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Bernoulli Trial
Binomial Distribution
Conditions for Binomial Distribution
Mean and Variance of Binomial Distribution
Poisson Distribution
Attributes of a Poisson Experiment
Applications of Binomial Distribution
A random experiment which has only two mutually exclusive outcomes, namely ‘success’ and ‘not success’ or ‘failure’ is called Dichotomous Experiment.
If a dichotomous experiment is repeated n times and if in each trial the probability of success p (0 < p < 1) is constant, then all such trials are called Bernoulli trails.
This term is named after Jacob Bernoulli, a 17^{th} century Swiss Mathematician.
Since, Bernoulli trial has only 2 possible outcomes, it can be framed as ‘yes’ or ’no’ question.
Example
Is the top card of the shuffled deck an ace?
Was the new born child a girl?
An experiment of n repeated independent Bernoulli trails is called a Binomial Experiment.
Consider a Binomial experiment. Let the outcomes in each trial be a success (S) or Failure (F) with probability p and (1-p) = q respectively.
Let a random variable X = No. of successes
0 ≤ X ≤ n.
We want to find the probability of an event that the n trial result in exactly x successes and (n-x) failures is
Let A = {S S S…. S x times}
B= {F F F……F (n-x) times }
Since trials are independent,
P( A∩B) = P(a) . P(b)
= P(S).P(S)…..x times. P(F). P(F)…. (n-x) times
= p.p.p…..x times . q.q.q ….. (n-x) times
= p^{x }(q)^{n-x}
Now, the number of possible orders of x successes and (n-x) failures in independent trials is ^{n}C_{x }and these are mutually disjoint orders. Hence, probability that n trials result in exactly x successes and (n-x) failures in any order is given by
This is known as Binomial Distribution
A discrete random variable X is said to follow a Binomial distribution with parameters n and p if its probability mass function (pmf) if given by
P[X = x] = p(x) = ^{n}C_{x }p^{x }q^{n-x }, x = 0,1 , 2….n
0< p< 1, q = 1 – p,
n = 1, 2, 3….
Otherwise
P[X = x] for every x is xth term in the expansion of (q+p)^{n }
Binomial distribution can be represented in tabular form as follows:
From the above table we can see that,
P(x) ≥ 0 for all x
∑ p(x) = ∑ ^{n}C_{x }p^{x }q^{n-x}
= (q + p)^{n}
= 1
Fig: Probability Mass function of Binomial Distribution
The binomial distribution is used whenever each of the following is satisfied:
Trials of experiment are Bernoulli trials
The performance of a Bernoulli trial results in an outcome that can be classified either as a success or a failure
The number of trials (n) is finite
All n trials are independent
The probability of success is constant from trial to trial, i.e. the trials are repeated under identical conditions. Further, if P(success) = p and P(failure) = q, then p + q =1, 0 < p < 1
The random variable X denotes the number of successes x in n independent Bernoulli trials.
Notations:
X ~ B(n,p) denotes that X follows Binomial Distribution with parameters n and p.
If X ~ B(n,p), then Y= (n – X) ~ B(n,q)
Fig: Types of Probability Distribution
The mean of random variable X denoted as E(X) is given by E(X) = np
The variance of random variable X denoted by Var (X) is given by Var(X) = npq
The standard derivation of random variable X is defined as the positive random variable X denoted by SD(X) is given by SD(X)= √Var(X) = √npq
Note: Mean > Variance that is, np > npq
The Poisson binomial distribution, which describes the number of successes in a series of independent Yes/No experiments with different success probabilities.
A Poisson distribution is the probability distribution that results from a Poisson experiment.
A Poisson experiment is a statistical experiment that has the following properties:
The experiment results in outcomes that can be classified as successes or failures.
The average number of successes (μ) that occurs in a specified region is known.
The probability that a success will occur is proportional to the size of the region.
The probability that a success will occur in an extremely small region is virtually zero.
Note that the specified region could take many forms. For instance, it could be a length, an area, a volume, a period of time, etc.
A Poisson Random Variable is the number of successes that result from a Poisson experiment. The probability distribution of a Poisson random variable is called a Poisson Distribution.
