section 1.6 suppose that s = b 1 b 2 … b m is an outcome space with b i b j = for i j, that...

Section 1.6

Suppose that S = B1 B2 … Bm is an outcome space with Bi Bj = for i j , that is, B1 , B2 , … , Bm are mutually exclusive and exhaustive. We call B1 , B2 , … , Bm a partition of the outcome space S.

S

B1 B2 Bm• • •

If P(A) > 0, then for each k = 1, 2, …, m,

S

B1 B2 Bm• • •

A

P(Bk | A) =P(Bk A)———— = P(A)

P(Bk A)———————————————— =P[(B1 A)(B2 A)…(Bm A)]

P(Bk A)—————— =

P(Bi A)i = 1

m

If P(A) > 0, then for each k = 1, 2, …, m,

P(Bk | A) = P(Bk) P(A | Bk)———————— P(Bi) P(A | Bi)

i = 1

m

Bayes’ Rule

Bayes’ Theorem is stated as follows:

P(Bk) P(A | Bk)———————— P(Bi) P(A | Bi)

i = 1

m

P(Bk A)———————————————— =P[(B1 A)(B2 A)…(Bm A)]

P(Bk A)—————— =

P(Bi A)i = 1

m

1.

(a)

Boxes of a certain type of cereal are produced at three factories, one located at Limesville, one at Middletown, and one at Transburg; these factories respectively are responsible for 20%, 30%, and 50% of the total production. Among the boxes of cereal produced at Limesville, 5% are underweight; among those produced at Middletown, 2% are underweight; among those produced at Transburg, 1% are underweight. Suppose one box of cereal is to be randomly selected among the total production, and the following events are defined:

L = one randomly selected box comes from LimesvilleM = one randomly selected box comes from MiddletownT = one randomly selected box comes from TransburgU = one randomly selected box is underweight

Find each of the probabilities listed, observe that L, M, T form a partition of the sample space, and complete the Venn diagram to display the events L, M, T, and U.

TML

U

P(L) = P(M) = P(T) =

P(U | L) = P(U | M) = P(U | T) =

0.2 0.3 0.5

0.05 0.02 0.01

(b) Find the probability that the randomly selected box is underweight.P(U) = P((U L) (U M) (U T)) =

P(U L) + P(U M) + P(U T) =

P(L) P(U | L) + P(M) P(U | M) + P(T) P(U | T) =

(0.2)(0.05) + (0.3)(0.02) + (0.5)(0.01) = 0.021

(c)

(d)

(e)

Find the probability that the randomly selected box is from the Limesville factory given that it is underweight.

Find the probability that the randomly selected box is from the Middletown factory given that it is underweight.

Find the probability that the randomly selected box is from the Transburg factory given that it is underweight.

P(L | U) =P(L U)———— = P(U)

(0.2)(0.05)———— = 0.021

10—21

P(M | U) =P(M U)———— = P(U)

(0.3)(0.02)———— = 0.021

6— =21

2— 7

P(T | U) =P(T U)———— = P(U)

(0.5)(0.01)———— = 0.021

5—21

2.

(a)

The player in a particular lottery chooses five distinct integers from 1 to 20, which are then printed on a lottery ticket. Friday at 6:00pm, five distinct integers from 1 to 20 are randomly selected, and any ticket which matches at least four of these five integers (in any order) is a winning ticket.

Sam buys one ticket with the integers 3, 4, 7, 8, 13. What is the probability that Sam has a winning ticket?

20

5

5

4

15

1

+20

5

5

5

15

0 1 = —— = 0.0049 204

(b)

20

5

5

4

15

1

+20

5

5

5

15

0

2

Sally buys two tickets, one with the integers 3, 4, 7, 8, 13 and one with the integers 1, 5, 9, 15, 18. What is the probability that Sally has a winning ticket?

2 = —— = 0.0098 204

(c) George buys two tickets, one with the integers 3, 4, 7, 8, 13 and one with the integers 3, 4, 7, 14, 20. What is the probability that George has a winning ticket?

20

5

5

4

15

1

+20

5

5

5

15

0

2 –

20

5

3

3

2

1

2

1

13

0

= 0.0095

(d)

3.

Linda buys two tickets, one with the integers 3, 4, 7, 8, 13 and one with the integers 3, 4, 7, 8, 13. What is the probability that Linda has a winning ticket?

20

5

5

4

15

1

+20

5

5

5

15

0

Consider a lottery where a player chooses m distinct integers from 1 to n, where m < 2n, on a lottery ticket. In order for the player to win, at least k, where m/2 < k < m, of the m integers on the ticket must match (in any order) m randomly selected integers. Suppose a player buys two lottery tickets, where exactly s of integers are the same on both tickets. What is the probability that the player has a winning ticket if s = 0, and how large can s be without changing this probability?

1 = —— = 0.0049 204

If s = 0, then both tickets cannot be winners, and the probability of one winning ticket is

n

m

m

k

n–m

m–k

+ n

m

2

m

k+1

n–m

m–k–1

+ … + n

m

m

m

n–m

0

Suppose n > 20, m = 10, k = 7, s = 10.Ticket #1: 01 02 03 04 05 06 07 08 09 10

Ticket #2: 01 02 03 04 05 06 07 08 09 10Both tickets can be winners.

k = 7 of the m = 10 winning numbers can be on ticket #1.



n

m

m

k

n–m

m–k

+ n

m

2

m

k+1

n–m

m–k–1

+ … + n

m

m

m

n–m

0






n

m

m

k

n–m

m–k

+ n

m

2

m

k+1

n–m

m–k–1

+ … + n

m

m

m

n–m

0




(k1)+(ks) = k = 7 of the m = 10 winning numbers can be on ticket #2.


n

m

m

k

n–m

m–k

+ n

m

2

m

k+1

n–m

m–k–1

+ … + n

m

m

m

n–m

0






n

m

m

k

n–m

m–k

+ n

m

2

m

k+1

n–m

m–k–1

+ … + n

m

m

m

n–m

0


Ticket #2: 11 12 13 14 15 16 17 08 09 10Both tickets cannot be winners.


(k4)+(ks) = k = 7 of the m = 10 winning numbers can NOT be on

ticket #2.


n

m

m

k

n–m

m–k

+ n

m

2

m

k+1

n–m

m–k–1

+ … + n

m

m

m

n–m

0


Ticket #2: 11 12 13 14 15 16 17 08 09 10Both tickets cannot be winners.

k + (k – s) m

This probability stays the same as long as the tickets cannot both be winners. It is possible for both tickets to be winners if 2k – s m but not if 2k – s > m . Consequently, in order for the probability to stay the same, we must have

s

2k – m – 1.


n

m

m

k

n–m

m–k

+ n

m

2

m

k+1

n–m

m–k–1

+ … + n

m

m

m

n–m

0

section 1.6 suppose that s = b 1 b 2 … b m is an outcome space with b i b j = for i j, that...

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