Download 5-2A Lecture

Section 5 – 2A:
Binomial Probability Distributions
Binomial Trials
A Binomial Trial is a procedure or action that has only 2 possible outcomes if the action is
performed one time. One of the two possible outcomes will be selected as the desired outcome
and called a Success. All other outcomes will be labeled a Failure.
P(success) = p
The variable p is used to denote the probability of getting a success if a procedure is performed
one time.
p = .35 means
If a procedure is performed one time the probability of getting a success is .35
P(failure ) = q
The value of q is the probability of NOT getting a success (failure) if a procedure is performed one
time. The probability of getting a success and the probability of getting a failure must
total one.
p+q =1
If P(success) = p = .35 then
P(failure) = q = .65
The P(success) and P(failure) for a Binomial Trial are complements of each other.
Examples
A) You chose one customer in the bookstore and record if they buy a statistics book or not.
success (p): the selected customer bought a statistics book.
failure (q): the selected customer did not buy a statistics book.
If the probability of one customer buying a statistics book is .05 then p = .05
and the probability of one customer NOT buying a statistics book is
q = .95
B) A person roles a die one time and wants to roll a 6.
success (p): the person rolls a 6 on the one roll.
failure (q): the person does not roll a 6 on the one roll.
The probability of one roll of the die being a 6 is 1/ 6
p = 1/ 6
The probability of one roll of the die NOT being a 6 is 5/ 6 q = 5/ 6
Stat 300 5 – 2A Lecture
Page 1 of 15
© 2012 Eitel
Binomial Experiments
A Binomial Experiment is a procedure where you conduct a binomial trail a fixed number of
times (n). The outcome from each trial is counted as a success or a failure.
The number of successes in the n trails will be the variable x. The value for x could be as low
as 0 (no successes) or as high as n (every trial was a success).
Requirements for a process that results in a Binomial Experiment
1. A fixed number of trails is conducted (n).
2. Each outcome can be classified into two categories: Success or Failure After you decide on a
clear definition of what a success will be then any other outcome is labeled a failure.
3. The trails are independent. The outcome for one trail does not affect the outcome of other trails.
4. The probability of a success (p) for 1 trail must be known and remain constant during all the trials
An Example of a process that will result in a Binomial Probability Experiment
Trial: You roll a die one time and record if you get a 3.
Experiment: Conduct 30 trials.
1. There are 30 trials. n = 30
2. Each trial consists of rolling a die and recording if the number rolled was a 3 or not. Success is
getting a 3 and any other number is a failure.
3. One roll of the die does not affect the next roll.
4. The probability of getting a 3 on each roll stays at
P(success) =
1
5
and P(failure) =
6
6
1
the entire time.
6
The outcome from each trial is a success if you rolled a 3.
After you have conducted the 10 trials you will have 10 outcomes.
The variable x will be used to describe the “possible number of successes”.
This number could be as low as 0 (no 3ʼs in 10 rolls) or as high as 10 (all 10 rolls were a 3).
Stat 300 5 – 2A Lecture
Page 2 of 15
© 2012 Eitel
A Bernoulli Probability Distribution Table
A travel magazine states that 40% of all adults have been to Hawaii. You randomly select 5 adults
and ask each of them if they have been to Hawaii. The results from asking 5 people that question will
vary from 0 people that have been there to all 5 having been there. The possible values of x will be
0, 1, 2,3,4 or 5. We will make x the random variable that represents the number of people who said
they had been to Hawaii out of the 5 people questioned. These values are put in the left column.
The values in the right column represent P(x). P(x) is the probability of getting x people out of
5 saying that they have been to Hawaii.
x: number
of persons who have been
to Hawaii out of 5 people
questioned
P(x)
0 of 5
.08
1 of 5
.26
2 of 5
.34
3 of 5
.23
4 of 5
.08
5 of 5
.01
P(0) = .08 This means that if I ask 5 randomly selected people, the probability that exactly 0 of
the five have been to Hawaii is .08.
P(1) = .26 This means that if I ask 5 randomly selected people, the probability that exactly 1 of
the five have been to Hawaii is .28.
The values for each P(x) in the table represent the expected values for each value of x based
on a Theoretical Model of the entire population of adults given that 40% of the adult
population has been to Hawaii.
How are these values calculated?
