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

144 PROBABILITY [CHAP. 7

7.27. Prove Theorem 7.7: The probability of exactlyksuccesses in a binomial experimentB(n, p)is given by

P(k)=p(ksuccesses)=

(

n

k

)

pkqn−k

The probability of one or more successes is 1−qn.

The sample space of thenrepeated trials consists of alln-tuples (i.e.,n-element sequences) whose components

are eitherS(success) orF(failure). LetAbe the event of exactlyksuccesses. ThenAconsists of alln-tuples of which

kcomponents areSandn−kcomponents areF. The number of suchn-tuples in the eventAis equal to the number of

ways thatklettersScan be distributed among thencomponents of ann-tuple; henceAconsists ofC(n, k)=

(

n

k

)

sample points. The probability of each point inAispkqn−k; hence

P (A)=

(

n

k

)

pkqn−k

In particular, the probability of no successes is

P( 0 )=

(

n

0

)

p^0 qn=qn

Thus the probability of one or more successes is 1−qn.

RANDOM VARIABLES, EXPECTATION

7.28.A player tosses two fair coins. He wins $2 if two heads occur, and $1 if one head occurs. On the other

hand, he loses $3 if no heads occur. Find the expected valueEof the game. Is the game fair? (The game

is fair, favorable, or unfavorable to the player according asE=0,E>0orE<0.)

The sample spaceS={HH,HT,TH,TT}, and each sample point has probability 1/4. For the player’s gain,

we have

X(HH)=$2,X(HT)=X(T H)=$1, X(TT)=−$3

Hence the distribution ofXfollows:

xi 21 − 3

pi 1 / 42 / 41 / 4

ThusE=E(X)= 2 ( 1 / 4 )+ 1 ( 2 / 4 )− 3 ( 1 / 4 )=$0.25. SinceE(X) >0, the game is favorable to the player.

7.29.You have won a contest. Your prize is to select one of three envelopes and keep what is in it. Each of two

of the envelopes contains a check for $30, but the third envelope contains a check for $3000. Find the

expectationEof your winnings (as a probability distribution).

LetXdenote your winnings. ThenX=30 or 3000, andP( 30 )=^23 andP( 3000 )=^13. Hence

E=E(X)= 30 ·^23 + 3000 ·^13 = 20 + 1000 = 1020

7.30.A random sample with replacement of sizen=2 is drawn from the set{ 1 , 2 , 3 }, yielding the following

9-element equiprobable sample space:

S={( 1 , 1 ), ( 1 , 2 ), ( 1 , 3 ), ( 2 , 1 ), ( 2 , 2 ), ( 2 , 3 ), ( 3 , 1 ), ( 3 , 2 ), ( 3 , 3 )}

(a) LetXdenote the sum of the two numbers. Find the distributionfofX, and find the expected value

E(X).

(b) LetYdenote the minimum of the two numbers. Find the distributiongofX, and find the expected

valueE(Y).

(a) The random variableXassumes the values 2, 3, 4, 5, 6. We compute the distributionfofX:

(i) One point (1, 1) has sum 2; hencef( 2 )=^19.