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Probability
2/25/2012
Basic Probability
We are going to study probability so that we can use it later during the study
of inferential statistics.
Rare Event Rule for Inferential Statistics
• If, under a given assumption, the probability of a particular observed
event is extremely small, we conclude that the assumption is probably not
correct.
Basic Probability
• An event is any collection of results or outcomes of a procedure.
• A simple event is an outcome or an event that cannot be further broken
down into simpler components.
• The sample space for a procedure consists of all possible simple events.
That is, the sample space consists of all outcomes that cannot be broken
down any further.
Basic Probability
Flipping a coin
Let h denote heads and t denote tails
Procedure
Example of Event
Complete Sample
space
Flip 1 coin
Heads
{h,t}
Basic Probability
Flipping a coin
Let h denote heads and t denote tails
Procedure
Example of Event
Complete Sample space
Flip 1 coin
Heads
{h,t}
Flip 2 coins
Basic Probability
Flipping a coin
Let h denote heads and t denote tails
Procedure
Example of Event
Complete Sample space
Flip 1 coin
Heads
{h,t}
Flip 2 coins
1 heads and 1 tails
Basic Probability
Flipping a coin
Let h denote heads and t denote tails
Procedure
Example of Event
Complete Sample space
Flip 1 coin
Heads
{h,t}
Flip 2 coins
1 heads and 1 tails {hh, ht, th, tt}
Basic Probability
Flipping a coin
Let h denote heads and t denote tails
Procedure
Example of Event
Complete Sample space
Flip 1 coin
Heads
{h,t}
Flip 2 coins
1 heads and 1 tails {hh, ht, th, tt}
Flip 3 coins
Basic Probability
Flipping a coin
Let h denote heads and t denote tails
Procedure
Example of Event
Complete Sample space
Flip 1 coin
Heads
{h,t}
Flip 2 coins
1 heads and 1 tail
(ht, th)
{hh, ht, th, tt}
Flip 3 coins
2 heads and 1 tail
(hht, hth, thh)
Basic Probability
Flipping a coin
Let h denote heads and t denote tails
Procedure
Example of Event
Complete Sample space
Flip 1 coin
Heads
{h,t}
Flip 2 coins
1 heads and 1 tail
(ht, th)
{hh, ht, th, tt}
Flip 3 coins
2 heads and 1 tail
(hht, hth, thh)
{hhh, hht, hth, thh, tth,
tht, htt, ttt}
Basic Probability
Notation for Probabilities
• P denotes a probability
Basic Probability
Notation for Probabilities
• P denotes a probability
• A, B, and C denote specific events.
Basic Probability
Notation for Probabilities
• P denotes a probability
• A, B, and C denote specific events.
• P(A) denotes the probability of an event A occurring.
Basic Probability
Notation for Probabilities
• P denotes a probability
• A, B, and C denote specific events.
• P(A) denotes the probability of an event A occurring.
Ex. When flipping a coin; if A={heads}, then P(A)=0.5
Basic Probability
Finding Probabilities
Basic Probability
Finding Probabilities
1. Relative Frequency Approximation of Probability Conduct or observe a
procedure, and count the number of times that event A actually occurs.
Based on these actual results, P(A) is approximated as follows:
Basic Probability
Finding Probabilities
1. Relative Frequency Approximation of Probability Conduct or observe a
procedure, and count the number of times that event A actually occurs.