By the use of the Binomial Probability Distribution Formula
Stat 300 5 – 2A Lecture
Page 3 of 15
© 2012 Eitel
Notation for the Binomial Probability Formula
n = the total number of trails
x = the number of of successes in n trails.
n – x = number of failures in n trails.
p is the probability of success for one of the n trails.
q is the probability of failure for one of the n trails.
Note: p + q =1
P(x) is the probability of getting exactly x successes among n trails.
Binomial Probability Distribution Formula
If the requirements to produce a Binomial Probability Distribution are met then
the probability of getting exactly x successes out of n trails is given by
P( x) = n C x • p x • qn− x
n is the
total
number
of trials
P(x) =
nC x
x is the
number of
successes
Stat 300 5 – 2A Lecture
n – x is the
x is the
number
number
of
of
failures
successes
•
px
•
p is the
probability
of a
success
Page 4 of 15
qn- x
q is the
probability
of a
failure
© 2012 Eitel
How to use the calculator to calculate n C x • p x • q n−x
Find the value of 7 C5 • .65 • .4 2
1. Key 7 into the calculator.
2. Press the PRB key. Press the right arrow key
until
nC x
is underlined then press the = key.
3. Key 5 into the calculator.
4. Press the multiplication key.
5. Key .6^5 into the calculator.
6. Press the multiplication key.
7. Key .4^2 into the calculator.
8. Press = to get .2612736 Round this off to .26
Practice Examples
A) What is the probability of getting
7 successes out of 7 trails
with p = .6 and q = .4
7 C7 •
.6 7 • .4 0 = __________
C) What is the probability of getting
12 successes out of 16 trails
with p = .82 and q = .18
16 C12 •
.8212 • .184 = __________
B) What is the probability of getting
1 success out of 8 trails
with p = .2 and q = .8
8 C1 •
.21 • .87 = __________
D) What is the probability of getting
8 successes out of 20 trails
with p = .12 and q = .88
20C 8 •
.128 • .8812 = __________
Note: The answer for a given P(x) may be a very small number . 0000347 will be displayed as
3.47 E − 6 or 3.47 × 10−6 in scientific notation with a negative exponent for the power of 10. When
you round this number off it rounds to 0. To show that the value for P(x) is very close to 0 but
not exactly 0 we write the value of P(x) as 0+
Answers:
A) .02799 ≈ .03
Stat 300 5 – 2A Lecture
B) . 3355 ≈ .34
C) .17657 ≈ .18
Page 5 of 15
D) .0011 = 0 +
© 2012 Eitel
Example 1 : Creating a Binomial Probability Distribution Table
A travel magazine states that 40% of all adults have been to Hawaii. You randomly select 5
adults and ask each of them if they have been to Hawaii. Complete a Binomial Probability
Distribution Table for the number of people who say they been to Hawaii out of any 5 randomly
selected people.
Notation for the Binomial Probability Formula
n = the total number of trails . We sampled 5 people. n = 5
p is the probability of success for one of the n trails. 40% of adults have been to Hawaii p = .40
q is the probability of failure for one of the n trails.
If 40% have been to Hawaii then 60% have NOT been to Hawaii. q = .60
The middle column shows the calculations. This column will not appear on the test. The
right column shows the values of p(X) for each x.
x: number
of persons who said they
# that have not been to Hawaii
# that have not been to Hawaii
P(x)
• .60
have been to Hawaii out of 5 P(x) = 5 C x • .40
people questioned
5 have not been to Hawaii
0 of 5 have been to Hawaii
P(x) = 5 C0 • .40 0 have been to Hawaii• .60
1 of 5 have been to Hawaii
P(x) = 5 C1 • .40 1 has been to Hawaii• .60
2 of 5 have been to Hawaii
P(x) = 5 C2 • .40 2 have been to Hawaii• .60
3 of 5 have been to Hawaii
P(x) = 5 C3 • .40 3 have been to Hawaii• .60 2 have not been to Hawaii
4 of 5 have been to Hawaii
P(x) = 5 C4 • .40 4 have been to Hawaii• .60
5 of 5 have been to Hawaii
P(x) = 5 C5 • .40 5 have been to Hawaii• .60
4 have not been to Hawaii
.08
.26
3 have not been to Hawaii
1 has not been to Hawaii
.34
.23
.08
0 have not been to Hawaii
.01
A) P( x ≤ 1) = P(0) + P(1) = .34
B. P( x ≥ 3 ) = P(3) + P(4) + P(5) = .32
C) P(no less than 4) = P(4 ) + P(5) = .09
D. P(no more than 2) = P(0) + P(1) + P(2) = .48
Stat 300 5 – 2A Lecture
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© 2012 Eitel
Example 2 : Creating a Binomial Probability Distribution Table
A industry magazine states that 80% of all adults own a cell phone You randomly select 7
adults and ask each of them if they own a cell phone. Complete a Binomial Probability
Distribution Table for the number of people who say they own a cell phone out of any 7 randomly
selected people.
