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Probability and Statistics
Binomial Distributions
Introduction to Binomial Distributions
Let X = the number of right handed students out of 4
randomly selected. Fill in the following table.
X Sample points
0
1
P(X)
Number of
ways that X
can occur
3
  =
0
Probability of one sample
point corresponding to X
(number of ways that X
can occur) x (probability
of each sample point)
 3
  =
1 
2
3
Letting p = the probability that one observes a right-handed student, q =
the probability that one observes a left handed student, and n = the total
number of students observed, can you create a general formula for P(X)?
P(X) = _____________________________________
Introduction to Binomial Distributions
Let X = the number of right handed students out of 4 randomly selected.
X Sample points
0
1
P(X)
LLLL
.134 =
0.0003
RLLL
LRLL
LLRL
LLLR
(.87)(.133 )
= 0.0019
(Times 4)
0.0076
(.872)(.132 )
LLRR RRLL = 0.0128
2 RLLR LRRL
(Times 6)
RLRL LRLR 0.0768
LRRR
(.873)(.13)
RLRR
= 0.0856
3
RRLR
(Times 4)
RRRL
0.3424
4=
.87
RRRR
4
0.5729
Number of
Probability of one sample
ways that X
point corresponding to X
can occur
 4
0x
4
  = 1
.87
.13
0
 4
  =
1 
(number of ways that X
can occur) x (probability
of each sample point)
4 .870 x 4
.13
0
(.87)(.133 )
4 (.87)(.133 )
1
4
=6
2
(.872 )(.132 )
4
(.872 )(.132 )
2
4
=4
3
(.873 )(.13)
4 (.873 )(.13)
3
4
=1
4
.874 𝑥 .130
4 .874 𝑥 .130
4
4
The Binomial Probability
Distribution Formula
(number of ways that x can occur)
x (probability of each sample point)
P(x = 0) =
4 .870 x 4
.13
0
4 (.87)(.133 )
P(x = 1) =
1
P(x = 2) =
4
(.872 )(.132 )
2
4 (.873 )(13)
P(x = 3) = 3
P(x = 4) =
4 .874 𝑥 .130
4
Letting p = the probability that one
observes a right-handed student, q
= the probability that one observes
a left handed student, and n = the
total number of students observed,
can you create a general formula
for P(X)?
In General:
𝑛
P(x) =
(𝑝 𝑥 )(𝑞 𝑛−𝑥 )
𝑥
The Binomial Probability
Distribution Formula
𝑛
𝑥
𝑛−𝑥
P(x) =
(𝑝 )(𝑞
)
𝑥
This same probability can be found on the calculator:
You can get this same value on the TI-84 by the following steps:
1) 2nd VARS
2) Down to binompdf (and press enter)
3) binompdf(n,p,x), so fill in n, p and x.
or with the newer calculators:
trials: n
p: p
x value: value of x that you want P(x) for.
Example 1: Use the calculator to find at least 4 of the probabilities
above.
X Sample points
0
1
P(X)
LLLL
.134 =
0.0003
RLLL
LRLL
LLRL
LLLR
(.87)(.133 )
= 0.0019
(Times 4)
0.0076
(.872)(.132 )
LLRR RRLL = 0.0128
2 RLLR LRRL
(Times 6)
RLRL LRLR 0.0768
LRRR
(.873)(13)
RLRR
= 0.0856
3
RRLR
(Times 4)
RRRL
0.3424
4=
.87
RRRR
4
0.5729
Number of
Probability of one sample
ways that X
point corresponding to X
can occur
 4
0x
4
  = 1
.87
.13
0
 4
  =
1 
(number of ways that X
can occur) x (probability
of each sample point)
4 .870 x 4
.13
0
(.87)(.133 )
4 (.87)(.133 )
1
4
=6
2
(.872 )(.132 )
4
(.872 )(.132 )
2
4
=4
3
(.873 )(.13)
4 (.873 )(.13)
3
4
=1
4
.874 𝑥 .130
4 .874 𝑥 .130
4
4
Example 2: Use the formula to find the
probability that one observes 5 H’s and 5 tails
when flipping a coin 10 times. In other words,
let X = the number of heads observed out of
10 tosses. Find P(5).
