Schaum's Outline of Discrete Mathematics, Third Edition (Schaum's Outlines)

146 PROBABILITY [CHAP. 7

(a) μ=xipi= 2

( 1

3

)

+ 3

( 1

2

)

+ 11

( 1

6

)

= 4

E(X^2 )=x^2 ipi= 22

( 1

3

)

+ 32

( 1

2

)

+ 112

( 1

6

)

= 26

σ^2 =Var(X)=E(X^2 )−μ^2 = 26 − 42 = 10

σ=

√

Var(X)=

√

10 = 3. 2

(b) μ=xipi= 1 ( 0. 4 )+ 3 ( 0. 1 )+ 4 ( 0. 2 )+ 5 ( 0. 3 )= 3

E(X^2 )=x^2 ipi= 1 ( 0. 4 )+ 9 ( 0. 1 )+ 16 ( 0. 2 )+ 25 ( 0. 3 )= 12

σ^2 =Var(X)=E(X^2 )−μ^2 = 12 − 9 = 3

σ=

√

Var(X)=

√

3 = 1. 7

7.33.A fair die is tossed yielding the equiprobable sample spaceS={ 1 , 2 , 3 , 4 , 5 , 6 }wheren(S)=6 and each

point has probability 1/6.

(a) LetXbe the random variable which denotes twice the number that occurs. Find the distributionfof

Xand its expectationE(X).

(b) LetYbe the random variable which assigns 1 or 3 according as an odd or even number occurs. Find the

distributiongofYand its expectationE(Y).

(a) Here the range spaceRX={ 2 , 4 , 6 , 8 , 10 , 12 }since

X( 1 )= 2 ,X( 2 )= 4 ,X( 3 )= 6 ,X( 4 )= 8 ,X( 5 )= 10 ,X( 6 )= 12

Also, each number occurs with probability 1/6. Thus the distributionfofXfollows:

x 24681012

f(x) 1 / 61 / 61 / 61 / 61 / 61 / 6

Hence

E(X)=

∑

xf (x)=

2

6

+

4

6

+

6

6

+

8

6

+

10

6

+

12

6

= 7

(b) Here the range spaceRY={ 1 , 3 }since

Y( 1 )= 1 ,Y( 2 )= 3 ,Y( 3 )= 1 ,Y( 4 )= 3 ,Y( 5 )= 1 ,Y( 6 )= 3

We compute the distributiongofYusing the fact thatn(S)=6:

(i) Three points 1, 3, 5 are odd and have image 1; henceg( 1 )= 3 /6.

(ii) Three points 2, 4, 6 are even and have image 3; henceg( 3 )= 3 /6.

Thus the distributiongofYis:

y 13

g(y) 3 / 63 / 6

Hence

E(Y)=

∑

yg(y)=

3

6

+

9

6

= 2

7.34. LetZ=X+YwhereXandYare the random variables in Problem 7.33. Find the distributionhofZ,

and findE(Z). Verify thatE(X+Y)=E(X)+E(Y).

The sample space is stillS={ 1 , 2 , 3 , 4 , 5 , 6 }and each point still has probability 1/6. We obtain using

Z(s)=(X+Y )(s)=X(s)+Y(s)

Z( 1 )=X( 1 )+Y( 1 )= 2 + 1 = 3 ; Z( 4 )=X( 4 )+Y( 4 )= 8 + 3 = 11 ,

Z( 2 )=X( 2 )+Y( 2 )= 4 + 3 = 7 ; Z( 5 )=X( 5 )+Y( 5 )= 10 + 1 = 11 ,

Z( 3 )=X( 3 )+Y( 3 )= 6 + 1 = 7 ; Z( 6 )=X( 6 )+Y( 6 )= 12 + 3 = 15.

Thustherange spaceRz={ 3 , 7 , 11 , 15 }. We compute the distributionhofZusing the fact thatn(S)=6:

(i) One point has image 3, soh( 3 )= 1 /6; (iii) Two points have image 11, soh( 11 )= 2 /6;

(ii) Two points have image 7, soh( 7 )= 2 /6; (iv) One point has image 15, soh( 15 )= 1 /6.