Notation for the Binomial Probability Formula
n = the total number of trails . We sampled 5 people. n = 7
p is the probability of success for one of the n trails. 40% of adults have been to Hawaii p = .8
q is the probability of failure for one of the n trails.
If 80%own a cell phone then 20% do NOT own a cell phone. q = .2
The middle column shows the calculations. This column will not appear on the test. The
right column shows the values of p(X) for each x.
x: number
of persons who say they
own a cell phone out of 7
people questioned
P(x) = 7 C x • .8 # that own a cell phone • .2 # that do not own a cell phone
P(x)
0 of 7 own a cell phone
P(x) = 7 C 0 • .80 own a cell phone • .27 do not own a cell phone
0+
1 of 7 own a cell phone
P(x) = 7 C1 • .81 owns a cell phone • .26 do not own a cell phone
0+
2 of 7 own a cell phone
P(x) = 7 C2 • .82 own a cell phone • .25 do not own a cell phone
0+
3 of 7 own a cell phone
P(x) = 7 C 3 • .83
• .2 4 do not own a cell phone
.03
4 of 7 own a cell phone
P(x) = 7 C 4 • .8 4 own a cell phone • .2 3 do not own a cell phone
.11
5 of 7 own a cell phone
P(x) = 7 C 5 • .85 own a cell phone • .22
.28
6 of 7 own a cell phone
P(x) = 7 C 6 • .86 own a cell phone • .21 does not own a cell phone
.37
7 of 7 own a cell phone
P(x) = 7 C7 • .87 own a cell phone • .20 do not own a cell phone
.21
A) P( x ≤ 4) = P(0) + P(1) + P(2) + P(3) + P( 4) = .14
C) P(no less than 6) = P(6) + P( 7) = .58
Stat 300 5 – 2A Lecture
own a cell phone
do not own a cell phone
B. P( x ≥ 5 ) = P(5) + P(6) + P(7) = .86
D. P(no more than 2) = P(0) + P(1) + P(2) = 0+
Page 7 of 15
© 2012 Eitel
Example 3
A multiple choice test has 5 possible choices A, B, C, D, E. This means that if you guess on a
single question you have a 20% chance of getting that single answer correct. Complete a
Binomial Probability Distribution Table for the number of questions you can expect to get correct on a
5 question quiz if you guess on every question. Round P(x) to 2 decimal places. Answer the
additional questions using the calculated values from the table.
n = 5 (5 questions)
p = .20 (1 out of 5) and q = .8
Solution:
x: answers
correct
set up
P(x)
0
P(0) = 5 C 0 • .20 • .8 5
.33
1
A. P( x ≤ 1) = .74
B. P( x ≥ 3) = .06
1
P(1) = 5 C1 • .2 • .8
4
.41
2
P(2) = 5 C 2 • .22 • .8 3
.20
3
P(3) = 5 C 3 • .23 • .8 2
.05
4
P(4 ) = 5 C 4 • .2 4 • .81
.01
5
P(5) = 5 C 5 • .25 • .8 0
.00 +
C. P(no less than 4) = .0 +
D. P(no more than 2) = .94
Explanations to the answers
A) P( x ≤ 1) = P(0) + P(1)
B. P( x ≥ 3 ) = P(3) + P(4) + P(5)
C) P(no less than 4) = P(4 ) + P(5)
D. P(no more than 2) = P(0) + P(1) + P(2)
Stat 300 5 – 2A Lecture
Page 8 of 15
© 2012 Eitel
Example 4
The procedure is to spin the spinner 4 times. Create a Probability
Distribution Table for the expected number of times out of 4 that we
can expect to land on Green
Solution
The variable x will stand for the possible number of times we may land on green out of 4 spins. x can
be any number from 0 to 4. P(x) stands for the probability that x out of 4 spins will land on green.
p is the probability of landing on green:
p = 1/ 6
q is the probability of NOT landing on green.
q = 5/ 6
n = the total number of trails n = 4
x = number of of successes in n trails.
n – x = number of of successes in n trails.