𝑛
P(x) =
(𝑝 𝑥 )(𝑞 𝑛−𝑥 )
𝑥
P(x=5) = 10 (𝑝 𝑥 )(𝑞 𝑛−𝑥 )
5
Means flip coin
Each having
10 times and
observe 5 heads
HHTTHHTHTT
or
HHHTTTHTHT
or
how many other
ways?
252 ways
= 252 x 0.000977
= 0.246
probability of
(.55 )(.55 ) =
0.000977
P(5) = binompdf(10, .5, 5)
= 0.246
Binomial Probability Distributions
Many experiments only have TWO possible outcomes. Which of the
experiments have only two outcomes for each trial?
a) Flipping a coin 20 times
Two outcomes for each trial:
Each flip can only result in Heads or Tails
Binomial Probability Distributions
Many experiments only have TWO possible outcomes. Which of the
experiments have only two outcomes for each trial?
b) Rolling a single die 5 times, recording the result
SIX outcomes for each trial:
Each roll can result in 1, 2, 3, 4, 5 or 6.
c) Rolling a single die 5 times and counting how many
times it landed on one.
Two outcomes for each trial:
Each roll can only result in 1 or not
Binomial Probability Distributions
Many experiments only have TWO possible outcomes. Which of the
experiments have only two outcomes for each trial?
d) Guess at 6 multiple choice questions with 4 possible
answers, record answer A, B, C, or D
FOUR outcomes for each trial:
Each questions can result in A, B, C, D
e) Guess at 6 multiple choice questions with 4 possible
answers, record correct or incorrect
Two outcomes for each trial:
Each question is correct or incorrect
The Binomial probability distribution can be applied if:
n
1) The experiment consists of _____
identical trials.
only two possible
2) There are ____________________outcomes
for each
trial.
success
s = ______________
or f =_______________
failure
probability of success/failure remains the same
3) The _____________
for
each trial.
independent
4) The trials are ____________________
WHY?
The __________________________________
is the
Binomial random variable
number of successes in n trials.
Example – If we consider the random variable X = the number of
right-handed students out of 4 randomly chosen, then let’s see if we
can use the binomial formula to find the probability distribution of
X.
Asking 4 students the identical question
1)______________________________________________
2 possible outcomes, left or right
2)______________________________________________
P(s) = .87
P(f) = .13 remains the same for each trial
3)______________________________________________
The occurrence of one student being left or right handed
doesn’t affect the probability of the next student being left
or right handed, therefore independent.
4)______________________________________________
Do you get it? – If we consider the random variable X = the
number of Aces out of 3 cards (without replacement), can we use
the Binomial probability distribution?
The trials are not identical, one card will be drawn from a
full deck and others will not
1)______________________________________________
2 possible outcomes, either an Ace or not
2)______________________________________________
P(ACE) = ? P(Not and ACE) = ? Changes for each trial
3)______________________________________________
The occurrence of one card drawn does affect the probability
of the next card, since without replacement, therefore
NOT independent.
4)______________________________________________
Notation for Binomial Probability Distributions
s and f
_______________
(success and failure) are the two possible
outcomes.
p and q
_______________
denotes the probabilities or s and f.
P(s) = ________
= probability of success in one of the n trials
p
q
P(f) = ________
= probability of failure in one of the n trials
n = number of trials
____
____
x = number of successes
n!
P( x) 
p x q n x
(n  x)! x!
P(x) = binompdf(n, p, x)
is the probability of getting
exactly x successes in n trials.
is the probability of getting
exactly x successes in n trials.
Mean and Standard Deviation for the Binomial Random Variable
  n p
  n pq
Example 3 Overbooking Flights: Air America has a policy of
booking as many as 42 people on an airplane that can seat
only 40. Past studies have revealed that only 85% of the
booked passengers actually arrive for the flight.
a) Find the probability that if Air America books 42 people,
not enough seats will be available. Interpret this result by
indicating what percent of the time we expect one of these
flights to be overbooked.
The plane will be overbooked if 41 or 42 of the booked
passengers show up. We need to find the probability of
41 or 42 passengers showing up. Find P(41) and P(42)
and add them together.
P(41) = binompdf(42, .85, 41)
.008 + .001 = .009
P(42) = binompdf(42, .85, 42)
Less than a 1% chance.
