Intro to Probability (Solutions Manual) by Charles M. Grinstead & J. Laurie Snell

Author:Charles M. Grinstead & J. Laurie Snell [Grinstead, Charles M. & Snell, J. Laurie]
Language: eng
Format: epub
Published: 2011-02-19T05:00:00+00:00

Let U have distribution

p =

0

π/2

π

3π/2

U

1/4 1/4 1/4

1/4

.

Then let X = cos(U) and Y = sin(U). X and Y have distributions p =

1

0

−1

0

X

1/4 1/4 1/4 1/4

,

p =

0

1

0

−1

Y

1/4 1/4 1/4 1/4

.

Thus E(X) = E(Y ) = 0 and E(XY ) = 0, so Cov(X, Y ) = 0. However, since sin2(x) + cos2(x) = 1, X and Y are dependent.

25.

(a) The expected value of X is

5000

µ = E(X) =

iP (X = i) .

i=1

The probability that a white ball is drawn is

n

P (white ball is drawn) =

P (X = i) i

5000 .

i=1

Thus

P (white ball is drawn) =

µ

5000 .

(b) To have P (white,white) = P (white)2 we must have 5000

n

( i )2

i

5000 P (X = i) = (

5000P (X = i))2 .

i=1

i=1

But this would mean that E(X2) = E(X)2, or V (X) = 0. Thus we will have independence only if X takes on a specific value with probability 1.

(c) From (b) we see that

1

P (white,white) =

E(X2) .

50002

Thus

(σ2 + µ2)

V (X) =

5000

.

2

27.

The number of boxes needed to get the j’th picture has a geometric distribution with (2n − k + 1)

p =

2

.

n

Thus

2n(k − 1)

V (Xj) = (2

.

n − k + 1)2

Therefore, for a team of 26 players the variance for the number of boxes needed to get the first half of the pictures would be

13

26(k − 1) = 7

(26

.01 ,

− k + 1)2

k=1

23

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