Based on these actual results, P(A) is approximated as follows:
number of times 𝐴 𝑜𝑐𝑐𝑢𝑟𝑟𝑒𝑑
𝑃 𝐴 =
number of times the procedure was repeated
Basic Probability
Finding Probabilities
2.Classical Approach to Probability (Requires Equal Likely Outcomes)
Assume that a given procedure has n different simple events and that each of
those simple events has an equal chance of occurring. If event A can occur in
s of these n ways, then:
Basic Probability
Finding Probabilities
2. Classical Approach to Probability (Requires Equal Likely Outcomes)
Assume that a given procedure has n different simple events and that each of
those simple events has an equal chance of occurring. If event A can occur in
s of these n ways, then:
number of times 𝐴 𝑜𝑐𝑐𝑢𝑟𝑟𝑒𝑑
𝑠
𝑃 𝐴 =
=
number of different simple events 𝑛
Basic Probability
Finding Probabilities
2. Classical Approach to Probability (Requires Equal Likely Outcomes)
Assume that a given procedure has n different simple events and that each of
those simple events has an equal chance of occurring. If event A can occur in
s of these n ways, then:
number of times 𝐴 𝑜𝑐𝑐𝑢𝑟𝑟𝑒𝑑
𝑠
𝑃 𝐴 =
=
number of different simple events 𝑛
3. Subjective Probabilities P(A), the probability of event A is estimated by
using knowledge of the relevant circumstances
Basic Probability
• Law of Large Numbers tells us that relative frequency approximations
tend to get better with more observations.
Basic Probability
• Law of Large Numbers tells us that relative frequency approximations
tend to get better with more observations.
• Probability and Outcomes That are not Equally Likely Don’t assume
outcomes are equally as likely when you know nothing about the
likelihood of each outcome.
Basic Probability
Ex. Given that there were 6,511,100 cars that crashed among 135,670,000
registered cars in the U.S. find the Probability that a random car will crash this
year in the U.S.
Basic Probability
Ex. Given that there were 6,511,100 cars that crashed among 135,670,000
registered cars in the U.S. find the Probability that a random car will crash this
year in the U.S.
𝑃 𝑐𝑟𝑎𝑠ℎ =
𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑐𝑎𝑟𝑠 𝑡ℎ𝑎𝑡 𝑐𝑟𝑎𝑠ℎ𝑒𝑑
6,511,100
=
= .048
𝑡𝑜𝑡𝑎𝑙 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑐𝑎𝑟𝑠
135,670,000
Basic Probability
Ex. When studying the affect of heredity on height, we can express each
individual genotype, AA, Aa, aA, and aa, on an index card and shuffle the four
cards and randomly select one of them. What is the probability that we
select a genotype in which the two components are different?
Basic Probability
Ex. When studying the affect of heredity on height, we can express each
individual genotype, AA, Aa, aA, and aa, on an index card and shuffle the four
cards and randomly select one of them. What is the probability that we
select a genotype in which the two components are different?
2
𝑃 outcome with different components = = 0.5
4
Basic Probability
Ex. Given that the population of Alaska is 0.2% of the total U.S. population,
find the probability that the next President of the United States is from
Alaska.
Basic Probability
Ex. Given that the population of Alaska is 0.2% of the total U.S. population,
find the probability that the next President of the United States is from
Alaska.
The Books estimate is 0.1% since Alaska’s remoteness presents a
challenge to politicians there.
Basic Probability
Ex. What is the probability of Drawing a King from a standard deck of 52
cards?
Basic Probability
Ex. What is the probability of Drawing a King from a standard deck of 52
cards?
𝑃 king =
# of kings
4
=
= 0.08
# of cards in deck 52
Basic Probability
Ex. Find the probability that when a couple has 3 kids, that exactly 1 will be a
girl. Assume that boys and girls are equally as likely for each birth.
Basic Probability
Ex. Find the probability that when a couple has 3 kids, that exactly 1 will be a
girl. Assume that boys and girls are equally as likely for each birth.
𝑃 1 girl in 3 births =
3
= .375
8
Basic Probability
Ex. When Mendel conducted his famous genetics experiments with peas, one
sample of offspring consisted of 428 green peas and 152 yellow peas. Based
on those results, estimate the probability of getting an offspring pea that is
green.
Basic Probability
Ex. When Mendel conducted his famous genetics experiments with peas, one
sample of offspring consisted of 428 green peas and 152 yellow peas. Based
on those results, estimate the probability of getting an offspring pea that is
green.
