1. Multiplication Rules for Probability
and Independent Events
P(event A and event B)
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2. Independent Events Important question to ask yourself: This means:
“Are these two events Independent?”
This means:
“If one of these events happens, does it affect the probability of the other event happening?”
“Or are they Dependent events?”
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3. Independent/Dependent Examples
Experiment: Rolling a single die, two trials
Are these events Independent?
Event #1: Rolling an odd number on the first roll
Event #2: Rolling an even number on the second
Event #2: Rolling a “high” number (4, 5, or 6) on the second roll.
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4. Independent/Dependent Examples
Experiment: Pick one card, then pick a second card
Are these events Independent or Dependent?
Event #1: Your first card is an 8, then you replace the card and the deck is reshuffled.
Event #2: Your second card is a face card
Event #1: Your first card is an 8 and you hang onto it
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5. Independent/Dependent Examples
Experiment: Draw two Scrabble tiles and (this is a key phrase) without replacement
Are these events independent?
Event #1: You get an “A” on the first pick
Event #2: You get a “B” on the second pick
Are these events Independent?
Event #1: You get one of the blank tiles on the first pick
Event #2: And the other blank tile on the second.
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6. Multiplication Rule to find 𝑃(𝐴 𝑎𝑛𝑑 𝐵)
If the events are
Independent events…
If the events are
Not independent events…
𝑃 𝐴 𝑎𝑛𝑑 𝐵 =𝑃 𝐴 ∙𝑃(𝐵)
The probability of event A
Times the probability of event B
𝑃 𝐴 𝑎𝑛𝑑 𝐵 =𝑃 𝐴 ∙𝑃 𝐵|𝐴
The probability of event A
Times the probability of event B, given that event A has already occurred.
The Vertical bar in 𝐵|𝐴 is means “assuming that”
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7. Example: Rolling a single die twice
P(odd and then a 4)
P(odd and odd again)
Independent events
P(odd) is ______
P(4) is ______
P(odd and 4) = P(odd) ∙ P(4) = _____ ∙ _____ = _____
Independent events
P(odd) is _____
P(odd and odd) = P(odd) ∙ P(odd) = _____ ∙ _____ = ______
Extend it: P(odd) ten times in a row: ______________
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8. Example: Pick two cards With and Without Replacement
Draw an ace, replace it in the deck, then draw a ten
Draw an ace, keep it, then draw a ten
With Replacement
Independent Events
P(ace) = _____
P(ten) = _____
P(ace and ten) = P(ace) ∙ P(ten) = ______ ∙ ______ = ______
Without Replacement
Dependent Events
P(ace) = _____
P(ten | ace) = _____
P(ace and ten) = P(ace) ∙ P(ten | ace) = _____ ∙ ______ = _____
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9. Example: How many passwords are possible? If repetition allowed or not
An 8-character password made of A-Z, 0-9 if repetition allowed
If you aren’t allowed to repeat any characters
“Repetition allowed” is another way of saying “with replacement”
36 8
What does E12 mean?
28,211,099,070,000, approximately.
Same idea as “Without Replacement”
36∙35∙34∙33∙32∙31∙30∙29=30,260,340
The possibilities with repetition allowed are about ____ times as many.
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10. Example: How many passwords are possible? If repetition allowed or not
An 8-character password made of A-Z, 0-9 if repetition allowed
If you aren’t allowed to repeat any characters
36 8 possible passwords
What is the probability that a randomly generated password will begin with a digit?
What is the probability that a randomly generated password will be all digits?
30,260,340
What is the probability that a randomly generated password will begin with a digit?
What is the probability that a randomly generated password will be all digits?
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11. Word Problems
Sometimes P(A and B) problems are presented in narrative form.
Example: Fair Deal Frank’s car lot has 3 new trucks, 4 used trucks, 5 new cars, and 6 used cars. Choose two keys at random
Find P(two new trucks)
Find P(two used vehicles)
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12. Fair Deal Frank, continued
P(two new trucks)
P(two used vehicles)
Dependent events
P(new truck) = _______
P(new truck|new truck) = ______
P(two new trucks) = P(new truck) ∙ P(new truck | new truck) = ________ ∙ ________ = ________
Dependent events
P(used) = _______
P(used | used) = ______
P(two used vehicles) = P(used)+P(used | used) = ______ ∙ ______ = ______
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13. In a large population
With 52 cards or a dozen keys or 100 Scrabble® tiles, we have to pay attention to “with replacement” or “without replacement”
But in a large population, like choosing three people at random, we don’t worry about it as much.
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14. Example 4-26: Survey on Stress from Bluman 5th Ed. © McGraw Hill
A Harris poll found that 46% of Americans say they suffer great stress at least once a week. If three people are selected at random, find the probability that all three will say that they suffer great stress at least once a week.
We are taking three people at random, without replacement.
But because the population is so large, we may treat the three selections as independent events.
The calculation is a lot simpler this way.
Bluman, Chapter 4

15. Example 4-26: Survey on Stress from Bluman 5th Ed. © McGraw Hill
A Harris poll found that 46% of Americans say they suffer great stress at least once a week. If three people are selected at random, find the probability that all three will say that they suffer great stress at least once a week.
Bluman, Chapter 4

