Despite the fact, numerous distributions fall in the category of ‘Continuous Probability Distributions’ Binomial and Poisson set examples for the ‘Discrete Probability Distribution’ and among widely used as well. Beside this common fact, significant points can be brought forward to contrast these two distributions and one should identify at which occasion one of this has been rightly chosen.

Binomial Distribution

‘Binomial Distribution’ is the preliminary distribution used to encounter, probability and statistical problems. In which a sampled size of ‘n’ is drawn with replacement out of ‘N’ size of trials out of which yields a success of ‘p’. Mostly this has been carried out for, experiments which provides two major outcomes, just like ‘Yes’, ‘No’ results. On the contrary to this, if the experiment is done without replacement, then model will be met with ‘Hypergeometric Distribution’ that to be independent from its every outcome. Though ‘Binomial’ comes into play at this occasion as well, if the population (‘N’) is far greater compared to the ‘n’ and eventually said to be the best model for approximation.

However, at most of the occasions most of us get confused with the term ‘Bernoulli Trials’. Nevertheless, both the ‘Binomial’ and ‘Bernoulli’ are similar in meanings. Whenever ‘n=1’ ‘Bernoulli Trial’ is especially named, ‘Bernoulli Distribution’

The following definition is a simple form of bringing the exact picture between, ‘Binomial’ and ‘Bernoulli’:

‘Binomial Distribution’ is the sum of independent and evenly distributed ‘Bernoulli Trials’. Below mentioned are some important equations comes under category of ‘Binomial’

Probability Mass Function (pmf): (ⁿ_k) p^k(1-p)^n-k ; (ⁿ_k) = [n !] / [k !] [(n-k) !]

Mean: np

Median: np

Variance: np(1-p)

At this particular example,

‘n’- The whole population of the model

‘k’- Size of the which is drawn and replaced from ‘n’

‘p’- Probability of success for every set of experiment which consists only two outcomes

Poisson Distribution

On the other hand this ‘Poisson distribution’ has been chosen at the event of most specific ‘Binomial distribution’ sums. In other words, one could easily say that ‘Poisson’ is a subset of ‘Binomial’ and more of a less a limiting case of ‘Binomial’.

When an event occurs within a fixed time interval and with a known average rate then it is common that case can be modeled using this ‘Poisson distribution’. Besides that, the event must be ‘independent’ as well. Whereas it is not the case in ‘Binomial’.

‘Poisson’ is used when problems arise with ‘rate’. This is not always true, but more often than not it is true.

Probability Mass Function (pmf): (_λ^k /k!) _e^-λ

Mean: λ

Variance: λ

What is the difference between Binomial and Poisson?

As a whole both are examples of ‘Discrete Probability Distributions’. Adding to that, ‘Binomial’ is the common distribution used more often, however ‘Poisson’ is derived as a limiting case of a ‘Binomial’.

According to all these study, we can arrive at a conclusion saying that regardless of the ‘Dependency’ we can apply ‘Binomial’ for encounter the problems as it is a good approximation even for independent occurrences. In contrast, the ‘Poisson’ is used at questions/problems with replacement.

At the end of the day, if a problem is solved with both the ways, which is for ‘dependent’ question, one must find the same answer at each instance

The Binomial Distribution

"Bi" means "two" (like a bicycle has two wheels) ... ... so this is about things with two results.

		Tossing a Coin: Did we get Heads (H) or Tails (T)
We say the probability of the coin landing H is ½ And the probability of the coin landing T is ½

		Throwing a Die: Did we get a four ... ? ... or not?
We say the probability of a four is 1/6 (one of the six faces is a four). And the probability of not four is 5/6 (five of the six faces are not a four)

Let's Toss a Coin!

Toss a fair coin three times ... what is the chance of getting two Heads?

We will use these terms:

Outcome: the result of three coin tosses
Event: "Two Heads" out of three coin tosses

Tosing a coin three times could get any one of these outcomes (H is for heads and T for Tails):

HHH
HHT
HTH
HTT
THH
THT
TTH
TTT

Which outcomes do we want?

"Two Heads" could be in any order: "HHT", "THH" and "HTH" all have two Heads (and one Tail).

So 3 of the outcomes produce "Two Heads".

What is the probability of each outcome?

Each outcome is equally likely, and there are 8 of them. So each has a probability of 1/8

So the probability of event "Two Heads" is:

Number of outcomes we want		Probability of each outcome
3	×	1/8	= 3/8

Let's Calculate Them All:

The calculations are (P means "Probability of"):

P(Three Heads) = P(HHH) = 1/8
P(Two Heads) = P(HHT) + P(HTH) + P(THH) = 1/8 + 1/8 + 1/8 = 3/8
P(One Head) = P(HTT) + P(THT) + P(TTH) = 1/8 + 1/8 + 1/8 = 3/8
P(Zero Heads) = P(TTT) = 1/8

We can write this in terms of a Random Variable, X, = "The number of Heads from 3 tosses of a coin":

P(X = 3) = 1/8
P(X = 2) = 3/8
P(X = 1) = 3/8
P(X = 0) = 1/8

And we can also draw a Bar Graph:

It is symmetrical!