𝑃 green =
# of green peas
428
=
= 0.73
# of green peas + # of yellow peas 428 + 152
Basic Probability
• The probability of an impossible event is 0.
• The probability of an event that is certain to occur is 1.
• For any event A, the probability of A is between 0 and 1 inclusive. That
is, 0 ≤ 𝑃 𝐴 ≤ 1.
• The complement of event A, denoted by 𝐴, consists of all outcomes in
which event A does not occur.
Basic Probability
Ex. A typical question on an SAT test requires the test taker to select one of
five possible choices: A, B, C, D, or E. The probability of correctly answering a
question when guessing is 1/5 or 0.2 Find the probability of making a random
guess and not being correct i.e. being incorrect.
Basic Probability
Ex. A typical question on an SAT test requires the test taker to select one of
five possible choices: A, B, C, D, or E. The probability of correctly answering a
question when guessing is 1/5 or 0.2 Find the probability of making a random
guess and not being correct i.e. being incorrect.
4
𝑃 not guessing correctly = 𝑃 correct = 𝑃 incorrect = = 0.8
5
Basic Probability
Rounding Probabilities
Give either the exact fraction or decimal or round off final decimal answer to
three significant digits.
• 0.04799219 → 0.0480
• 1/3 → 1/3 or 0.333
• 2/4 → ½ or 0.5
• 1941/3405 → .570
Addition Rule
We are going to learn the addition rule for probabilities, which allows us to
find the probabilities of events that can be expressed as P(A or B), which
denote the probability of A occurring or B occurring or both occurring.
Addition Rule
We are going to learn the addition rule for probabilities, which allows us to
find the probabilities of events that can be expressed as P(A or B), which
denote the probability of A occurring or B occurring or both occurring.
A compound event is any event combining two or more simple events.
Addition Rule
Consider the following table.
Did the Subject Actually Lie
No (Did Not Lie)
Yes (Lied)
Positive test result
(Test indicates that subject lied)
15
(false positive)
42
(true positive)
Negative test result
(Test indicates that subject did not lie)
32
(true negative)
9
False (negative)
Addition Rule
Consider the following table.
Did the Subject Actually Lie
No (Did Not Lie)
Yes (Lied)
Positive test result
(Test indicates that subject lied)
15
(false positive)
42
(true positive)
Negative test result
(Test indicates that subject did not lie)
32
(true negative)
9
False (negative)
If 1 subject is randomly selected from the 98 subjects given a polygraph test,
find the probability of selecting a subject who had a positive test result or
lied.
Addition Rule
Consider the following table.
Did the Subject Actually Lie
No (Did Not Lie)
Yes (Lied)
Positive test result
(Test indicates that subject lied)
15
(false positive)
42
(true positive)
Negative test result
(Test indicates that subject did not lie)
32
(true negative)
9
False (negative)
If 1 subject is randomly selected from the 98 subjects given a polygraph test,
find the probability of selecting a subject who had a positive test result or
lied.
Addition Rule
Consider the following table.
Did the Subject Actually Lie
No (Did Not Lie)
Yes (Lied)
Positive test result
(Test indicates that subject lied)
15
(false positive)
42
(true positive)
Negative test result
(Test indicates that subject did not lie)
32
(true negative)
9
False (negative)
If 1 subject is randomly selected from the 98 subjects given a polygraph test,
find the probability of selecting a subject who had a positive test result or
lied.
15 + 42 + 9 = 66
Addition Rule
Formal Addition Rule
𝑃 𝐴 𝑜𝑟 𝐵 = 𝑃 𝐴 + 𝑃 𝐵 − 𝑃(𝐴 𝑎𝑛𝑑 𝐵)
where 𝑃(𝐴 𝑎𝑛𝑑 𝐵) denotes the probability that A and B both occur at the
same time as an outcome.