Example 3 Overbooking Flights: Air America has a policy of
booking as many as 42 people on an airplane that can seat
only 40. Past studies have revealed that only 85% of the
booked passengers actually arrive for the flight.
b) Use the two standard deviation interval to
determine what is the least number of passengers we
could expect to (usually) arrive at one of these
flights?   n  p
  n pq
µ = np = 42 x 0.85 = 35.7
𝜎 = 𝑛𝑝𝑞 = 42 𝑥 0.85 𝑥 0.15 = 2.3
𝜇 ± 2𝜎 = 35.7 ± 2 2.3 = (31.1, 40.3)
The least number we expect to arrive is 32 and most is 40.
Example 4 A statistics quiz consists of 10 multiple-choice
questions, each with four possible answers. Let the random variable
x = the number of questions correct out of 10.
a) Find the following:
0, 1, 2, …10
Domain of X = __________________
0.25
p = _____
10
n = _______
b) Create a probability distribution for X in the form of a table.
X
P(X)
0
1
2
3
4
5
6
7
8
9
10
.06 .18 .28 .25 .15 .06 .02 .00 .00 .00 .00
You can find the entire distributions by binompdf(10, .25).
c) What is the probability that a student gets exactly 3 correct?
____________
X
P(X)
0
1
2
3
4
5
6
7
8
9
10
.06 .18 .28 .25 .15 .06 .02 .00 .00 .00 .00
P(x = 3) = .25 or a 25% chance.
d) For someone who makes random guesses for all of the answers,
find the probability of failing if the minimum passing grade is 70%.
X
P(X)
0
1
2
3
4
5
6
7
8
9
10
.06 .19 .28 .25 .15 .06 .02 .00 .00 .00 .00
P(x < 7) = 1 or a 100% chance.
e) If the entire class guesses at 10 questions, what is the Expected
value for the average number of questions one would get correct?
X
P(X)
0
1
2
3
4
5
6
7
8
9
10
.06 .19 .28 .25 .15 .06 .02 .00 .00 .00 .00
µ = np = 10 x 0.25 = 2.5
f) How many questions correct would be usual (find the 2 standard
deviation interval)?
X
P(X)
0
1
2
3
4
5
6
7
8
9
10
.06 .18 .28 .25 .15 .06 .02 .00 .00 .00 .00
µ = np = 10 x 0.25 = 2.5
𝜎 = 𝑛𝑝𝑞 = 10 𝑥 0.25 𝑥 0.75 = 1.37
𝜇 ± 2𝜎 = 2.5 ± 2 1.37 = (−0.24, 5.24)
The usual outcomes are 0, 1, 2, 3, 4, 5 questions correct.
µ = np = 10 x 0.25 = 2.5
𝜎 = 𝑛𝑝𝑞 = 10 𝑥 0.25 𝑥 0.75 = 1.37
𝜇 ± 2𝜎 = 2.5 ± 2 1.37 = (−0.24, 5.24)
The usual outcomes are 0, 1, 2, 3, 4, 5 questions correct.
X
P(X)
0
1
2
3
4
5
6
7
8
9
10
.06 .19 .28 .25 .15 .06 .02 .00 .00 .00 .00
We will now do this experiment and verify the above.
1) ___ 2) __ 3) ___ 4) ____ 5) ___ 6) ___ 7) ____8) __ 9) ___ 10)___
Example 5 Rotator cuff surgery has a 90% chance of
success. The surgery is performed on five patients.
Out of the five patients, three of them were successful. Is
this usual given the 90% success rate, or should we steer
clear of this doctor?
X
0
P(x)
.00001
1
.00045
2
.0081
3
.0729
4
5
.32805
.59049
P(x = 3) = .0729, so not unusual, but very unlikely
μ = 4.5
σ = 0.67
2 standard deviations = (3.16, 5.84)
Using the two standard deviation interval, x = 3 is
unusual.
Example 1 Find the probability distribution of X = the number
of heads observed if you toss a coin 4 times.
