Chapter 7 Probability - Arizona State University

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7-8: Probability and Statistics

Mat 211

Dr. Firoz

Chapter 7 Probability

Definition: Probability is a real valued set function P that assigns to each event A in the sample space S a number P(A), called the probability of A, such that the following properties hold:

a) 1 P(A) 0 b) P(S) 1 c) P(S) P(A) P(A ) P(A) 1 P(A ) , where A is the complement of A. d) For events A,and B we have P(A B) P(A) P(B) P(A B) e) If A and B are mutually exclusive (means A B , P(A B) P( ) 0 , then

P(A B) P(A) P(B) f) For three events A, B, and C verify that

P(A B C) P(A) P(B) P(C) P(A B) P(B C) P(C A) P(A B C)

Example 1. Roll a die once. The total outcomes are namely 1, 2, 3, 4, 5, and 6. The sample space is the set S {1, 2, 3, 4, 5, 6}. The event E rolling a 3 is the singleton set

E {3}. Now the probability of the event E for rolling a 3 is p(E) p(rolling a 3) 1 , 6

because there are 6 total outcomes and 1 desired outcome.

Example 2. Roll a die once. Write the sample space. Find the following probabilities:

a) P(E) P(rolling a 3 or a 5)

b) P(E) P(rolling a 3 or more)

c) P(E) P(rolling a number greater than 10)

d) P(E) P(rollingan even number)

e) o(E) = odds for the event that the roll is a 3 or a 5

Answer: a) 1/3

b) 2/3

c) 0

d) ? e) 2 : 4

Example 3. Roll a die twice. Write the sample space. Find the following probabilities:

a) P(E) P(rolling a sum 3 or a sum 5) b) P(E) P(rolling a sum 3 or more) c) P(E) P(rolling a sum greater than 10) d) P(E) P(rolling a sum greater than 12)

The sample space of the event of rolling a die twice:

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11 12 13 14 15 16 21 22 23 24 25 26 31 32 33 34 35 36 41 42 43 44 45 46 51 52 53 54 55 56 61 62 63 64 65 66

Answer: a) 6/36

b) 35/36

c) 3/36

d) 0

Example 4. Roll a die twice. Write the sample space. Find the following probabilities: a) P(E) P(all two rollsare either a1or a 3) b) P(E) P(all two rollsare not 2) c) P(E) P(all two rollsare above 3) d) P(E) P(all are a1or all are a 5)

The sample space of the event of rolling a die twice:

11 12 13 14 15 16 21 22 23 24 25 26 31 32 33 34 35 36 41 42 43 44 45 46 51 52 53 54 55 56 61 62 63 64 65 66

Answer: a) Look at the sample space, we have all three rolls either a 1 or a 3 are 11, 13, 31, and 33. Thus P(E) P(all two rollsare either a1or a 3) = 4/36.

We could find this probability using independent event and multiplication principle as the probability of getting first outcome either 1 or 3 is 2/6, then the probability of getting second outcome either 1 or 3 is again 2/6. By multiplication principle the probability of getting all two outcomes either a 1 or a 3 is 2 / 6 2 / 6 4 / 36

b) P(E) P(all two rollsare not 2) = 25/36, you may verify looking at the sample space,

that we have 12, 21, 22, 23, 24, 25, 26, 32, 42, 52, 62 with outcome 2's. There are 25 outcomes with no 2's. Using independent event and multiplication principle we have P(E) P(all two rollsare not 2) = 5/ 6 5/ 6 25/ 36

c) P(E) P(all two rollsare above 3) = 9/36 Using independent event and multiplication principle we have P(E) P(all two rollsare above 3) = 3/ 6 3/ 6 9 / 36 d) P(E) P(all are a1or all are a 5) = 1/ 6 1/ 6 1/ 6 1/ 6 2 / 36 . You look at the sample space we have only two outcomes 11 and 55.

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Example 5. Roll a die three times. How many outcomes do we have in the sample space? Find the following probabilities:

a) P(E) P(all three rolls are either a 1 or a 3) b) P(E) P(all three rolls are not 2) c) P(E) P(all three rolls are above 3)

d) P(E) P(all are a1or all are a 5)

Answer: a) 2 / 6 2 / 6 2 / 6 8/ 216 c) 3/ 6 3/ 6 3/ 6 27/ 216

b) 5/ 6 5/ 6 5/ 6 125/ 216 d) 1/ 6 1/ 6 1/ 6 1/ 6 1/ 6 1/ 6 2 / 216

Example 6. Roll a die five times. How many outcomes do we have in the sample space. Find the following probabilities:

a) P(E) P(all five rolls are either a 1 or a 3) b) P(E) P(all five rolls are not 2) c) P(E) P(all five rolls are above 3)

d) P(E) P(all are a1or all are a 5)

Answer: a) 25 / 65 b) 55 / 65

c) 35 / 65

d) 2 / 65

Methods of Enumeration

1. A true/false test contains 10 questions. In how many ways can a student answer

the questions? If a student makes random guesses, what is the probability that the

student will make exactly 5 questions correct?

