FINITE STOCHASTIC PROCESSES

4.18.

A box contains three coins; one coin is fair, one coin is two-headed, and one coin is weighted so that the probability of heads appearing is 9. A coin is selected at random and tossed. Find the probability

p

that heads appears.

Construct the tree diagram as shown in Figure (a)below. Note that I refers to the fair coin, I1 to the two-headed coin, and I11 to the weighted coin. Now heads appears along three of the

paths; hence

L . L + L . 1 + 1 . 1

= -11

p = 3 2 3 3 3 18

4.19.

We are given three urns as follows:

Urn A contains 3 red and

5

white marbles.

Urn B contains 2 red and 1white marble.

Urn C contains

2

red and 3 white marbles.

An urn is selected a t random and a marble is drawn from the urn. If the marble is red, what is the probability that it came from urn A ?

Construct the tree diagram as shown in Figure (b) above.

We seek the probability that A ^was selected, given that the marble is red; that is, P(A

I

^R).

In order to find P ( A

I

R), it is necessary first to compute P ( AnR) and P(R).

The probability that urn A is selected and a red marble drawn is +*:

= i;

^{that is,}

P(AnR) =

5.

Since there are three paths leading to a red marble, P(R)=13 ii8

+

^{3 3 + I * 3 5 =173 360'}

Thus

65

CHAP. 41 CONDITIONAL PROBABILITY AND INDEPENDENCE

Alternately, by Bayes’ theorem,

4.20.

Box A contains nine cards numbered 1 through 9, and box B contains five cards numbered

1

through

5.

A box is chosen a t random and a card drawn. If the number is even, find the probability that the card came from box A.

The tree diagram of the process is shown in Figure (a)below.

We seek P(A IE), the probability that A was selected, given that the number is even. The

$;

^{that is, P(AnE)}

i.

probability that box A and an even number is drawn is :*: = = Since there are two paths which lead to an even number, P(E)=^{1 9}

+

2 * 5 - 4 5 .

-

Thus

2

4.21.

An urn contains 3 red marbles and 7 white marbles. A marble is drawn from the urn and a marble of the other color is then put into the urn. A second marble is drawn from the urn.

(i) Find the probability

p

that the second marble is red.

(ii) If both marbles were of the same color, what is the probability p that they were both white?

Construct the tree diagram as shown in Figure (b) above.

(i) Two paths of the tree lead to a red marble: ^p

=

$j

&+&

^$j=

E.

(ii) The probability that both marbles were white is $j*

5 =%.

The probability that both marbles were of the same color, i.e. the Probability of the reduced sample space, is

+ $ 5

⁼

E.

Hence the conditional probability p =

%/$

=

5.

$,

4.22.

We are given two urns as follows:

Urn A contains 3 red and 2 white marbles.

Urn B contains

2

red and

5

white marbles.

An urn is selected at random; a marble is drawn and put into the other urn; then a marble is drawn from the second urn. Find the probability

p

that both marbles drawn are of the same color.

Construct the following tree diagram:

66

CONDITIONAL PROBABILITY AND INDEPENDENCE [CHAP. 4

Note that if urn A is selected and a red marble drawn and put into urn B, then urn B has 3 red marbles and 5 white marbles.

Since there are four paths which lead to two marbles of the same color,

1 3 3 901

--.-.-

_{2 7 2 1680}

INDEPENDENCE

4.23.

Let A

=

event that a family has children of both sexes, and let B = event that a family has a t most one boy. (i) Show that A and B are independent events if a family has three children. (ii) Show that A and B are dependent events if a family has two children.

(i) We have the equiprobable space S = { b b b , bbg, bgb, b g g , g b b , g b g , g g b , g g g ) . Here A = W g , b s b , bgg7 g b b , g b g , g g b ) and so P ( A ) = ^-6 = ^-3

8 4

Since P ( A )P(B) =

2 -&

^{= Q = P ( AnB), A and B} are independent.

(ii) We have the equiprobable space Here

A = P ( A ) = -1

2

B = P(B) = -3

4

A n B = P ( A n B ) = 1^-₂

Since P ( A )P(B) ^# P ( AnB), A

4.24.

Prove: If

A

and B are independent events, then

Ac

and Bc are independent events.

P(AcnBC) = P ( ( AuB)c) = 1

-

P ( AuB) = 1

-

P ( A )

-

P ( B )

+

^{P ( AnB)}

= 1

-

P ( A )

-

P(B)

+

P ( A ) P(B) = [l

-

P ( A ) ] [ l

-

P(B)] = P(Ac)P(Bc) 4.25.