Intuitive Addition Rule
To find 𝑃 𝐴 𝑜𝑟 𝐵 , find the sum of the # of ways event A can occur and the #
of ways event B can occur, adding in such a way as to not double count. The
𝑃 𝐴 𝑜𝑟 𝐵 is equal to the sum divided by the total number of outcomes
Addition Rule
Events A and B are disjoint ( or mutually exclusive) if they cannot occur at
the same time. (That is disjoint events do not overlap.)
Note: If A and B are disjoint, then 𝑃 𝐴 𝑎𝑛𝑑 𝐵 = 0
Addition Rule
Events A and B are disjoint ( or mutually exclusive) if they cannot occur at
the same time. (That is disjoint events do not overlap.)
Note: If A and B are disjoint, then 𝑃 𝐴 𝑎𝑛𝑑 𝐵 = 0
Complementary Events
𝑃 𝐴 +𝑃 𝐴 =1
𝑃 𝐴 =1−𝑃 𝐴
𝑃 𝐴 = 1 − 𝑃(𝐴)
Addition Rule
FBI data show that 62.4% of murders are cleared by arrests. We can express
the probability of a murder being cleared by an arrest as P(cleared)=0.624.
For a randomly selected murder, find P(cleared).
Addition Rule
FBI data show that 62.4% of murders are cleared by arrests. We can express
the probability of a murder being cleared by an arrest as P(cleared)=0.624.
For a randomly selected murder, find P(cleared).
𝑃 cleared = 1 − 𝑃 cleared = 1 − 0.624 = 0.376
Multiplication Rule
In this section we present the basic multiplication rule, which is used to find
𝑃 𝐴 𝑎𝑛𝑑 𝐵 , the probability that event A occurs in a first trial and event B
occurs in a second trial.
Multiplication Rule
In this section we present the basic multiplication rule, which is used to find
𝑃 𝐴 𝑎𝑛𝑑 𝐵 , the probability that event A occurs in a first trial and event B
occurs in a second trial.
Notation
𝑃 𝐴 and 𝐵 = 𝑃( 𝐴 occurs in a first trial and 𝐵 occurs in a second trial)
Multiplication Rule
In this section we present the basic multiplication rule, which is used to find
𝑃 𝐴 𝑎𝑛𝑑 𝐵 , the probability that event A occurs in a first trial and event B
occurs in a second trial.
Notation
𝑃 𝐴 and 𝐵 = 𝑃( 𝐴 occurs in a first trial and 𝐵 occurs in a second trial)
Ex. What is the probability of flipping a fair coin and getting a heads, and then
rolling a six sided die and getting a 4.
Multiplication Rule
In this section we present the basic multiplication rule, which is used to find
𝑃 𝐴 𝑎𝑛𝑑 𝐵 , the probability that event A occurs in a first trial and event B
occurs in a second trial.
Notation
𝑃 𝐴 and 𝐵 = 𝑃( 𝐴 occurs in a first trial and 𝐵 occurs in a second trial)
Ex. What is the probability of flipping a fair coin and getting a heads, and then
rolling a six sided die and getting a 4.
𝑃 ℎ 𝑎𝑛𝑑 4 = 0.5 ∙ 0.167 = 0.83
Multiplication Rule
Ex. Find the probability or randomly selecting two individuals without
replacement, where the first person had a positive test result and the second
person had a negative rest result test.
Did the Subject Actually Lie
No (Did Not Lie)
Yes (Lied)
Positive test result
(Test indicates that subject lied)
15
(false positive)
42
(true positive)
Negative test result
(Test indicates that subject did not lie)
32
(true negative)
9
False (negative)
57
𝑃 positive =
98
Multiplication Rule
Ex. Find the probability or randomly selecting two individuals without
replacement, where the first person had a positive test result and the second
person had a negative rest result test.
Did the Subject Actually Lie
No (Did Not Lie)
Yes (Lied)
Positive test result
(Test indicates that subject lied)
15
(false positive)
42
(true positive)
Negative test result
(Test indicates that subject did not lie)
32
(true negative)
9
False (negative)
57
41
𝑃 positive =
; 𝑃 𝑛𝑒𝑔𝑎𝑡𝑖𝑣𝑒 =
98
97
Multiplication Rule
Ex. Find the probability or randomly selecting two individuals without
replacement, where the first person had a positive test result and the second
person had a negative rest result test.
