Binomial distribution

Binomial Probability Distributions
A coin-tossing experiment is a simple example of an
important discrete random variable called the binomial
random variable.
Example- A sociologist is interested in the proportion of
elementary school teachers who are men or women
Example- A soft-drink marketer is interested in the
proportion of drinkers who prefer the brand or not

Example- A sales person interested in sale of a policy, if
he visits 10 people in a day what is the probability of
his selling policy to one person, or 2 person, ….
Example- If ten persons enter the shop, what is the
probability that he will purchase the Mobile, or 2 will
purchase the mobile and so on

Definition: A binomial experiment is one that has these
four characteristics:
1. The experiment consists of n identical results/
trials/ observation.
2. Each trial results in one of two outcomes: one
outcome is called a success, S, and the other a failure,
F.
3. The probability of success on a single trial is equal
to p and remains the same from trial to trial. The
probability of failure is equal to (1 - p) = q.
4. The trials are independent.

The Binomial Probability function
( )!
( ) (1 )
!( )!
x n xn
f x p p
x n x
-
= -
-
x = the number of successes
p = the probability of a success on one trial
n = the number of trials
f(x) = the probability of x successes in n trials

Mean and Standard Deviation for the Binomial
Random Variable:
Mean: m = np
Variance: s 2 = npq
Standard deviation: npq=s

Binomial Formula. Suppose a binomial
experiment consists of n trials and results
in x successes. If the probability of success on
an individual trial is P, then the binomial
probability is:
b(x; n, P) = nCx * Px * (1 - P)n – x
b(x; n, P) = nCr * Pr * (1 - P)n – r
Where r is the value which the random variable takes

EXAMPLE
Suppose a coin is tossed 2 times. What is the probability of getting
(a) 0 head
Solution: This is a binomial experiment in which the number of trials is
equal to 2, the number of successes is equal to 0, and the probability
of success on a single trial is 1/2. Therefore, the binomial probability
is:
b(0; 2, 0.5) = 2C0 * (0.5)0 * (0.5)2
= 1/4
(b) 1 head
b(1; 2, 0.5) = 2C1 * (0.5)1 * (0.5)1
= 2/4
(c) 2 head
b(2; 2, 0.5) = 2C2 * (0.5)2 * (0.5)0
= 1/4

EXAMPLE
(a) 0 head
Solution: This is a binomial experiment in which the number of trials is
equal to 10, the number of successes is equal to 0, and the
probability of success on a single trial is 1/2. Therefore, the binomial
probability is:
b(0; 10, 0.5) = 10C0 * (0.5)0 * (0.5)10
= 1/1024
Number of
Trial
1 2 3 4 5 6 7 8 9 10
outcome T T T T T T T T T T
Probability 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5

EXAMPLE
(b) 1 head
Number of
Trial
1 2 3 4 5 6 7 8 9 10
Outcome 1 H T T T T T T T T T

EXAMPLE
(b) 1 head
Number of
Trial
1 2 3 4 5 6 7 8 9 10
Outcome 2 T H T T T T T T T T

EXAMPLE
(b) 1 head
Number of
Trial
1 2 3 4 5 6 7 8 9 10
Outcome 3 T T H T T T T T T T

EXAMPLE
(b) 1 head
Number of
Trial
1 2 3 4 5 6 7 8 9 10
Outcome 4 T T T H T T T T T T

EXAMPLE
(b) 1 head
Number of
Trial
1 2 3 4 5 6 7 8 9 10
Outcome 5 T T T T H T T T T T
Outcome 6 T T T T T H T T T T
Outcome 7 T T T T T T H T T T
Outcome 8 T T T T T T T H T T
Outcome 9 T T T T T T T T H T
Outcome 10 T T T T T T T T T H

EXAMPLE
(b) 1 head
Number of
Trial
1 2 3 4 5 6 7 8 9 10
Outcome 5 T T T T H T T T T T
Outcome 6 T T T T T H T T T T
Outcome 7 T T T T T T H T T T
Outcome 8 T T T T T T T H T T
Outcome 9 T T T T T T T T H T
Outcome 10 T T T T T T T T T H
Probability 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5