P(x) is the probability of getting exactly x successes among n trails.
P( x) = n C x • p x • qn− x
or
P(S) = n CS • pS • qF
Use your calculator to find each P(x)
x: number of
green out of 4
spins
P(successes and failures)
P(0 greens and 4 not greens)
⎛ 1⎞
4 C 0 greens • ⎝ ⎠
6
0 greens
0 greens
1 greens
1 greens
P(1 green and 3 not greens)
⎛ 1⎞
•
⎝ 6⎠
P(2 greens and 2 not greens)
⎛ 1⎞
4 C 2 greens • ⎝ ⎠
6
2 greens
2 greens
P(3 greens and 1 not green)
⎛ 1⎞
4 C 3 greens • ⎝ ⎠
6
3 greens
3 greens
P(4 greens and 0 not greens)
⎛ 1⎞
⎝ 6⎠
4 greens
4 greens
P(S) = n CS • pS • qF
4 C1 green
4 C 4 greens •
⎛ 5⎞
•
⎝ 6⎠
⎛ 5⎞
•
⎝ 6⎠
P(x)
4 not greens
.48
3 not greens
.39
⎛ 5⎞
•
⎝ 6⎠
2 not greens
⎛ 5⎞
•
⎝ 6⎠
1 not greens
⎛ 5⎞
⎝ 6⎠
0 not greens
•
.12
.02
.00 +
∑ P(x) = 1.01
Note the total not 1 due to round off
Stat 300 5 – 2A Lecture
Page 9 of 15
© 2012 Eitel
Optional Lecture
A explanation of how the Binomial Probability Formula can be developed
A Bernoulli Distribution Table found using an entire sample space
Create a Probability Distribution Table for the expected number of
times out of 3 that we can expect to land on blue. P(x) stands for the
probability that x out of 3 spins will land on blue. The probabilities for
0 blues , 1 blue , 2 blue and 3 blues with 3 spins are calculated below.
The Sample Space = { BBB , BBR , BRB , BRR , RBB , RBR , RRB, RRR }
Second
Spin
1/ 3
B
Third
Spin
1/ 3
3
0
2
1
2
1
B P( B B B) =⎛ 1 ⎞ • ⎛ 3⎞
⎝ 3⎠ ⎝ 3⎠
B
0
2/ 3
R P(B B R)= ⎛ 1⎞ • ⎛ 2⎞
⎝ 3⎠ ⎝ 3⎠
1/ 3
3
⎛ 1⎞ ⎛ 2⎞
P(3 B, 0 R) = 1•
•
=.04
⎝ 3⎠ ⎝ 3⎠
1/ 3
2/ 3
R
B P(B R B) = ⎛ 1⎞ • ⎛ 2⎞
⎝ 3⎠ ⎝ 3⎠
2/ 3
P(2 B, 1 R ) =
1
2
2
1
R P(B R R) = ⎛ 1⎞ • ⎛ 2⎞
⎝ 3⎠ ⎝ 3⎠
First
Spin
2
1
⎛ 1⎞ ⎛ 2⎞
3•
•
=.22
⎝ 3⎠ ⎝ 3⎠
1/ 3
1/ 3
B
B P(R B B) = ⎛ 1⎞ ⎛ 2⎞
•
⎝ 3⎠
⎝ 3⎠
1
R P(R B R) = ⎛ 1⎞ ⎛ 2⎞
•
2/ 3
⎝ 3⎠
R
1/ 3
2/ 3
R
Second
Spin
Stat 300 5 – 2A Lecture
1
2/ 3
B P(R R B)
=
2/ 3
2
2
⎛ 1⎞ ⎛ 2⎞
P(1 B , 2 R ) = 3•
•
=.44
⎝ 3⎠ ⎝ 3⎠
⎝ 3⎠
1
2
0
1
⎛ 1⎞ ⎛ 2⎞
•
⎝ 3⎠ ⎝ 3⎠
⎛ 1⎞ ⎛ 3⎞
•
R P(R R R ) =
⎝ 3⎠ ⎝ 3⎠
0
3
⎛ 1⎞ ⎛ 2⎞
P(0 B, 3 R) = 1•
•
=.30
⎝ 3⎠ ⎝ 3⎠
Third
Spin
Page 10 of 15
© 2012 Eitel
The Probability Distribution Table for the model above
