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Chapter 4: Q.4.35 (page 166)

A box contains 5red and 5blue marbles. Two marbles are withdrawn randomly. If they are the same color, then you win \(1.10; if they are different colors, then you win -\)1.00. (That is, you lose $1.00.) Calculate

(a) the expected value of the amount you win;

(b) the variance of the amount you win.

Short Answer

Expert verified

(a) The expected value of the amount you win will be -0.0667.

(b) The variance of the amount you win will be1.089.

Step by step solution

01

Given information (Part a)

A box contains 5red and 5blue marbles. Two marbles are withdrawn randomly. If they are the same color, then you win $1.10; if they are different colors, then you win−$1.00.

02

Solution (Part b)

Let,

x=1.1

p(1.1)=52102+52102

So,P(1.1)is the probability that either both are red or both are blue =49

X=−1.0

p(−1.0)=5151102

=59

E(X)=1.1×49−1.0×59

=−0.0667

03

Final answer (Part a)

The expected value of the amount you win will be−0.0667

04

Given information (Part b)

A box contains 5red and 5blue marbles. Two marbles are withdrawn randomly. If they are the same color, then you win $1.10; if they are different colors, then you win-$1.00

05

Solution (Part b)

Let the calculation will be,

EX2=(1.1)2×49+(−1)2×59

=0.537778+0.555556

=1.0933

Var⁡(x)=EX2−[E(X)]2

=1.0933−(−0.0667)2

=1.089

06

Final answer (Part b)

The variance of the amount you win will be 1.089.

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