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

136 PROBABILITY [CHAP. 7

Thus we can conclude that the probability that a value ofXlies between 65 and 85 is at least 1−( 1 / 2 )^2 = 3 /4;
that is:
P( 65 ≤X≤ 85 )≥ 3 / 4

Similarly, by lettingk=3, we can conclude that the probability that a value ofXlies between 60 and 90 is at

least 1−( 1 / 3 )^2 = 8 /9.

Sample Mean and Law of Large Numbers

Consider a finite number of random variableX,Y,...,Zon a sample spaceS. They are said to beindependent
if, for any valuesxi,yj,...,zk,

P(X=xi,Y=yj,...,Z=zk)≡P(X=xi)P (Y=yj)...P(Z=zk)

In particular,XandYare independent if

P(X=xi,Y=yj)≡P(X=xi)P (Y=yj)

Now letXbe a random variable with meanμ. We can consider the numerical outcome of each ofn
independent trials to be a random variable with the same distribution asX. The random variable corresponding
to theith outcome will be denoted byXi(i= 1 , 2 ,...,n). (We note that theXiare independent with the same
distribution asX.) The average value of allnoutcomes is also a random variable which is denoted byXnand
called thesample mean. That is:

Xn=

X 1 +X 2 +···+Xn

n

The law of large numbers says that asnincreases the value of the sample meanXnapproaches the mean valueμ.
Namely:

Theorem 7.11 (Law of Large Numbers): For any positive numberα, no matter how small, the probability that
the sample meanXnhas a value in the interval[μ−α, μ+α]approaches 1 asnapproaches
infinity. That is:

P([μ−α≤X≤μ+α])→1asn→∞.

EXAMPLE 7.19 Suppose a die is tossed 5 times with outcomes:

x 1 = 3 ,x 2 = 4 ,x 3 = 6 ,x 4 = 1 ,x 5 = 4

Then the corresponding valuexof the sample meanX 5 follows:

x=

3 + 4 + 6 + 1 + 4

5

= 3. 6

For a fair die, the meanμ= 3 .5. The law of large numbers tells us that, asngets larger, there is a greater
likelihood thatXnwill get closer to 3.5.

SolvedProblems

SAMPLE SPACES AND EVENTS

7.1. Let a coin and a die be tossed; and let the sample spaceSconsists of the 12 elements:

S={H 1 ,H 2 ,H 3 ,H 4 ,H 5 ,H 6 ,T 1 ,T 2 ,T 3 ,T 4 ,T 5 ,T 6 }