x: number of
blues in 3 spins
Formula
0
P(x)
3
0 of 3 are blue
⎛ 1⎞ ⎛ 2⎞
1•
•
⎝ 3⎠ ⎝ 3⎠
1 of 3 are blue
⎛ 1⎞ ⎛ 2⎞
3•
•
⎝ 3⎠ ⎝ 3⎠
⎛ 1⎞ ⎛ 2⎞
3•
•
⎝ 3⎠ ⎝ 3⎠
1
2 of 3 are blue
3
0
3 of 3 are blue
1•
1
2
⎛ 1⎞ ⎛ 2⎞
•
⎝ 3⎠ ⎝ 3⎠
.30
2
.44
.22
.04
There is 1 way that you can spin 0 blues in three spins. RRR
If you spin 0 blues there must be 3 reds.
P(0B) = the number of ways that 0 blues in three spins can happen • P(R ) • P(R ) •P(R )
0
P(0B)= 1•
3
⎛ 1⎞ ⎛ 2⎞
•
= .30
⎝ 2 ⎠ ⎝ 3⎠
There are 3 ways that you can spin 1 blues in three spins. BRR or RBR or RRB
If you spin 1 blue there must be 2 reds.
P(1B) = the number of ways that 1 blue in three spins can happen • P(B ) • P(R ) •P(R )
1
2
⎛ 1⎞ ⎛ 2⎞
P(1B)= 1•
•
= .44
⎝ 3⎠ ⎝ 3⎠
There are 3 ways that you can spin 2 blues in three spins. BBR or BRB or RBB
If you spin 1 blue there must be 2 reds.
P(2B) = the number of ways that 2 blues in three spins can happen • P(B ) • P(B ) •P(R )
2
P(1B)= 3•
1
⎛ 1⎞ ⎛ 2⎞
•
= .22
⎝ 3⎠ ⎝ 3⎠
There is 1 way that you can spin 3 blues in three spins. BBB
If you spin 3 blues there must be 0 reds.
P(3B) = the number of ways that 3 blues in three spins can happen • P(B ) • P(B ) •P(B )
3
0
⎛ 1⎞ ⎛ 2⎞
P(0B)= 1•
•
= .04
⎝ 3⎠ ⎝ 3⎠
Stat 300 5 – 2A Lecture
Page 11 of 15
© 2012 Eitel
Finding the number of ways an event can occur
We found the number of ways x blues can occur in n events by creating the entire population
and then counting the number of each type of occurrence. There is a formula that calculates the number
of ways that x things can occur in n events. n C x is the number of ways that x items can occur
when they are taken from a population of n items.
3 C 0 finds the number of ways 0 Blues can occur in 3 events.
3 C 1 finds the number of ways 1 Blues can occur in 3 events.
3 C 2 finds the number of ways 2 Blues can occur in 3 events.
3 C 3 finds the number of ways 3 Blues can occur in 3 events.
3C 0 = 1
3C 1 =3
3C 2 =3
3C 1 =1
RRR
BRR
RBR
RRB
BBR
BRB
RBB
BBB
Each of these calculator answers check the results we got by creating the population and counting each
occurrence of the event. It is much easier to use the n C x formula.