The probability that a man will live 10 more years is

$,

and the probability that his

wife will live

10

more years is 9. Find the probability that (i) both will be alive in

10

years, (ii) a t least one will be alive in

10

years, (iii) neither will be alive in

10

years, (iv) only the wife will be alive in

10

years.

Let A = event that the man is alive in 10 years, and B = event that his wife is alive in 10 years; then P ( A )=

#,

and P(B)=Q.

(i) We seek P(A nB). Since A and B a r e independent, P(A nB) =P(A)P(B)=

+

⁼

^A.

67

CHAP. 41 CONDITIONAL PROBABILITY AND INDEPENDENCE

(ii) We seek P ( AU B ) . P ( AUB)_{= P ( A )}

+

^P(B)

^-

^{P ( AnB) =}

⁵ +

^Q

^-

₌

+

(iii) We seek P(AcnBc). Now P(Ac) = 1

-

P(A)= 1

- t

=

8

^and P(Bc)= 1-P(B)= 1

-

^Q =

3.

Furthermore, since Ac and Bc are independent, P(AcnBc:,= P(Ac)P(Bc) =

2 8

=

a.

Alternately, since ( AuB)c= ACnBC, P(AcnBC)= P ( ( AuB)c)= 1 -P ( AU B ) = 1

-

₌

4.

(iv) We seek P(AcnB). Since P(Ac) = 1

-

P ( A ) =

0

and Ac and B are independent (see Problem 4.56), P(AcnB)= P(Ac)P(B)=

%.

4.26.

Box A contains 8 items of which 3 are defective, and box B contains

5

items of which

2

are defective. An item is drawn a t random from each box.

(i) What is the probability

p

that both items are nondefective?

(ii) What is the probability

p

that one item is defective and one not?

(iii) If one item is defective and one is not, what is the probability

p

that the defec- tive item came from box A

?

(i) The probability of choosing a nondefective item from .A is and from B is

t.

Since the events are independent, p =

i-! =#.

(ii) Method 1. The probability of choosing two defective items is _{: o f =}

$.

From (i) the probability that both are nondefective is

5.

^{Hence p = 1 - -}

&

^{= E .40}

Method 2. The probability p 1 of choosing a defective item from A and a nondefective item

3 3 9

from B is 5 - 5 = 40. The probability p , of choosing a nondefective item from A and a

9 1 19

defective item from B is

f -$

=

i.

^{Hence p = pl}

+

^{p , = ; ~ i i}

+

^{4 =}

^z.

(iii) Consider the events X = {defective item from A } and Y = {one item is defective and one nondefective}. We seek P(X

I

^{Y).By (ii), P ( X nY)= p1 =}

%

^{and P ( Y )=}

E.

^Hence

4.27.

The probabilities that three men hit a target are respectively Q, t and 4. ^Each

shoots once a t the target. (i) Find the probability

p

that exactly one of them hits the target. (ii) If only one hit the target, what is the probability that it was the first man

?

Consider the events A = {first man hits the target}, B = {second man hits the target), and C = {third man hits the target}; then P(A) =

Q,

P(B) =

&

and P(C) = Q. The three events are independent, and P(Ac)=

Q,

P(Bc) =

9,

P(Cc)=

8.

(i) Let E Z= {exactly one man hits the target}. Then

E = (AnBcnCC)^U (AcnBnCc:)^U (AcnBcnC)

In other words, if only one hit the target, then it was either only the first man, AnBcnCC, or only the second man, AcnBnCC, or only the third man, AcnBcnC. Since the three events are mutually exclusive, we obtain (using Problem 4.62)

p = P ( E ) = P(AnBCnCC)

+

P(AcnBnCc)

+

P(AcnBCnC)

= P ( A )P(Bc)P(Cc)

+

P(Ac)P ( B )P(Cc)

+

P(Ac)P(Bc)P(C)

- - 31

72

(ii) We seek P ( A

I

E), the probability that the first man hit the target given that only one man hit the target. Now A nE = A nBcn CC is the event that only the first man hit the target.

By (i), P ( A n E )= P(AnBCnCC)=

&

^{and P(E)=}

z;

^hence

68

CONDITIONAL PROBABILITY AND INDEPENDENCE [CHAP. 4