Making a Formula

Now ... what are the chances of 5 heads in 9 tosses ... to list all outcomes (512) would take a long time!

So let's make a formula.

In our previous example, how could we get the values 1, 3, 3 and 1 ?

They are actually in the third row of Pascal’s Triangle ... !

Can we make them using a formula?

Sure we can, and here it is:
		n = total number k = number we want
It is often called "n choose k" and you can read more about it at Combinations and Permutations. Note: the "!" means "factorial", for example 4! = 1×2×3×4 = 24

Let's use it:

Example: 3 tosses getting 2 Heads

We have n=3 and k=2

n!	=	3!	=	3×2×1	= 3
k!(n-k)!		2!(3-2)!		2×1 × 1

So there are 3 outcomes for "2 Heads"

(We knew that already, but now we have a formula for it.)

Let's use it for a harder question:

Example: what are the chances of 5 heads in 9 tosses?

We have n=9 and k=5

n!	=	9!	=	9×8×7×6×5×4×3×2×1	= 126
k!(n-k)!		5!(9-5)!		5×4×3×2×1 × 4×3×2×1

And for 9 tosses there are 2⁹ = 512 total outcomes, so we get the probability:

Number of outcomes we want		Probability of each outcome
126	×	1	=	126
126	×	512	=	512

P(X=5) =	126	=	63	= 0.24609375
	512		256

About a 25% chance.

(Easier than listing them all.)

Bias!

So far the chances of success or failure have been equally likely.

But what if the coins are biased (land more on one side than another) or choices are not 50/50.

Example: You sell sandwiches. 70% of people choose chicken, the rest choose pork.

What is the probability of selling 2 chicken sandwiches to the next 3 customers?

This is just like the heads and tails example, but with 70/30 instead of 50/50.

Let's draw a tree diagram:

The "Two Chicken" cases are highlighted.

Notice that the probabilities for "two chickens" all work out to be 0.147, because we are multiplying two 0.7s and one 0.3 in each case.

Can we get the 0.147 from a formula? What we want is "two 0.7s and one 0.3"

0.7 is the probability of each choice we want, call it p
2 is the number of choices we want, call it k

Probability of "choices we want" (two chickens) is: p^k

And

The probability of the opposite choice is: 1-p
The total number of choices is: n
The number of opposite choices is: n-k

Probability of "opposite choices" (one pork) is: (1-p)^(n-k)

So all choices together is:

p^k(1-p)^(n-k)

Example: (continued)

p = 0.7 (chance of chicken)
n = 3
k = 2

So we get:

p^k(1-p)^(n-k) = 0.7²(1-0.7)^(3-2) = 0.7²(0.3)⁽¹⁾ = 0.7 × 0.7 × 0.3 = 0.147

which is the probability of each outcome.

And the total number of those outcomes is:

n!	=	3!	=	3×2×1	= 3
k!(n-k)!		2!(3-2)!		2×1 × 1

And we get:

Number of outcomes we want		Probability of each outcome
3	×	0.147	=	0.441

So the probability of event "2 people out of 3 choose chicken" = 0.441

OK. That was a lot of work for something we knew already, but now we can answer harder questions.

Example: You say "70% choose chicken, so 7 of the next 10 customers should choose chicken" ... what are the chances you are right?

p = 0.7
n = 10
k = 7

So we get:

p^k(1-p)^(n-k) = 0.7⁷(1-0.7)^(10-7) = 0.7⁷(0.3)⁽³⁾ = 0.0022235661

That is the probability of each outcome.

And the total number of those outcomes is:

n!	=	10!	=	10×9×8	= 120
k!(n-k)!		7!(10-7)!		3×2×1

And we get:

Number of outcomes we want		Probability of each outcome
120	×	0.0022235661	=	0.266827932

In fact the probability of 7 out of 10 choosing chicken is only about 27%

Moral of the story: even though the long-run average is 70%, don't expect 7 out of the next 10.

Putting it Together

Now we know how to calculate how many:

k!(n-k)!

And the probability of each:

p^k(1-p)^(n-k)

We can multiply them together:

Probability of k out of n ways:

P(k out of n) =	n!		p^k(1-p)^(n-k)
	k!(n-k)!

The General Binomial Probability Formula

Important Notes:

The trials are independent,
There are only two possible outcomes at each trial,
The probability of "success" at each trial is constant.

Quincunx

Have a play with the Quincunx (then read Quincunx Explained) to see the Binomial Distribution in action.

Throw the Die

A fair die is thrown four times. Calculate the probabilities of getting:

0 Twos
1 Two
2 Twos
3 Twos
4 Twos

In this case n=4, p = P(Two) = 1/6

X is the Random Variable ‘Number of Twos from four throws’.

Substitute x = 0 to 4 into the formula:

P(k out of n) =	n!	p^k(1-p)^(n-k)
	k!(n-k)!

