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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 Two marbles are withdrawn randomly.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 Two marbles are withdrawn randomly.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)=1.1×491.0×59

=0.0667

03

Final answer (Part a)

The expected value of the amount you win will be0.0667

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 Two marbles are withdrawn randomly.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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