7.4 Basic Rules for Finding Probabilities

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7.4 Basic Rules for Finding Probabilities
Probability an Event Does Not Occur
Rule 1 (for not the event) P(AC) 1 P(A)
Example 7.9 Probability a Stranger Does Not
Share Your Birth DateP(next stranger you meet
will share your birthday) 1/365.P(next
stranger you meet will not share your birthday)
1 1/365 364/365 .9973.
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Probability That Either of Two Events Happen
Rule 2 (addition rule for either/or) Rule 2a
(general) P(A or B) P(A) P(B) P(A and
B) Rule 2b (for mutually exclusive events) If
A and B are mutually exclusive events, P(A or
B) P(A) P(B)
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Example 7.10 Roommate Compatibility
Brett is off to college. There are 1000 male
students. Brett hopes his roommate will not like
to party and not snore.
A likes to party P(A) 250/1000 .25 B
snores P(B) 350/1000 .35
Probability Brett will be assigned a roommate who
either likes to party or snores, or both is P(A
or B) P(A) P(B) P(A and B) .25
.35 .15 .45 So the probability his roommate
is acceptable is 1 .45 .55
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Example 7.11 Probability of Two Boys or Two
Girls in Two Births
What is the probability that a woman who has two
children has either two girls or two boys?
Recall that the probability of a boy is .512
and probability of a girl is .488. Then we have
(using Rule 3b)
Event A two girls P(A) (.488)(.488)
.2381 Event B two boys P(B) (.512)(.512)
.2621
Note Events A and B are mutually exclusive
(disjoint).
Probability woman has either two boys or two
girls is P(A or B) P(A) P(B) .2381
.2621 .5002
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Probability That Two or More Events Occur
Together
Rule 3 (multiplication rule for and) Rule 3a
(general) P(A and B) P(A)P(BA) Rule 3b (for
independent events) If A and B are independent
events, P(A and B) P(A)P(B) Extension of Rule
3b (for 2 indep events) For several
independent events,P(A1 and A2 and and An)
P(A1)P(A2)P(An)
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Example 7.8 Probability of Male and Gambler
(cont)
For 9th graders, 22.9 of the boys and 4.5 of
the girls admitted they gambled at least once a
week during the previous year. The population
consisted of 50.9 girls and 49.1 boys.
Event A male Event B weekly gambler P(A)
.491 P(BA) .229
P(male and gambler) P(A and B) P(A)P(BA)
(.491)(.229) .1124
About 11 of all 9th graders are males and weekly
gamblers.
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Example 7.12 Probability Two Strangers
Both Share Your Birth Month
Assume all 12 birth months are equally
likely.What is the probability that the next two
unrelated strangers you meet both share your
birth month?
Event A 1st stranger shares your birth month
P(A) 1/12 Event B 2nd stranger shares your
birth month P(B) 1/12
Note Events A and B are independent.
P(both strangers share your birth month) P(A
and B) P(A)P(B) (1/12)(1/12) .007
Note The probability that 4 unrelated strangers
all share your birth month would be (1/12)4.
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Determining a Conditional Probability
Rule 4 (conditional probability) P(BA) P(A
and B)/P(A) P(AB) P(A and B)/P(B)
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Example 7.13 Alicia Answering
If we know Alicia is picked to answer one of the
questions, what is the probability it was the
first question?
A Alicia selected to answer Question 1, P(A)
1/50 B Alicia is selected to answer any one
of the questions, P(B) 3/50 Since A is a
subset of B, P(A and B) 1/50
P(AB) P(A and B)/P(B) (1/50)/(3/50) 1/3
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In Summary
Students sometimes confuse the definitions of
independent and mutually exclusive events.

• When two events are mutually exclusive and one
  happens, it turns the probability of the other
  one to 0.
• When two events are independent and one happens,
  it leaves the probability of the other one alone.

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In Summary
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Sampling with and without Replacement

• A sample is drawn with replacement if individuals
  are returned to the eligible pool for each
  selection.
• A sample is drawn without replacement if sampled
  individuals are not eligible for subsequent
  selection.

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7.5 Strategies for Finding Complicated
Probabilities
Example 7.2 Winning the LotteryEvent A
winning number is 956. What is P(A)?Method 1
With physical assumption that all 1000
possibilities are equally likely, P(A)
1/1000.Method 2 Define three events,B1 1st
digit is 9, B2 2nd digit is 5, B3 3rd digit
is 6Event A occurs if and only if all 3 of these
events occur. Note P(B1) P(B2) P(B3) 1/10.
Since these events are all independent, we have
P(A) (1/10)3 1/1000.
Can be more than one way to find a probability.
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Hints and Advice for Finding Probabilities

• P(A and B) define event in physical terms and
  see if know probability. Else try multiplication
  rule (Rule 3).
• Series of independent events all happen multiply
  all individual probabilities (Extension of Rule
  3b)
• One of a collection of mutually exclusive events
  happens add all individual probabilities (Rule
  2b extended).
• Check if probability of complement easier, then
  subtract it from 1 (applying Rule 1).

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Hints and Advice for Finding Probabilities

• None of a collection of mutually exclusive events
  happens find probability one happens, then
  subtract that from 1.
• Conditional probability define event in physical
  terms and see if know probability. Else try Rule
  4 or next bullet as well.
• Know P(BA) but want P(AB) Use Rule 3a to find
  P(B) P(A and B) P(AC and B), then use Rule 4.

