Mat 142 College MathematicsDepartment of Mathematics and Statistics
Chapter 2 Probability
This lecture note is based on the text book Mathematical Literacy in a numerate Society – by Matthew A. Isom and Jay Abramson
The probability an event E will occur is the number of desired outcomes divided by the total possible number of outcomes. We write .
Example 1. Roll a die once. The total outcomes are namely 1, 2, 3, 4, 5, and 6. Now the probability of the event for rolling a 3 is , because there are 6 total outcomes and 1 desired outcomes.
Example 2. Roll a die once. Write the sample space. Find the following probabilities:
a)
b)
c)
d)
Answer: a) 1/3b) 2/3c) 0d) ½
Example 3. Roll a die twice. Write the sample space. Find the following probabilities:
a)
b)
c)
d)
The sample space of the event of rolling a die twice:
11 / 12 / 13 / 14 / 15 / 1621 / 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) 6/36b) 35/36c) 3/36d) 0
Example 4. Roll a die twice. Write the sample space. Find the following probabilities:
a)
b)
c)
d)
The sample space of the event of rolling a die twice:
11 / 12 / 13 / 14 / 15 / 1621 / 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 = 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
b) = 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 =
c) = 9/36
Using independent event and multiplication principle we have=
d) = . You look at the sample space we have only two outcomes 11 and 55.
Example 5. Roll a die three times. How many outcomes do we have in the sample space? Find the following probabilities:
a)
b)
c)
d)
Answer: a) b)
c) d)
Example 6. Roll a die five times. How many outcomes do we have in the sample space. Find the following probabilities:
a)
b)
c)
d)
Answer: a) b) c) d)
Example 7. 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
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 8. 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 spaceb) 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 9. 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
2)The probability that all three people have same birthday is
Example 10. Given the binomial probability formula . Evaluate the following. 1) b(10, 7, 0.7) 2) b(5, 3, 1/36) c) b(8, 8, 0.5)
Solution: 1) ? Use calculator for simplification.
Try yourself for 2) and 3).
Example 11. A fair coin is tossed 7 times. 1) What is the probability that of getting 7 tails? 2) What is the probability of getting at most 6 tails?
Solution: 1) The probability of getting 7 tails is
b(7, 7, 0.5)
2) the probability of getting at most 6 tails =
(One can solve this problem without considering binomial probability, try yourself)
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
Example 12. 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 / TotalMen (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) d) p(W | N)e)
f) g) p(Y)h) p(M)i) p(Y | M) j) p(M | Y)
k) l) m) n)
Answer: a) 0.23b) 0.53c) 0.14d) 0.32e) 0.08
f) 0.08g) 0.70h) 0.47i) 0.58j) 0.39k) 0.27
l) 0.27m) 0.43n) 0.61
Expected Value: The standard way of finding expected value of an experiment is to multiply the value of each outcome of the experiment by the probability of that outcome and add the results.
We write expected value E =
Example 13. Your midterm exam grades are 88, 90. Average midterm exam is 50 percent of your grade. Your homework average is 80, it is 15 percent of your grade, your quiz average is 85, which is 10 percent of your grade and you earned a 78 on the final, which is 25 percent of your grade. What is your expected grade? What should be your average midterm grade to have your expected grade 90?
Solution: The expected grade is E = (88+90)/2(0.5)+80(0.15)+85(0.10)+78(0.25) = 84.5
Try yourself to find the next answer.
Independent Events: Two events A and B are called independent if and only if , otherwise A and B are dependent.
Example 14. 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 , the event with a head on the first toss and an event with a head on the second toss .
Now show that .
Bayes’ Formula: Let A, B, C are mutually exclusive events whose union is the sample space S. Let E be the arbitrary event in S such that , then
,,
where and so on.
Example 15. A company produces 1,000 refrigerators a week at three plants. Plant A produces 350 refrigerators a week, plant B produces 250 refrigerators a week, and plant C produces 400 refrigerators a week. Production records indicate thatb5% of the refrigerators at plant A will be defective, 3% of those produced at plant B will be defective, and 7% of those produced at plant C will be defective. All refrigerators are shipped to a central warehouse. If a refrigerator at the warehouse is found to be defective, what is the probability that it was produced a) at plant A? b) at plant B? c) at plant C?
We consider D as defective and D’ as non defective.
0.05 D
A
0.95 D’
0.35
0.03 D
Start 0.25 B
0.97 D’
0.40
0.07 D
C
0.93 D’
We now answer all questions from the tree diagram.
a)
You now try to find b)
c)
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