57
41
𝑃 positive =
; 𝑃 𝑛𝑒𝑔𝑎𝑡𝑖𝑣𝑒 =
98
97
So
57 41
𝑃 positive 1st and negative 2nd =
∙
= 0.246
98 97
Multiplication Rule
Notation
P(B│A) represents the probability of event B occuring after it is
assumed that event A has already occurred.
Multiplication Rule
Notation
P(B│A) represents the probability of event B occuring after it is
assumed that event A has already occurred.
Two events A and B are independent if the occurrence of one does not affect
the probability of the occurrence of the other. If A and B are not
independent, the are said to be dependent.
Multiplication Rule
Notation
P(B│A) represents the probability of event B occuring after it is
assumed that event A has already occurred.
Two events A and B are independent if the occurrence of one does not affect
the probability of the occurrence of the other. If A and B are not
independent, the are said to be dependent.
Formal Multiplication Rule
𝑃 𝐴 and 𝐵 = 𝑃 𝐴 ∙ 𝑃 𝐵 𝐴
If A and B are independent events then 𝑃 𝐵 𝐴 = 𝑃 𝐵 .
Multiplication Rule
Ex. What is the probability of 1st drawing a King of Hearts and then without
replacement drawing a Face card(Kings, Queens, Jacks)?
Multiplication Rule
Ex. What is the probability of 1st drawing a King of Hearts and then without
replacement drawing a Face card(Kings, Queens, Jacks)?
1 11
𝑃 KH and FC =
∙
= 0.00415
52 51
Multiplication Rule
Ex. What is the probability of 1st drawing a King of Hearts and then without
replacement drawing a Face card(Kings, Queens, Jacks)?
1 11
𝑃 KH and FC =
∙
= 0.00415
52 51
Ex. What is the probability of 1st drawing a King of Hearts and then with
replacement drawing a Face card(Kings, Queens, Jacks)?
Multiplication Rule
Ex. What is the probability of 1st drawing a King of Hearts and then without
replacement drawing a Face card(Kings, Queens, Jacks)?
1 11
𝑃 KH and FC =
∙
= 0.00415
52 51
Ex. What is the probability of 1st drawing a King of Hearts and then with
replacement drawing a Face card(Kings, Queens, Jacks)?
1 12
𝑃 KH and FC =
∙
= 0.231
52 52
Multiplication Rule
The 5% guideline for Cumbersome Calculations
If calculations are very cumbersome and if the sample size is no more than 5%
of the size of the population, treat the selections as being independent.
Multiplication Rule
Ex. Assume that we have a batch of 100,000 heart pacemakers, including
99,950 that are good (G) and 50 that are defective (D).
a. If two of those 100,000 pacemakers are randomly selected without
replacement, find the probability that they are both good.
b. If 20 of those 100,000 pacemakers are randomly selected without
replacement, find the probability that they are all good.
Multiplication Rule
Ex. Assume that we have a batch of 100,000 heart pacemakers, including
99,950 that are good (G) and 50 that are defective (D).
a. If two of those 100,000 pacemakers are randomly selected without
replacement, find the probability that they are both good.
b. If 20 of those 100,000 pacemakers are randomly selected without
replacement, find the probability that they are all good.
Solution to a. is 𝑃 first G and second G =
Lets try b.
99,950 99,949
∙
100,000 99,999
= .999
Multiplication Rule
Ex. Assume that we have a batch of 100,000 heart pacemakers, including
99,950 that are good (G) and 50 that are defective (D).
a. If two of those 100,000 pacemakers are randomly selected without
replacement, find the probability that they are both good.
b. If 20 of those 100,000 pacemakers are randomly selected without
replacement, find the probability that they are all good.
Solution to a. is 𝑃 first G and second G =
99,950 99,949
∙
100,000 99,999
= .999
Lets try b.