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Steps for Finding Probabilities
Step 1 List each separate random circumstance
involved in the problem. Step 2 List the
possible outcomes for each random
circumstance. Step 3 Assign whatever
probabilities you can with the knowledge you
have. Step 4 Specify the event for which you
want to determine the probability. Step 5
Determine which of the probabilities from step 3
and which probability rules can be combined to
find the probability of interest.
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Example 7.17 Alicia Is Probably Healthy
What is the probability that Alicia has the
disease given that the test was positive?
Steps 1 to 3 Random circumstances, outcomes,
probabilities. Random circumstance 1 Alicias
disease status Possible Outcomes A disease AC
no disease Probabilities P(A) 1/1000 .001
P(AC) .999 Random circumstance 2 Alicias
test results Possible Outcomes B test is
positive, BC test is negative Probabilities
P(BA) .95 (positive test given
disease) P(BCA) .05 (negative test given
disease) P(BAC) .05 (positive test given no
disease) P(BCAC) .95 (negative test given no
disease)
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Example 7.17 Alicia Healthy? (cont)
Step 4 Specify event you want to determine the
probability. P(disease positive test)
P(AB). Step 5 Determine which probabilities
and probability rules can be combined to find
the probability of interest. Note we have P(BA)
and we want P(AB). Hints tell us to use P(B)
P(A and B) P(AC and B). Note P(A and B)
P(BA)P(A), similarly for P(AC and B). So P(AC
and B) P(BAC)P(AC) (.05)(.999) .04995 P(A
and B) P(BA)P(A) (.95)(.001) .00095 P(B)
.04995 .00095 .0509
There is less than a 2 chance that Alicia has
the disease, even though her test was positive.
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Example 7.8 Teens and Gambling (cont)
Two-Way Table Hypothetical Hundred Thousand
Sample of 9th grade teens 49.1 boys, 50.9
girls. Results 22.9 of boys and 4.5 of girls
admitted they gambled at least once a week
during previous year.
Start with hypothetical 100,000 teens
(.491)(100,000) 49,100 boys and thus 50,900
girls Of the 49,100 boys, (.229)(49,100) 11,244
would be weekly gamblers. Of the 50,900 girls,
(.045)(50,900) 2,291 would be weekly gamblers.
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Example 7.8 Teens and Gambling (cont)
P(boy and gambler) 11,244/100,000 .1124 P(boy
gambler) 11,244/13,535 .8307 P(gambler)
13,535/100,000 .13535
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Tree Diagrams
Step 1 Determine first random circumstance in
sequence, and create first set of branches for
possible outcomes. Create one branch for each
outcome, write probability on branch. Step 2
Determine next random circumstance and append
branches for possible outcomes to each branch in
step 1. Write associated conditional
probabilities on branches. Step 3 Continue this
process for as many steps as necessary. Step 4
To determine the probability of following any
particular sequence of branches, multiply the
probabilities on those branches. This is an
application of Rule 3a. Step 5 To determine the
probability of any collection of sequences of
branches, add the individual probabilities for
those sequences, as found in step 4. This is an
application of Rule 2b.
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Example 7.18 Alicias Possible Fates
P(Alicia has D and has a positive test)
.00095. P(test is positive) .00095 .04995
.0509. P(Alicia has D positive test)
.00095/.0509 .019
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Example 7.8 Teens and Gambling (cont)
P(boy and gambler) (.491)(.229) .1124 P(girl
and not gambler) (.509)(.955)
.4861 P(gambler) .1124 .0229 .1353 P(boy
gambler) .1124/.1353 .8307
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7.6 Using Simulation to Estimate Probabilities
Some probabilities so difficult or time-consuming
to calculate easier to simulate. If you
simulate the random circumstance n times and the
outcome of interest occurs in x out of those n
times, then the estimated probability for the
outcome of interest is x/n.
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Example 7.19 Getting All the Prizes
Cereal boxes each contain one of four prizes.Any
box is equally likely to contain each of the four
prizes. If buy 6 boxes, what is the probability
you get all 4 prizes?
Shown above are 50 simulations of generating a
set of 6 digits, each equally likely to be 1, 2,
3, or 4. There are 19 bold outcomes in which
all 4 prizes were collected. The estimated
probability is 19/50 .38. (Actual probability
is .3809.)
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7.7 Coincidences Intuitive Judgments about
Probability
Confusion of the Inverse Example Diagnostic
Testing Confuse the conditional probability have
the disease given a positive test result --
P(Disease Positive),with the conditional
probability of a positive test result given
have the disease -- P(Positive Disease), also
known as the sensitivity of the test. Often
forget to incorporate the base rate for a disease.
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Specific People versus Random Individuals
The chance that your marriage will end in divorce
is 50.
Does this statement apply to you personally? If
you have had a terrific marriage for 30 years,
your probability of ending in divorce is surely
less than 50.
Two correct ways to express the aggregate divorce
statistics

• In long run, about 50 of marriages end in
  divorce.
• At the beginning of a randomly selected marriage,
  the probability it will end in divorce is about
  .50.

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Coincidences
A coincidence is a surprising concurrence of
events, perceived as meaningfully related, with
no apparent causal connection.
Example 7.23 Winning the Lottery Twice
In 1986, Ms. Adams won the NJ lottery twice in a
short time period. NYT claimed odds of one
person winning the top prize twice were about 1
in 17 trillion. Then in 1988, Mr. Humphries won
the PA lottery twice.
1 in 17 trillion probability that a specific
individual who plays the lottery exactly twice
will win both times.
Millions of people play the lottery. It is not
surprising that someone, somewhere, someday
would win twice.
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The Gamblers Fallacy
The gamblers fallacy is the misperception of
applying a long-run frequency in the short-run.

• Primarily applies to independent events.
• Independent chance events have no memory.

Example Making ten bad gambles in a row doesnt
change the probability that the next gamble will
also be bad.