EXAMPLE
(b) 1 head
b(1; 10, 0.5) =
10C1 * (0.5)1 * (0.5)9
= 10/1024

EXAMPLE
(c) 2 head
b(2; 10, 0.5) = 10C2 * (0.5)2 * (0.5)8
= 45/1024

As a Sales Manager you analyze the sales
records for all sales persons under your
guidance.
Ram has a success rate of 75% and average
10 sales calls per day. Shyam has a success
rate of 45% but average 16 calls per day
What is the probability that each sales
person makes 6 sales on any given day

A manufacturer is making a
product with a 20% defective rate.
if we select 5 randomly chosen
items at the end of the assembly
line, what is the probability of
having 1 defective items in our
sample?

EXAMPLE-
Suppose you take a multiple choice test with
10 questions, and each question has five
answer choices, what is the probability you
get exactly 4 questions correct just by
guessing
Number of trails= n= 10, event is correct
answer, P(Success)=1/5
Interested in 4 correct answers, r= 4

In San Francisco, 30% of workers take public transportation daily
1. In a sample of 10 workers, what is the probability that exactly
three workers take public transportation daily
2. In a sample of 10 workers, what is the probability that at least
three workers take public transportation daily
3. How many workers are expected to take public transportation
daily?
4. Compute the variance of the number of workers that will take the
public transport daily.
5. Compute the standard deviation of the number of workers that
will take the public transportation daily.

In San Francisco, 30% of workers take public transportation daily
1. In a sample of 10 workers, what is the probability that exactly
three workers take public transportation daily= 0.2668
2. In a sample of 10 workers, what is the probability that at least
three workers take public transportation daily= 0.6172
3. How many workers are expected to take public transportation
daily? = 3
4. Compute the variance of the number of workers that will take the
public transport daily. = 2.10
5. Compute the standard deviation of the number of workers that
will take the public transportation daily.= 1.449

Twelve of the top twenty finishers in the 2009 PGA Championship at
Hazeltine National Golf Club in Chaska, Minnesota, used a Titleist
brand golf ball (Golf Ball Test website, November 12, 2009). Suppose
these results are representative of the probability that a randomly
selected PGA Tour player uses a Titleist brand golf ball. For a sample of
15 PGA Tour players, make the following calculation
Compute the probability that exactly 10 of the 15 PGA Tour players use
a Titleist brand golf ball
Compute the probability that more than 10 of the 15 PGA Tour players
use a Titleist brand golf ball
For a sample of 15 PGA Tour players, compute the expected number of
players who use a Titleist brand golf ball
For a sample of 15 PGA Tour players, compute the variance and
standard deviation of the number of players who use a Titleist brand
golf ball

Using the 20 golfers in the Hazeltine PGA Championship, the
probability that a PGA professional golfer uses a Titleist brand golf ball
is p = 14/20 = .6
For the sample of 15 PGA Tour players, use a binomial distribution with
n = 15 and p = .6
F(10)= 0.1859
P(x > 10) = f (11) + f (12) + f (13) + f (14) + f (15)
.1268 + .0634 + .0219 + .0047 + .0005 = .2173
E(x) = np = 15(.6) = 9
Var(x) = s2 = np(1 - p) = 15(.6)(1 - .6) = 3.6
SD= 1.8974

A brokerage survey reports that 30 percent of
individual investors have used a discount broker (i.e
one which does not charge the full commission). In a
random sample of 10 individuals, what is the
probability that
a. Exactly two of the sampled individuals have used a
discount broker
b. Not more than three have used a discount broker
c. At least three of them have used a discount broker.

Example-
12 out of 20 players in IPL used MRF Bat and won the match. For a
randomly selected sample of 20 such players of IPL. Find the
probability that
1. exactly 10 out of 15 players use MRF bat.
2. More then 10 players use MRF bat
3. Find the expected number of players using MRF bat
4. Find the SD and variance of the number of players using MRF bat

Binomial distribution

Recommended

Recommended

More Related Content

Similar to Binomial distribution

Similar to Binomial distribution (20)

Recently uploaded

Recently uploaded (20)

Binomial distribution