This leads to the final Probability Distribution Table with the complete formula
We complete table like these without models using only the formula
and the given value for P(B) =1/ 3
x: number of
blues in 3 spins
Formula
P(x)
0 blues
0 of 3 are blue
⎛ 1⎞
P(0 B in 3 spins) = 3 C0 blues •
⎝ 3⎠
⎛ 1⎞
P(1 B in 3 spins) = 3 C1 blues •
⎝ 3⎠
1 blue
1 of 3 are blue
2 of 3 are blue
P(2 B in 3 spins) = 3 C2 blues •
3 of 3 are blue
⎛ 1⎞
P(3 B in 3 spins) = 3 C3 blues •
⎝ 3⎠
⎛ 1⎞
⎝ 3⎠
⎛ 2⎞
•
⎝ 3⎠
⎛ 2⎞
•
⎝ 3⎠
2 blues
3 blues
•
3 reds
.30
2 reds
⎛ 2⎞
⎝ 3⎠
⎛ 2⎞
•
⎝ 3⎠
.44
1 red
.22
0 reds
.04
The values for each P(x) are based on the model. If you perform a sample of 50 sets of 3 spins, the
computed P(x) based on this fifty sets of spins will not match the table. As the numbers of sets of
spins gets very large the values for each P(x) will approach the vales based on the theoretical model.
Stat 300 5 – 2A Lecture
Page 12 of 15
© 2012 Eitel
A Probability Distribution Table
for the expected number of times out of 3 that we can expect to land on blue
if P(B ) = 1/ 3 P(R ) = 2/ 3
x: number of
blues in 3 spins
Formula
P(x)
⎛ 1⎞
P(0 B in 3 spins) = 3 C0 blues •
⎝ 3⎠
0 blues
0 of 3 are blue
P(1 B in 3 spins) = 3 C1 blues •
⎛ 1⎞
⎝ 3⎠
1 blue
1 of 3 are blue
2 of 3 are blue
⎛ 1⎞
P(2 B in 3 spins) = 3 C2 blues •
⎝ 3⎠
3 of 3 are blue
⎛ 1⎞
P(3 B in 3 spins) = 3 C3 blues •
⎝ 3⎠
3C 0 blues •
P( 0 Blues in 3 spins) =
⎛ 2⎞
•
⎝ 3⎠
•
2 blues
3 blues
⎛ 2⎞
⎝ 3⎠
3 reds
.30
2 reds
.44
⎛ 2⎞
•
⎝ 3⎠
1 red
⎛ 2⎞
•
⎝ 3⎠
0 reds
.22
.04
(1 / 3)0 blues • (2 / 3) 3 reds = .30
The number of ways 0 Blues in 3 spins can happen = 3 C 0
P(B ) = 1/ 3 is raised to a power equal to the number of blues, 0 Blues (1/ 3) 0
P(R ) = 2/ 3 is raised to a power equal to the number of reds, 3 Reds (2 / 3) 3
P( 1 Blue in 3 spins) =
3C1 blues
• (1/ 3)1 blue • (2 / 3) 2 reds = .44
The number of ways 1 Blue in 3 spins can happen = 3 C 1
P(B ) = 1/ 3 is raised to a power equal to the number of blues, 1 Blue (1/ 3)1
P(R ) = .75 is raised to a power equal to the number of reds, 2 Reds (2 / 3) 2
P( 2 Blues in 3 spins) =
3C 2 blues •
(1 / 3)2 blues • (2 / 3)1 red = .22
The number of ways 2 Blues in 3 spins can happen = 3 C 2
P(B ) = 1/ 3 is raised to a power equal to the number of blues, 2 Blues (1/ 3) 2
P(R ) = 2/ 3 is raised to a power equal to the number of reds, 1 Red (2 / 3)1
P( 3 Blues in 3 spins) = 3 C3 blues • .253 blues • .750 reds = .02
The number of ways 3 Blues in 3 spins can happen = 3 C 3
P(B ) = 1/ 3 is raised to a power equal to the number of blues, 3 Blue (1/ 3) 3
P(R ) = 2/ 3 is raised to a power equal to the number of reds, 0 Reds (2 / 3) 3
Stat 300 5 – 2A Lecture
Page 13 of 15
© 2012 Eitel
Additional Notes
P(0B in 3 spins) =
P(1B in 3 spins) =
Number Number Number
of ways
of
of
0B can blue= 0 red = 3
happen
0
P( 0B,3R) )= 3 C0 •
Number Number Number
of ways
of
of
1B can blue= 1 red = 2
happen
3
⎛ 1⎞
⎛ 2⎞
•
= .30
⎝ 3⎠
⎝ 3⎠
1
P( 1B,2R) )= 3 C1 •
P(2B in 3 spins) =
P(3B in 3 spins) =
Number Number Number
of ways
of
of
2B can blue= 2 red = 1
happen
2
1
⎛ 1⎞
⎛ 2⎞
P( 2B,1R) )= 3 C2 •
•
= .22
⎝ 3⎠
⎝ 3⎠
Stat 300 5 – 2A Lecture
2
⎛ 1⎞
⎛ 2⎞
•
= .44
⎝ 3⎠
⎝ 3⎠
Number Number Number
of ways
of
of
3B can blue= 3 red = 0
happen
3
0
⎛ 1⎞
⎛ 2⎞
P( 3B,0R) )= 3 C3 •
•
= .04
⎝ 3⎠
⎝ 3⎠
Page 14 of 15
© 2012 Eitel
P(0B)
There is 1 way that you can spin 0 blues in three spins. If you spin 0 blues there must be 3 reds.
