Introduction to Probability 1st Edition Anderson Solutions Manual
richard@qwconsultancy.com
1|Pa ge
Introduction to Probability Detailed Solutions to Exercises David F. Anderson Timo Seppäläinen Benedek Valkó
c
David F. Anderson, Timo Seppäläinen and Benedek Valkó 2018
Solutions to Chapter 2
51
1) = P (N 9). We can check this by using the geometric sum
(c) We have P (X formula to get
9 X
k=1
p(1
p)k 1 = p
1 (1 p)9 ) =1 1 (1 p)
(1
p)9 .
Here is another way to see this, without any algebra. Imagine that we draw balls with replacement infinitely many times. Think of X as the number of green balls in the first 9 draws. N is still the number of draws needed for the first green. Now if X 1, then we have at least one green within the first 9 draws, which means that the first green draw happened within the first 9 draws. Thus X 1 implies N 9. But this works in the opposite direction as well: if N 9 then the first green draw happened within the first 9 draws, which means that we must have at least one green within the first 9 picks. Thus N 9 implies X 1. This gives the equality of event: {X 1} = {N 9}, and hence the probabilities must agree as well. 2.62. Regard the drawing of three marbles as one trial, with success probability p given by p = P (all three marbles blue) =
9 3 13 3
=
7 · 8 · 9 · 10 42 = . 10 · 11 · 12 · 13 143
42 X ⇠ Bin(20, 143 ). The probability mass function is ✓ ◆ 20 42 k 101 20 k P (X = k) = for k = 0, 1, 2, . . . , 20. 143 143 k
2.63. The number of heads in n coin flips has distribution Bin(n, 1/2). Thus the probability of winning if we choose to flip n times is ✓ ◆ n 1 n(n 1) fn = P (n flips yield exactly 2 heads) = = . 2 2n 2n+1
We want to find the n which maximizes fn . Let us compare fn and fn+1 . We have n(n 1) (n + 1)n < () 2(n 1) < n + 1 () n+1 2 2n+2 Similarly, fn > fn+1 if and only if n > 3, and f3 = f4 . Thus fn < fn+1 ()
n < 3.
f 2 < f 3 = f 4 > f5 > f 6 > . . . . This means that the maximum happens at n = 3 and n = 4, and the probability 3 of winning at those values is f3 = f4 = 3·2 24 = 8 . 2.64. Let X be the number of correct answers. X is the number of successes in 20 independent trials with success probability p + 12 r. P (X
19) = P (X = 19) + P (X = 20) = 20 p + 12 r
19
q + 12 r + p + 12 r
20
.
2.65. Let A be the event that at least one die lands on a 4 and B be the event that all three dice land on di↵erent numbers. Our sample space is the set of all triples (a1 , a2 , a3 ) with 1 ai 6. All outcomes are equally likely and there are 216 outcomes. We need P (A|B) = PP(AB) (B) . There are 6 · 5 · 4 = 120 elements in B. To count the elements of AB, we first consider Ac B. This is the set of triples where
52
Solutions to Chapter 2
the three numbers are distinct and none of them is a 4. So #Ac B = 5 · 4 · 3 = 60. Then #AB = #B #Ac B = 120 60 = 60 and 60 P (AB) 1 = 216 = . 120 P (B) 2 216
P (A|B) = 2.66. Let
fn = P (n die rolls give exactly two sixes) =
✓ ◆ n 1 2 5 n 2 n(n 1)5n 2 = . 6 6 2 2 · 6n
Next, 1)5n 2 (n + 1)n5n 1 < 2 · 6n 2 · 6n+1 () n < 11.
fn < fn+1 ()
n(n
() 6(n
1) < 5(n + 1)
By reversing the inequalities we get the equivalence fn > fn+1 () n > 11. By complementing the two equivalences, we get fn = fn+1 () fn
fn+1 and fn fn+1
() n
11 and n 11 () n = 11.
Putting all these facts together we conclude that the probability of two sixes is maximized by n = 11 and n = 12 and for these two values of n, that probability is 11 · 10 · 59 ⇡ 0.2961. 2 · 611
2.67. Since {X = n + k} ⇢ {X > n} for k P (X = n + k|X > n) =
1, we have
P (X = n + k, X > n) P (X = n + k) (1 p)n+k 1 p = = . P (X > n) P (X > n) P (X > n)
Evaluate the denominator: P (X > n) =
1 X
P (X = k) =
k=n+1
= p(1
1 X
(1
p)k 1 p
k=n+1
p)n
1 X
(1
p)k = p(1
k=0
p)n ·
1
1 (1
p)
= (1
Thus, P (X = n + k|X > n) =
(1 p)n+k 1 p (1 p)n+k 1 p = P (X > n) (1 p)n
= (1 2.68. For k
p)k 1 p = P (X = k).