Like this (to 4 decimal places):

P(X = 0) = (4!/0!4!) × (1/6)⁰(5/6)⁴ = 1 × 1 × (5/6)⁴ = 0.4823
P(X = 1) = (4!/1!3!) × (1/6)¹(5/6)³ = 4 × (1/6) × (5/6)³ = 0.3858
P(X = 2) = (4!/2!2!) × (1/6)²(5/6)² = 6 × (1/6)² × (5/6)² = 0.1157
P(X = 3) = (4!/3!1!) × (1/6)³(5/6)¹ = 4 × (1/6)³ × (5/6) = 0.0154
P(X = 4) = (4!/4!0!) × (1/6)⁴(5/6)⁰ = 1 × (1/6)⁴ × 1 = 0.0008

Summary: "for the 4 throws, there is a 48% chance of no twos, 39% chance of 1 two, 12% chance of 2 twos, 1.5% chance of 3 twos, and a tiny 0.08% chance of all throws being a two (but it still could happen!)"

This time the Bar Graph is not symmetrical:

It is not symmetrical!

It is skewed because p is not 0.5

Sports Bikes

Your company makes sports bikes. 90% pass final inspection (and 10% fail and need to be fixed).

What is the expected Mean and Variance of the 4 next inspections?

First, let's calculate all probabilities.

n = 4,
p = P(Pass) = 0.9

X is the Random Variable "Number of passes from four inspections".

Substitute x = 0 to 4 into the formula:

P(k out of n) =	n!	p^k(1-p)^(n-k)
	k!(n-k)!

Like this:

P(X = 0) = (4!/0!4!) × 0.9⁰0.1⁴ = 1 × 1 × 0.0001 = 0.0001
P(X = 1) = (4!/1!3!) × 0.9¹0.1³ = 4 × 0.9 × 0.001 = 0.0036
P(X = 2) = (4!/2!2!) × 0.9²0.1² = 6 × 0.81 × 0.01 = 0.0486
P(X = 3) = (4!/3!2!) × 0.9³0.1¹ = 4 × 0.729 × 0.1 = 0.2916
P(X = 4) = (4!/4!0!) × 0.9⁴0.1⁰ = 1 × 0.6561 × 1 = 0.6561

Summary: "for the 4 next bikes, there is a tiny 0.01% chance of no passes, 0.36% chance of 1 pass, 5% chance of 2 passes, 29% chance of 3 passes, and a whopping 66% chance they all pass the inspection."

Mean, Variance and Standard Deviation

Let's calculate the Mean, Variance and Standard Deviation for the Sports Bike inspections.

There are (relatively) simple formulas for them. They are a little hard to prove, but they do work!

The mean, or "expected value", is:

μ = np

For the sports bikes:

μ = 4 × 0.9 = 3.6

So we would expect 3.6 bikes (out of 4) to pass the inspection.
Makes sense really ... 0.9 chance for each bike times 4 bikes equals 3.6

The formula for Variance is:

Variance: σ² = np(1-p)

And Standard Deviation is the square root of variance:

σ = √(np(1-p))

For the sports bikes:

Variance: σ² = 4 × 0.9 × 0.1 = 0.36

Standard Deviation is:

σ = √(0.36) = 0.6

Note: we could also calculate them manually, by making a table like this:

X	P(X)	X × P(X)	X² × P(X)
0	0.0001	0	0
1	0.0036	0.0036	0.0036
2	0.0486	0.0972	0.1944
3	0.2916	0.8748	2.6244
4	0.6561	2.6244	10.4976
	SUM:	3.6	13.32

The mean is the Sum of (X × P(X)):

μ = 3.6

The variance is the Sum of (X² × P(X)) minus Mean²:

Variance: σ² = 13.32 − 3.6² = 0.36

Standard Deviation is:

σ = √(0.36) = 0.6

And we got the same results as before (yay!)

Summary

The General Binomial Probability Formula

P(k out of n) =	n!	p^k(1-p)^(n-k)
	k!(n-k)!

Mean value of X: μ = np

Variance of X: σ² = np(1-p)
Standard Deviation of X: σ = √(np(1-p))

Mathematics

Wednesday 24 December 2014

statistics for MBA

Difference Between Binomial and Poisson for MBA

Thursday 20 November 2014

for MBA 1st & BCA 5th sem

for MBA 1st & BCA 5th sem

for MBA 1st & BCA 5th sem

Friday 31 October 2014

Binomial Distribution for MBA 1st & BCA 5th

The Binomial Distribution

Let's Toss a Coin!

Which outcomes do we want?

What is the probability of each outcome?

Let's Calculate Them All:

Making a Formula

Example: 3 tosses getting 2 Heads

Example: what are the chances of 5 heads in 9 tosses?

Bias!

Example: You sell sandwiches. 70% of people choose chicken, the rest choose pork.

What is the probability of selling 2 chicken sandwiches to the next 3 customers?

Example: (continued)

Example: You say "70% choose chicken, so 7 of the next 10 customers should choose chicken" ... what are the chances you are right?

Putting it Together

Quincunx

Throw the Die

Sports Bikes

Mean, Variance and Standard Deviation

Summary