99,950 99,949
𝑃 all 20 pacemakers good =
∙ ⋯
100,000 99,999
Multiplication Rule
Ex. Assume that we have a batch of 100,000 heart pacemakers, including
99,950 that are good (G) and 50 that are defective (D).
a. If two of those 100,000 pacemakers are randomly selected without
replacement, find the probability that they are both good.
b. If 20 of those 100,000 pacemakers are randomly selected without
replacement, find the probability that they are all good.
Solution to a. is 𝑃 first G and second G =
99,950 99,949
∙
100,000 99,999
= .999
Lets try b.
99,950 99,949 99,948
𝑃 all 20 pacemakers good =
∙
∙
∙⋯
100,000 99,999 99,998
Multiplication Rule
Ex. Assume that we have a batch of 100,000 heart pacemakers, including
99,950 that are good (G) and 50 that are defective (D).
a. If two of those 100,000 pacemakers are randomly selected without
replacement, find the probability that they are both good.
b. If 20 of those 100,000 pacemakers are randomly selected without
replacement, find the probability that they are all good.
Solution to part a. is 𝑃 first G and second G =
99,950 99,949
∙
100,000 99,999
= .999
Lets try part b.
99,950 99,949 99,948
99,931
𝑃 all 20 pacemakers good =
∙
∙
∙⋯ ∙
100,000 99,999 99,998
99,981
Multiplication Rule
Ex. Assume that we have a batch of 100,000 heart pacemakers, including
99,950 that are good (G) and 50 that are defective (D).
a. If two of those 100,000 pacemakers are randomly selected without
replacement, find the probability that they are both good.
b. If 20 of those 100,000 pacemakers are randomly selected without
replacement, find the probability that they are all good.
Solution to part a. is 𝑃 first G and second G =
99,950 99,949
∙
100,000 99,999
= .999
Lets try part b.
99,950 99,949 99,948
99,931
𝑃 all 20 pacemakers good =
∙
∙
∙⋯ ∙
100,000 99,999 99,998
99,981
Hmm…
Multiplication Rule
Ex. Assume that we have a batch of 100,000 heart pacemakers, including
99,950 that are good (G) and 50 that are defective (D).
b. If 20 of those 100,000 pacemakers are randomly selected without
replacement, find the probability that they are all good.
Lets try part b.
99,950 99,949 99,948
99,931
𝑃 all 20 pacemakers good =
∙
∙
∙⋯ ∙
100,000 99,999 99,998
99,981
Hmm… Notice 5% of 100,000 is (0.05)(100,000) this counts as a cumbersome
calculation so lets treat it as independent.
Multiplication Rule
Ex. Assume that we have a batch of 100,000 heart pacemakers, including
99,950 that are good (G) and 50 that are defective (D).
b. If 20 of those 100,000 pacemakers are randomly selected without
replacement, find the probability that they are all good.
Lets try part b.
99,950 99,949 99,948
99,931
𝑃 all 20 pacemakers good =
∙
∙
∙⋯ ∙
100,000 99,999 99,998
99,981
Hmm… this counts as a cumbersome calculation so lets treat it as
independent.
99,950 99,950 99,950
99,950
𝑃 all 20 pacemakers good =
∙
∙
∙⋯ ∙
100,000 100,000 100,000
100,000
20
99,950
=
= .990
100,000
Multiplication Rule
Ex. Assume that we have a batch of 100,000 heart pacemakers, including
99,950 that are good (G) and 50 that are defective (D).
b. If 20 of those 100,000 pacemakers are randomly selected without
replacement, find the probability that they are all good.
Lets try part b.
99,950 99,949 99,948
99,931
𝑃 all 20 pacemakers good =
∙
∙
∙⋯ ∙
100,000 99,999 99,998
99,981
Hmm… this counts as a cumbersome calculation so lets treat it as
independent.