1, the assumed memoryless property gives P (X = k) = P (X = k + 1 | X > 1) =
P (X = k + 1) P (X > 1)
p)n .
Solutions to Chapter 3
73
R3 We could also compute this probability by evaluating the integral 2 f (x)dx. (c) Using the probability density function we can write Z 1 2 2X E[(1 + X) e ]= f (x)(1 + x)2 e 2x dx 0 Z 1 Z 1 2 2x 2 = (1 + x) e (1 + x) dx = e 2x dx 0
0
1
1 2x 1 e = . 2 2 x=0
=
3.38. (a) Since Z is continuous and the pd.f. is given, we can compute its expectation as Z 1 Z 1 z=1 5 6 E[Z] = zf (z)dz = z · 52 z 4 dz = 12 z = 0. 1
1
Z 1/2
Z 1/2
z= 1
(b) We have P (0 < Z < 1/2) =
f (z)dz =
0
0
z=1/2 5 4 1 5 2 z dz = 2 z
z=0
= 12
1 5 1 = 64 . 2
(c) We have P {Z < 12 | Z > 0} =
P (Z < 12 and Z > 0) P (0 < Z < 1/2) = . P (Z > 0) P (Z > 0)
1 The numerator is 64 . The denominator is Z 1 Z 1 z=1 5 4 z5 P (Z > 0) = f (z)dz = z dz = = 1/2. 2 z=0 0 0 2
Thus, 1 64
1 . 1/2 32 (d) Since Z is continuous and the pd.f. is given, we can compute E[Z n ] for n as follows Z 1 Z 1 Z 1 5 n+4 E[Z n ] = z n f (z)dz = z n · 52 z 4 dz = dz 2z P {Z < 12 | Z > 0} =
1
=
1
1
1
z=1
5 = 2(n+5) z n+5 5 = 2(n+5) 1
5 = 2(n+5) 1n+5
z= 1 n+5
( 1)
( 1)n+5
.
Note that ( 1)n+5 = 1 if n is odd and ( 1)n+5 = 1 if n is even. Thus ( 5 , if n is odd n E[Z ] = n+5 0, if n is even. 3.39. (a) One possible example: P (X = 1) =
1 , 3
P (X = 2) =
3 4
1 5 = , 3 12
P (X = 3) = 1 P (X = 1) P (X = 2) =
1 . 4
74
Solutions to Chapter 3
Then F (1) = P (X 1) = P (X = 1) = 13 , F (2) = P (X 2) = P (X = 1) + P (X = 2) =
3 4
and F (3) = P (X 3) = P (X = 1) + P (X + 2) + P (X = 3) = 1.
(b) There are a number of possible solutions. Here is one that can be checked easily using part (a): 81 0x1 > 3 > > <5 1<x2 f (x) = 12 1 > 2<x3 > > :4 0 otherwise.
1 3.40. Here is a continuous example: R 1 let f (x) = x2 for x 1 and 0 otherwise. This is a nonnegative function with 1 f (x)dx = 1, thus there is a random variable X with p.d.f. f . Then the cumulative distribution function of X is given by ( Z x 0, if x < 1 F (x) = f (y)dy = R x 1 dy = 1 1/x, if x 1. 1 1 y2 1 n for each positive integer n.
In particular, F (n) = 1
3.41. We begin by deriving the probability F (s) = P (X s) using the law of total probability. For s 2 (3, 4), F (s) = P (X s) =
6 X
k=1
P (X s | Y = k)P (Y = k) =
3 X 1
k=1
6
+
6 X s
k=4
1 · k 6
1 37s = + . 2 360 We can find the density function f on the interval (3, 4) by di↵erentiating this. Thus 37 f (s) = F 0 (s) = 360
for s 2 (3, 4).
3.42. (a) Note that 0 X 1 so FX (x) = 1 for x 1 and FX (x) = 0 for x < 0. For 0 x < 1 the event {X x} is the same as the event that the chosen point is in the trapezoid Dx with vertices (0, 0), (x, 0), (x, 2 x), (0, 2). The area of this trapezoid is 12 (2 + 2 x)x, while the area of D is (2+1)1 = 32 . Thus 2 P (X x) = Thus
1 (2 + 2 area(Dx ) = 2 3 area(D) 2
8 > <1,
FX (x) = 4x >3 :
0,
x2 3 ,
x)x
=
4x 3
x2 . 3
if x 1 if 0 x < 1 if x < 0.
To find FY we first note that 0 Y 2 so FY (y) = 1 for y for y < 0.
2 and FY (y) = 0