0
P(RRR) =
3
2 2 2 ⎛ 1⎞ ⎛ 2⎞
• • =
•
= .30
3 3 3 ⎝ 3⎠ ⎝ 3⎠
0
3
2 2 2
⎛ 1⎞ ⎛ 2⎞
P(0B,3R) = • • = 1•
•
= .02
⎝ 3⎠ ⎝ 3⎠
3 3 3
P(1B)
There are 3 ways that you can spin 1 blue in three spins. If you spin 1 blue there must be 2 reds.
We could find P(1B) by finding
the probability of each way that 1B
and 2R could occur and then add
We could also find P(1B) by finding the
up all the separate probabilities.
probability of one of the orders that 1B
can happen and then multiply that by the
1
2
number of different ways that 1B can
1 2 2 ⎛ 1⎞ ⎛ 2⎞
P(BRR)= • • =
•
= .148
happen
3 3 3 ⎝ 3⎠ ⎝ 3⎠
Number
Number
Number
1
2
of ways
of
of
1 2 2 ⎛ 1⎞ ⎛ 2⎞
P(RBR)= • • =
•
= .148
1B
successes
failures
3 3 3 ⎝ 3⎠ ⎝ 3⎠
can
1 blue
2 reds
1
2
happen
1 2 2 ⎛ 1⎞ ⎛ 2⎞
P(RRB)= • • =
•
= .148
3 3 3 ⎝ 3⎠ ⎝ 3⎠
If we add the 3 probabilities above we get
the probability of getting 1B in three spins
1
2
⎛ 1⎞ ⎛ 2⎞
P(1B, 2R) =
•
= .43
⎝ 3⎠ ⎝ 3⎠
P(1B, 2R) =
1
⎛ 1⎞
3 •
•
⎝ 3⎠
⎛ 2⎞
⎝ 3⎠
2
= .44
P(2B)
There are 3 ways that you can spin 2 blues in three spins. If you spin 2 blues there must be 1 red.
We could find P(2B) by finding
the probability of each way that 2B
and 1R could occur and then add
up all the separate probabilities.
We could also find P(2B) by finding the
2
1
1 1 2 ⎛ 1⎞ ⎛ 2⎞
probability of one of the orders that 2B
P(BBR)=
• • =
•
= .074
3 3 3 ⎝ 3⎠ ⎝ 3⎠
can happen and then multiply that by the
number of different ways that 2B can
2
1
1 1 2 ⎛ 1⎞ ⎛ 2⎞
happen
P(BRB)=
• • =
•
= .074
3 3 3 ⎝ 3⎠ ⎝ 3⎠
Number
Number
Number
2
1
of
1 1 2 ⎛ 1⎞ ⎛ 2⎞
of ways
of
P(RBB)=
• • =
•
= .074
failures
2B
successes
3 3 3 ⎝ 3⎠ ⎝ 3⎠
1 red
can
2 blues
If we add the 3 probabilities
happen
above we get the probability of
getting 2B in three spins
2
1
⎛ 1⎞ ⎛ 2⎞
P(2B,1R) =
•
= .22
⎝ 3⎠ ⎝ 3⎠
P( 2B,1R) )= 3 •
⎛ 1⎞
⎝ 3⎠
2
•
⎛ 2⎞
⎝ 3⎠
1
= .22
P(3B)
There is 1 way that you can spin 3 blues in three spins. If you spin 3 blues there must be 0 reds.
Stat 300 5 – 2A Lecture
Page 15 of 15
© 2012 Eitel