Given the mean number of successes (μ) that occur in a specified region, we can compute the Poisson probability based on the following formula:
Formula of Poisson
Suppose we conduct a Poisson experiment, in which the average number of successes within a given region is µ. Then, the Poisson probability is:
P(x; μ) = (e^{-μ}) (μ^{x}) / x!
Where x is the actual number of successes that result from the experiment and is approximately equal to 2.71828
The Poisson distribution has the following properties:
The mean of the distribution is equal to μ.
The variance is also equal to μ.
A fair coin is tossed 8 times. Find the probability that it shows heads
Exactly 5 times
A larger number of times than tails
At least once
Solution
Let X= Number of heads
P = Probability of getting head
P = 1/2
Q = 1- p = 1-1/2 = ½
Given: n = 8
X ~ B(8,1/2)
The p.m.f of X is given as
P(X = x) = p(x) = ^{n}C_{x }p^{x }q^{n-x }
that is, p(x) = ^{8}C_{x }(1/2)^{x }(1/2)^{8-x } , x=0,1,2….8
= P (5) = ^{8}C_{5 }(1/2)^{5 }(1/2)^{8-5}
^{= } ^{8}C_{3 }(1/2)^{5 }(1/2)^{3 }[since ^{n}C_{x = }^{n}C_{n-x }]
= 7/32
Hence, the probability of getting exactly 5 heads is 0.21875
= P[ X ≥ 5] = p(5) + p(6) + p(7) + p(8)
= ^{8}C_{5 }(1/2)^{5 }(1/2)^{8-5 }+ ^{8}C_{6 }(1/2)^{6 }(1/2)^{8-6 }+ ^{8}C_{7 }(1/2)^{7 }(½)^{8-7 }+ ^{8}C_{8 }(1/2)^{8 }(1/2)^{8-8}
= ^{8}C_{3 }(1/2)^{5 }(1/2)^{3 }+ ^{8}C_{2 }(1/2)^{6 }(1/2)^{2 }+ ^{8}C_{1 }(1/2)^{7 }(½) + ^{8}C_{8 }(1/2)^{8 }
= (½)^{8 }[56 + 28 + 8 + 1] = 93/256
= P[X ≥ 5] = 0.36328
= P [ X ≥ 1] = 1-P[X = 0]
= 1- p(0)
= 1- ^{8}C_{0 }(1/2)^{0 }(½) ^{8-0}
= 1 – (1/2)^{8 }= 1 - 1/256 = 255 / 256
P[ X ≥ 1] = 0.996
Hence, the probability of getting heads at least once is 0.996
Suppose that 80% of all families own a television set. If 10 families are interviewed at random, find the probability that
Seven families own a television set
At most three families own a television set
Let X = Number of families who own a television
P = Probability of family who own a television set
P = 80% = 80/100 = 4/5
Q = 1- p = 1-4/5 = 1/5
Given: n =10
X ~ B(10,4/5)
that is, p(x) = ^{10}C_{x }4/5^{x }1/5^{10-x }
^{= 10}C_{7 (}4/5)^{7 }(1/5)^{10-7}
^{= 10}C_{3 (}4/5)^{7 }(1/5)^{3 }[since ^{n}C_{x = }^{n}C_{n-x }]
= 0.2013
Hence, the probability that seven families own a television set is 0.2013
= P [X ≤ 3] = P[X = 0] + P[X = 1] + P[X = 2] + P[X = 3]
= p(0) + p(1) + p(2) + p(3)
^{= 10}C_{0} (4/5)^{0} (1/5)^{10-0} + ^{10}C_{1} (4/5)^{1} (1/5)^{10-1} + ^{10}C_{2} (4/5)^{2} (1/5)^{10-2} + ^{10}C_{3} (4/5)^{3} (1/5)^{10-3}
= [1 + 40 + 45 * 16 + 120 * 64](1/5)^{10}
= 8441/9765625
P [ X ≤ 3 ] = 0.00086
Hence, the probability that at most three families own a television set is 0.00086
To find the probability of getting x heads in the experiment of tossing a fair coin n times. Here, X ~ B (n, p = ½)
To find the probability of getting success in the experiment of throwing a fair die or a pair of fair dice n times
To find the probability of correct answers in a multiple choice test consisting of n questions
To find the probability of success in any dichotomous experiment.
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