99,950 99,950 99,950
99,950
𝑃 all 20 pacemakers good =
∙
∙
∙⋯ ∙
100,000 100,000 100,000
100,000
20
99,950
=
= .990
100,000
Multiplication Rule
Ex. Lets say a geneticist developed a procedure for increasing the likelihood of
female babies. In an initial test, 20 couples that used the method had 20
female babies. Assuming that the gender selection procedure had no effect
find the probability of having 20 female babies out of 20 babies by chance.
Multiplication Rule
Ex. Lets say a geneticist developed a procedure for increasing the likelihood of
female babies. In an initial test, 20 couples that used the method had 20
female babies. Assuming that the gender selection procedure had no effect
find the probability of having 20 female babies out of 20 babies by chance.
𝑃 all 20 babies female = 𝑃 female ∙ 𝑃 female ∙ ⋯ ∙ 𝑃(female) (20 times)
Multiplication Rule
Ex. Lets say a geneticist developed a procedure for increasing the likelihood of
female babies. In an initial test, 20 couples that used the method had 20
female babies. Assuming that the gender selection procedure had no effect
find the probability of having 20 female babies out of 20 babies by chance.
𝑃 all 20 babies female = 𝑃 female ∙ 𝑃 female ∙ ⋯ ∙ 𝑃(female) (20 times)
= 0.5 ∙ 0.5 ∙ ⋯ ∙ 0.5 (20 times)
Multiplication Rule
Ex. Lets say a geneticist developed a procedure for increasing the likelihood of
female babies. In an initial test, 20 couples that used the method had 20
female babies. Assuming that the gender selection procedure had no effect
find the probability of having 20 female babies out of 20 babies by chance.
𝑃 all 20 babies female = 𝑃 female ∙ 𝑃 female ∙ ⋯ ∙ 𝑃(female) (20 times)
= 0.5 ∙ 0.5 ∙ ⋯ ∙ 0.5 (20 times)
= (0.5)20
= 0.000000954
Multiplication Rule
Ex. Say the probability of an electrical system failure is 0.001.
a. If the engine in an aircraft has one electrical system, what is the
probability that it will work?
Multiplication Rule
Ex. Say the probability of an electrical system failure is 0.001.
a. If the engine in an aircraft has one electrical system, what is the
probability that it will work?
𝑃 𝑤𝑜𝑟𝑘𝑖𝑛𝑔 𝑠𝑦𝑠𝑡𝑒𝑚 = 𝑃 𝑠𝑦𝑠𝑡𝑒𝑚 𝑡ℎ𝑎𝑡 𝑑𝑜𝑒𝑠 𝑛𝑜𝑡 𝑓𝑎𝑖𝑙
Multiplication Rule
Ex. Say the probability of an electrical system failure is 0.001.
a. If the engine in an aircraft has one electrical system, what is the
probability that it will work?
𝑃 𝑤𝑜𝑟𝑘𝑖𝑛𝑔 𝑠𝑦𝑠𝑡𝑒𝑚 = 𝑃 𝑠𝑦𝑠𝑡𝑒𝑚 𝑡ℎ𝑎𝑡 𝑑𝑜𝑒𝑠 𝑛𝑜𝑡 𝑓𝑎𝑖𝑙
= 1 − 𝑃 𝑠𝑦𝑠𝑡𝑒𝑚 𝑓𝑎𝑖𝑙𝑢𝑟𝑒 = 1 − 0.001 = .999
Multiplication Rule
Ex. Say the probability of an electrical system failure is 0.001.
b. If the engine in an aircraft has two independent electrical systems, what
is the probability that the engine can function with an electrical system.
Multiplication Rule
Ex. Say the probability of an electrical system failure is 0.001.
b. If the engine in an aircraft has two independent electrical systems, what
is the probability that the engine can function with an electrical system.
𝑃 1st fails and 2nd fails = 0.001 ∙ 0.001 = 0.000001
Multiplication Rule
Ex. Say the probability of an electrical system failure is 0.001.
b. If the engine in an aircraft has two independent electrical systems, what
is the probability that the engine can function with an electrical system.