From Last section we would:
S = {HHHH THHH HTHH HHTH HHHT TTTT HTTT THTT
TTHT TTTH HHTT TTHH THHT HTTH THTH HTHT}
X = { 0, 1, 2, 3, 4}
X
P(X)
0
1/16 = .0625
1
4/16 = .25
2
6/16 = .375
3
4/16 = .25
4
1/16 = .0625
Using Section 4.4
n = _______
4 THE NUMBER OF TRIALS
p = _______ THE PROBABILITY OF SUCCESS IN ONE OF THE TRIALS
0.5
q = 1 – p = _______THE PROBABILITY OF FAILURE IN ONE OF THE TRIALS
0.5
If we want to find the probability that out of 4
tosses 2 are heads we calculate P(2) as follows
n!
p x q n x
Use P( x) 
(n  x)! x!
n = 4, p = .5, q=.5
= Binompdf(4, .5, 2)
P(2) =
__________________________________________
Or by the calculator
1) 2nd VARS
2) Down to Binompdf (and press enter)
3) Binompdf(4, .5, 2) = .375
x
0
P(0) =
4!
.50.540  .0625  binompdf(4 ,.5,0)
(4  0)!0!
1
4!
1 4 1
P(1) =
.5 .5  .25  binompdf(4 ,.5,1)
(4  1)!1!
2
P(2) =
4!
2 4 2
.5 .5  .375  binompdf(4 ,.5,2)
(4  2)!2!
4!
3 4 3
.5 .5  .25  binompdf(4 ,.5,3)
3 P(3) =
(4  3)!3!
4!
4 4 4
.5 .5  .0625  binompdf(4 ,.5,4)
4 P(4) =
(4  4)!4!
Mean and Standard Deviation for the
Binomial Random Variable
  n p
  n pq
Example 2 Suppose we have each student in the
class toss a coin 15 times. If we listed the sample
points like in chapter 3 we there would be
______________ of them. WOW! TOO MANY.
a) What is the mean and standard deviation?
b) How many head’s would be unusual to get
when this experiment is performed?
c) What is the probability of that you get 11 or more
heads when you toss a coin 15 times?
d) We will not do this experiment and verify the
above.
Example 3 Using Directory Assistance Use the binomial probability
formula to find the probability of getting exactly 3 correct responses
among 5 different requests from AT&T directory assistance. Assume
in that in general, AT&T is correct 90% of the time.
We use the following formula
n!
P( x) 
p x q n x = binompdf(n, p, x)
(n  x)! x!
5
n = __________
The number of correct responses out of 5
x = _______________________________________________________
0.9
p = ___________(
90% of the time directory assistance gives correct number)
0.1
q = ___________(
10% of the time directory assistance gives wrong number)
Find the probability distribution for
X
X
P(X)
0
1
.00001 .00045
2
3
.0081
.0729
4
5
.32805 .59049
X
P(X)
0
1
.00001 .00045
2
3
.0081
.0729
4
5
.32805 .59049
a) Find and interpret P(3).
P(3) = .0729 There is a
7.29% chance that AT&T will give exactly 3
correct responses out of 5, or in other words 7 out
of 100 times when calling directory assistance 5
times, we would get exactly 3 correct responses.
b) Is this low? Why? Yes this is low. AT&T is correct
90% of the time, 3 out of 5 means correct 60% of the
time, so not that likely, but NOT unusual since above 5%.
c)- What is the probability of getting at least 3 correct
responses. So either 3, 4, or 5 out of 5 calls?
P(at least 3 correct responses) = P(3) + P(4) + P(5) =
.0729 + .32805 + .59049 = .99144
d)What is the mean and standard deviation?
Why is the mean not 2.5, the middle of the values of the
random variable X?
How could the mean be 2.5?
μ = np = 5 (.9) = 4.5
  npq  5(.9)(.1)  .671
e)What are the usual values of X? Show two ways to answer
this question.
 Example 4 A statistics quiz consists of 10 multiple-choice questions, each
with five possible answers. For someone who makes ransom guesses for all
of the answers, find the probability of passing if the minimum passing
grade is 60%. Is the probability high enough to make it worth the risk of
trying to pass by making random guesses instead of studying?
Let X = ________________________________
p = _____________________
n = _________________
 Create a probability distribution for X in the form of a table and then
answer the question.
What is the average number of questions one would get correct if they
guessed at all 10 questions?
How many questions correct would be usual?
We will now do this experiment and verify the above.
Example 5
Suppose that I draw 5 cards from a deck of cards and
count the number of hearts drawn.
a)
Find the probability that 3 of the 5 cards are hearts.
b)
Can we use the binomial distribution to find the
above probability?