16. Problems given as Data Tables
Example: a gathering of high income people who signed up for a webinar on how to shelter income from the tax man
Men
Women
Total
Doctor
300
200
Lawyer
600
Neither
400
500
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17. High Income Earners Choose one person at random…
P(you picked a woman doctor)
Shortcut
Not Independent events
P(woman) = _______
P(doctor | woman) = ______
P(A and B) = P(A) ∙ P(B | A) = ____ ∙ ____ = _____
Go back to the basic definition of probability: 𝑃 𝑒𝑣𝑒𝑛𝑡 = 𝑚𝑦 𝑜𝑢𝑡𝑐𝑜𝑚𝑒𝑠 𝑡𝑜𝑡𝑎𝑙 𝑜𝑢𝑡𝑐𝑜𝑚𝑒𝑠
“My outcomes” is count of women doctors, right there in a cell.
“Total outcomes” in bottom right corner of the chart.
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18. Complementary Events example
Each widget that rolls off the assembly line has a 99% chance of passing inspection.
If five widgets are selected at random, what is the probability that at least one of them fails?
Are the events “Widget #1 is good”, “#2 is good”, etc. Independent or Dependent?
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19. Complementary Events in the widget inspection department
P(this one is good) = _____
P(all are good, all five out of five) = ____ ∙ ____ ∙ ____ ∙ ____ ∙ ____ = ( _____ )___ = _____
P(at least one is bad) means “Not all are good”.
“not all are good” is complement of “all good”
P(not all are good) = 1 - P(all are good) = 1 - ____ = _____
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20. Same as the widget problem with complementary events
“A single die is rolled 10 times. Find the probability of rolling at least one ‘6’.”
Complementary Event is ___________________.
“If the probability of twins is 5% and there are ten expectant moms in the maternity ward, find P(at least one set of twins).”
Complementary event is ___________________.
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21. Complementary Events example
Suppose that a railroad crossing lights and bells and bars have a 99.9% chance of operating successfully each time a train passes by.
If there are 5 trains a day x 365 days a year, what is the probability of no malfunctions? P(perfect operation) = (0.99)what exponent?
What is the probability of at least one error?
It’s the complement of the “perfect operation”.
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22. Conditional Probability
Suppose we know P(A) and we know P(A and B) and want to find the unknown P(B | A).
Someone took 𝑃 𝐴 ∙𝑃 𝐵 𝐴 =𝑃 𝐴 𝑎𝑛𝑑 𝐵
And divided each side by 𝑃 𝐴
Getting this result: 𝑃 𝐵 𝐴 = 𝑃(𝐴 𝑎𝑛𝑑 𝐵) 𝑃(𝐴)
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23. 4.3 Conditional Probability
Conditional probability is the probability that the second event B occurs given that the first event A has occurred. 23
Bluman, Chapter 4

24. Example 4-33: Parking Tickets Bluman 5th ed. © McGraw Hill
The probability that Sam parks in a no-parking zone and gets a parking ticket is 0.06,
and the probability that Sam cannot find a legal parking space and has to park in the no-parking zone is 0.20.
On Tuesday, Sam arrives at school and has to park in a no-parking zone. Find the probability that he will get a parking ticket.
Bluman, Chapter 4

25. Bluman’s Example 4-33: Parking Tickets
WRITE OUT DEFINITIONS!
Let T = gets a ticket event,
Let N = parks in a no-parking zone event.
What the problem says
What it means in algebra
Parking in a no-parking zone
Call it event N
Getting a ticket
Call it event T
P(no parking zone and ticket is 0.06)
P(N and T) = 0.06
P(he has to park in a no-parking zone) is 0.20
P(N) = 0.20
If he parks in a no-parking zone, find probability he will get a ticket
P(T | N) = ?
Bluman, Chapter 4

26. Example 4-33: Parking Tickets Bluman 5th ed. © McGraw Hill
What the problem says
What it means in algebra
Parking in a no-parking zone
Call it event N
Getting a ticket
Call it event T
P(no parking zone and ticket is 0.06
P(N and T) = 0.06
P(he has to park in a no-parking zone) is 0.20
P(N) = 0.20
If he parks in a no-parking zone, find probability he will get a ticket
P(T | N) = ?
Bluman, Chapter 4

27. Example 4-34: Women in the Military
A recent survey asked 100 people if they thought women in the armed forces should be permitted to participate in combat. The results of the survey are shown.
Bluman, Chapter 4

28. Example 4-34: Women in the Military
a. Find the probability that the respondent answered yes (Y), given that the respondent was a female (F).
Bluman, Chapter 4

29. Example 4-34: Women in the Military
When you have a data table problem like this, the shortcut is to have Event A determine one row or one column, and compute directly based on that one row or one column.
P(Y | F) = Look at Female row only: 8 Yes / 50 Total
Bluman, Chapter 4

30. Example 4-34: Women in the Military
b. Find the probability that the respondent was a male (M), given that the respondent answered no (N).
Shortcut
Shown
In red
boxes
Bluman, Chapter 4