𝑃 1st fails and 2nd fails = 0.001 ∙ 0.001 = 0.000001
So there is a 0.000001 chance both engines fail so;
Multiplication Rule
Ex. Say the probability of an electrical system failure is 0.001.
b. If the engine in an aircraft has two independent electrical systems, what
is the probability that the engine can function with an electrical system.
𝑃 1st fails and 2nd fails = 0.001 ∙ 0.001 = 0.000001
So there is a 0.000001 chance both engines fail so;
𝑃 𝑤𝑜𝑟𝑘𝑖𝑛𝑔 𝑠𝑦𝑠𝑡𝑒𝑚 = 1 − 0.000001 = 0.999999
Complements and Conditional
• Probability of “at least one”:
Find the probability that among several trials, we
get at least one of some specified event.
• “At least one” is equivalent to “one ore more”
• The complement of getting at least one of an
item of a particular type is that you get no items
of that type.
• P(at least one) = 1 – P(none)
Complements and Conditional
Find the probability of a couple having at least
one girl among 3 children. Assume that boys and
girls are equally likely and that the gender of a
child is independent of any other child.
Complements and Conditional
Find the probability of a couple having at least one girl
among 3 children. Assume that boys and girls are
equally likely and that the gender of a child is
independent of any other child.
• What’s the complement of getting at least one girl?
Complements and Conditional
Find the probability of a couple having at least one girl
among 3 children. Assume that boys and girls are
equally likely and that the gender of a child is
independent of any other child.
• What’s the complement of getting at least one girl?
• Getting no girls, so all 3 would be boys.
Complements and Conditional
Find the probability of a couple having at least one girl
among 3 children. Assume that boys and girls are
equally likely and that the gender of a child is
independent of any other child.
• What’s the complement of getting at least one girl?
• Getting no girls, so all 3 would be boys.
• The complement is boy and boy and boy P(boy and
boy and boy) = ½ x ½ x ½ = 1/8.
Complements and Conditional
Find the probability of a couple having at least one girl
among 3 children. Assume that boys and girls are
equally likely and that the gender of a child is
independent of any other child.
• What’s the complement of getting at least one girl?
• Getting no girls, so all 3 would be boys.
• The complement is boy and boy and boy P(boy and
boy and boy) = ½ x ½ x ½ = 1/8.
• So, the P(at least one girl) = 1 – P(all boys) = 1 – 1/8 =
7/8
Complements and Conditional
•
Assume that the probability of a defective Firestone
tire is 0.0003. If the retail outlet CarStuff buys 100
Firestone tires, find the probability that they get at
least one that is defective. If the probability is high
enough, plans must be made to handle defective
tires returned by those consumers. Should they
make those plans?
Complements and Conditional
What if somebody offered to bet that at least two
people in your math class had the same birthday?
Would you take the bet?
Complements and Conditional
Conditional Probability
• The Conditional probability of B given A can
be found by assuming that event A has
occurred, and, working under that
assumption, calculating the probability that
event B will occur.
• 𝑃 𝐵𝐴 =
𝑃(𝐴 𝑎𝑛𝑑 𝐵)
𝑃(𝐴)
Complements and Conditional
If we randomly select one Senator, what is the
probability of getting a male, given that a
Republican was selected?
Republican
Democrat
Independent
Male
46
39
1
Female
5
9
1
Complements and Conditional
If we randomly select one Senator, what is the
probability of getting a Democrat or
Independent, given that a male was selected?
Republican
Democrat
Independent
Male
46
39
1
Female
5
9
1
Complements and Conditional
What is the probability of drawing an Ace of
Spades given that a black card was drawn?
Homework!
•
•
•
•
4-2: 1-19 odd ,25 and 27. (12 questions)
4-3: 1-19 odd, 33 and 35. (12 questions)
4-4: 1-9 odd, 13-21 odd, and 27 (11 questions)
4-5: 1-33 odd