 − p) 2 (b) No, with this procedure P{X = 0} = P {first flip is a tail} = 1 − p 29. Each flip after the first will, independently, result in a changeover with probability 1/2. Therefore, P{k changeovers} =
30.
n−1 (1/2)n−1 k
P{X = i} e−λ λi /i! = −λ i−1 = λ/i P{X = i − 1} e λ /(i − 1)! Hence, P{X = i} is increasing for λ ≥ i and decreasing for λ < i.
32. (a) .394 (b) .303 (c) .091 1 33. c
−1
(1 − x 2 )d x = 1
1 x 3 c x− =1 3 −1 3 c= 4 3 1 F(y) = (1 − x 2 )d x 4 −1 y3 2 3 y− + , −1 < y < 1 = 4 3 3
Instructor’s Manual to Accompany
34. c
17
2 0
4x − 2x 2 d x = 1
c(2x 2 − 2x 3 /3) = 1 8c/3 = 1 3 c= 8 1 3 3 3/2 P <X< = 4x − 2x 2 d x 2 2 8 1/2 11 = 16 ∞ 10 1 dx = 35. P{X > 20} = 2 2 20 x area of disk of radius x 36. P{D ≤ x} = area of disk of radius 1 π x2 = x2 = π 37. P{M ≤ x} = P{max(X 1 , . . . , X n ) ≤ x} = P{X 1 ≤ x, . . . , X n ≤ x} n = P{X i ≤ x} i=1
= xn d p{M ≤ x} = nx n−1 dx 10 38. (a) Using that 10 i=1 X(i) = i=1 Xi , yield the solution: 10μ. (b) N is the number of the Xi that are less than x. Because each is independently less than x with probability F (x), it follows that N is binomial with parameters 10 and F (x). (c) The desired event is that the number of the Xi that are less than m is between 2 and 7. Thus, 7 10 (1/2)10 . P (X(2) < m < X(8) ) = i f M (x) =
i=2
8 12
39. (a)
7 6 20 13
(1/7)
(b) This is the probability that if all balls are removed then the specified blue ball is chosen last among itself and all the red balls, giving the result 1/9. (c) With B being the number of blue balls remaining in urn when one stops, then X = 20 − B. Defining an indicator for each blue ball now gives, upon using the result of (b), that E[X] = 20 − 12/9.
18
Introduction to Probability Models
40. Let X denote the number of games played. P{X = 4} = p 4 + (1 − p)4 P{X = 5} = P{X = 5, I wins 3 of first 4} + P{X = 5, II wins 3 of first 4} = 4 p3 (1 − p) p + 4(1 − p)3 p(1 − p) P{X = 6} = P{X = 6, I wins 3 of first 5} + P{X = 6, II wins 3 of first 5} = 10 p3 (1 − p)2 p + 10 p 2 (1 − p)3 (1 − p) P{X = 7} = P{first 6 games are split} = 20 p 3 (1 − p)3 E[X ] =
7
i P{X = i}
i=4
When p = 1/2, E[X] = 93/16 = 5.8125 41. Let X i equal 1 if a changeover results from the ith flip and let it be 0 otherwise. Then number of changeovers =
n
Xi
i=2
As, E[X i ] = P{X i = 1} = P{flip i − 1 = flip i} = 2 p(1 − p) we see that E[number of changeovers] =
n
E[X i ]
i=2
= 2(n − 1) p(1 − p) 42. Suppose the coupon collector has i different types. Let X i denote the number of additional coupons collected until the collector has i + 1 types. It is easy to see that the X i are independent geometric random variables with respective parameters (n − i)/n, i = 0, 1, . . . , n − 1. Therefore, n−1 n−1 n−1 [X i ] = Xi = n/(n − i) i=0
i=0
i=0 n
=n
j=1
1/ j