Binomial Distribution for MBA 1st & BCA 5th

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
		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:

n!

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!		pk(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!	pk(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!	pk(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!	pk(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))

BINOMIAL DISTRIBUTION (FOR MBA )

probability for M.B.A. 1st & B.C.A 5th

Probability: Types of Events

Life is full of random events!
You need to get a "feel" for them to be a smart and successful person.
The toss of a coin, throw of a dice and lottery draws are all examples of random events.

Events

When we say "Event" we mean one (or more) outcomes.

Example Events:

• Getting a Tail when tossing a coin is an event
• Rolling a "5" is an event.

An event can include several outcomes:

• Choosing a "King" from a deck of cards (any of the 4 Kings) is also an event
• Rolling an "even number" (2, 4 or 6) is an event

Events can be:

• Independent (each event is not affected by other events),
• Dependent (also called "Conditional", where an event is affected by other events)
• Mutually Exclusive (events can't happen at the same time)

Let's look at each of those types.

Independent Events

Events can be "Independent", meaning each event is not affected by any other events.
This is an important idea! A coin does not "know" that it came up heads before ... each toss of a coin is a perfect isolated thing.

Example: You toss a coin three times and it comes up "Heads" each time ... what is the chance that the next toss will also be a "Head"?

The chance is simply 1/2, or 50%, just like ANY OTHER toss of the coin.

What it did in the past will not affect the current toss!

Some people think "it is overdue for a Tail", but really truly the next toss of the coin is totally independent of any previous tosses.

Saying "a Tail is due", or "just one more go, my luck is due" is called The Gambler's Fallacy

Learn more at Independent Events.

Dependent Events

But some events can be "dependent" ... which means they can be affected by previous events.

Example: Drawing 2 Cards from a Deck

After taking one card from the deck there are less cards available, so the probabilities change!

Let's look at the chances of getting a King.
For the 1st card the chance of drawing a King is 4 out of 52
But for the 2nd card:

• If the 1st card was a King, then the 2nd card is less likely to be a King, as only 3 of the 51 cards left are Kings.
• If the 1st card was not a King, then the 2nd card is slightly more likely to be a King, as 4 of the 51 cards left are King.

This is because we are removing cards from the deck.

Replacement: When we put each card back after drawing it the chances don't change, as the events are independent.
Without Replacement: The chances will change, and the events are dependent.

You can learn more at Dependent Events: Conditional Probability

Tree Diagrams

When we have Dependent Events it helps to make a "Tree Diagram"

Example: Soccer Game

You are off to soccer, and love being the Goalkeeper, but that depends who is the Coach today:

• with Coach Sam your probability of being Goalkeeper is 0.5
• with Coach Alex your probability of being Goalkeeper is 0.3

Sam is Coach more often ... about 6 of every 10 games (a probability of 0.6).

Let's build the Tree Diagram!
Start with the Coaches. We know 0.6 for Sam, so it must be 0.4 for Alex (the probabilities must add to 1):

Then fill out the branches for Sam (0.5 Yes and 0.5 No), and then for Alex (0.3 Yes and 0.7 No):

Now it is neatly laid out we can calculate probabilities (read more at "Tree Diagrams").

Mutually Exclusive

Mutually Exclusive means we can't get both events at the same time.

It is either one or the other, but not both
Examples:

• Turning left or right are Mutually Exclusive (you can't do both at the same time)
• Heads and Tails are Mutually Exclusive
• Kings and Aces are Mutually Exclusive

What isn't Mutually Exclusive

• Kings and Hearts are not Mutually Exclusive, because we can have a King of Hearts!

Like here:

Aces and Kings are Mutually Exclusive		Hearts and Kings are not Mutually Exclusive