Answer:

210 1024, c(10,5) /1024 0.2461.

2. How many three letter words (without meaning) are possible when repetition of letters is not allowed? What is the probability that those words will not start with a vowel? Answer: P(26,3), 0.8077

3. How many three letter words (without meaning) are possible when repetition of letters is allowed? What is the probability that those words will not start with a vowel? Answer: 263, 0.8077

4. A coin is tossed 9 times. What is the probability of getting at least 2 heads? Answer: 0.9805

5. A company has 9 women and 8 men. What is the probability that a 7 person committee will have 4 men and 3 women? Answer: 0.3023

6. If the letters in the word POKER are rearranged, what is the probability that the word will begin with a K and ends with an O? Answer: 0.05

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7. A computer retail store has 12 personal computers in stock. A customer wants to purchase three of the computers. Assume that of the 12 computers, 4 are defective. If the computers are selected at random what is the probability that exactly one of the purchased computers is defective? Answer: 0.509

8. A computer retail store has 12 personal computers in stock. A customer wants to purchase three of the computers. Assume that of the 12 computers, 4 are defective. If the computers are selected at random what is the probability that at least one of the purchased computers is defective? Answer: 0.7455

Independent Probabilities:

Example 1. A card is chosen at random from a deck of 52 cards. It is then replaced, the deck reshuffled and a second card is chosen. What is the probability of getting a jack and an eight? Solution. The event is independent. The probability of drawing first card a jack is 4/52 and second card an eight is 4/52. Also drawing a first card an eight is 4/52 and second card a jack is 4/52. The probability of drawing a jack and an eight is 4 / 52 4 / 52 4 / 52 4 / 52 2 /169

Exercise: A card is chosen at random from a deck of 52 cards. It is then replaced, the deck reshuffled and a second card is chosen.

a) What is the probability of getting a jack and then an eight? Ans: 1/169 b) What is the probability of getting a diamond and then a heart? Ans: 1/16

Example 2. A family has two children. Using b to stand for boy and g for girl in ordered pairs, give each of the following.

a) the sample space

b) the event E that the family has exactly one daughter.

c) the event F the family has at least one daughter

d) the event G that the family has two daughters

e) p(E)

f) p(F)

g) p(G)

Example 3. A group of three people is selected at random. 1)What is the probability that all three people have different birthdays. 2) What is the probability that at least two of them have the same birthday?

1) The probability that all three people have different birthdays is 365P3 365 364 363 0.992 3653 365 365 365

2) The probability that at least two people have same birthday is 1 365 364 363 1 0.992 0.008 365 365 365

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Conditional probability. A conditional probability is a probability whose sample space has been limited to only those outcomes that fulfill a certain condition. A conditional probability of an event A, given event B is p(A | B) n(A B)

n(B)

Example 1. In a newspaper poll concerning violence on television, 600 people were

asked, "what is your opinion of the amount of violence on prime time television ? is there

too much violence on television?" Their responses are indicated in the table below.

Yes (Y) No (N)

Don't know

Total

Men (M)

162

95

23

280

Women (W) 256

45

19

320

Total

418

140

42

600

Suppose we label the events in the following manner: W is the event that a response is from a woman, M is the event that a response is from a man, Y is the event that a response is yes, and N is the event that a response is no, then the event that a woman responded yes would be written as Y | W and p(Y | W) = 256/320 = 0.8. Use the given table to answer following questions.

a) p(N)

b) p(W)

c) p(N | W)

f) p(W N) g) p(Y)

h) p(M)

k) p(Y M )

l) p(M Y)

d) p(W | N) e) p(N W)

i) p(Y | M) j) p(M | Y)

m) p(W Y)

n) p(W | Y)

Answer: a) 0.23

b) 0.53

c) 0.14

d) 0.32

e) 0.08

f) 0.08

g) 0.70

h) 0.47

i) 0.58

j) 0.39

k) 0.27

l) 0.27

m) 0.43

n) 0.61

Independent Events

Independent Events: Two events A and B are called independent if and only if P(A B) P(A)P(B) , otherwise A and B are dependent.

Example 1. In two tosses of a single fair coin show that the events "A head on the first toss" and "A head on the second toss" are independent.

Solution: The sample space S S {HH, HT, TH, TT}, the event with a head on the first toss A {HH, HT} and an event with a head on the second toss B {HH, TH}. Now show that P(A B) P(A)P(B) .

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