MTH5P2A: The Binomial Distribution

This unit explains The Binomial Distribution

Definition

The Binomial Distribution describes the behavior of a random variable(count variable)  X under the following conditions:

1. the number of trials(n) is fixed
2. trials have only 2 possible outcomes (success/failure)
3. each trial is independent of the other
4. probability of success(p) is constant throughout

X(the random variable) is a measure of the number of successes in n trials.

Example

A simple example is choosing 1 ball from a bag of 10 identical balls, each numbered (1-10). Once noted, the ball is returned to the bag.
A single ball is chosen on 3 separate occasions.

Success is in obtaining a ‘5‘ ball.

So the random variable X has values 0, 1, 2, 3

in other words, from our 3 tries we could have obtained:

0 fives,    1 five,   2 fives,   3 fives

On the first try, the probability of obtaining a 5 is 1/10 .
The probability of not getting a 5 is 9/10 .

Every time we dip into the bag of 10 balls, the probability of obtaining a ‘5’ is 1/10. The probability is constant.

Getting a ‘5’ or not getting a ‘5’ means that there are only 2 outcomes.

Every try is independent. Previous tries do not affect the result, since previously chosen balls are returned to the bag. Every try is taken from a bag of 10 balls.

Notation B(np)

The full notation describing a Binomial distribution is:

binomial distribution notation

where,

X a random variable (0, 1, 2, 3,…)
B   ‘is distributed Binomially
n number of trials
probability of single trial ‘success’

Example (continued from above)

Say that there are only 3 tries of attempting to take a 5-ball from a bag of 10 balls.

So n = 3 and p = 1/10.

The possible number of 5’s taken in the 3 trials is summarized by the values of the random variable X .

X = 0, 1, 2, 3

using the Binomial notation,

Binomial notation problem #1

Limits

The population size(n) of a Binomial Distribution must be much larger than the sample size(r).

The distribution only applies to trials from a simple random sample, where n is at least x10 times > r .

Outside this limit, results do not follow the equation.

Combinations   nCr

To appreciate the Binomial equation we must first have an understanding of combinations.

The definition of a Combination is: ‘the number of ways r items can be chosen from a set of nitems’

A short-hand way of writing this is nCr .

This can be written mathematically as:

combinations equation

*note ! means ‘factorial’ – eg !3 = 3 x 2 x 1

Though an easier method of calculation, especially with large values of n, is to use a calculator.

Example

With 52 cards in a deck, how many ways can 3 different cards be chosen?

deck of cards combination problem #1

Using a calculator,

52   SHIFT    nCr   3    =    22100

The Binomial Equation

the Biinomial equation - form #1                  (i

where the probability of failure, q, probabilty of failure

Since,                      nCr equivalent form

and           P(X=r) = P(r)

the Binomial equation   (i   is sometimes expressed as:

Binomial equation second form

Example

Using the problem first given in Binomial distribution part 1 and extending it to probability prediction:

1 ball is chosen from a bag of 10 identical balls, each numbered (1-10). Once noted, the ball is returned to the bag.
A single ball is chosen on 3 separate occasions.
Success is in obtaining a ‘5‘ ball.

i) What is the probability of obtaining a single ‘5’ ball in 3 choices?

ii) Draw a tree-diagram to find the probability of obtaining a single ‘5’ ball in the last choice.

i) let X be the random variable for obtaining a ‘5’ ball.

X has possible values (0, 1, 2, 3).

The number of trials n = 3

The number of required successes r = 1

probability of single successful trial (ie a ‘5’ ball) = 1/10

picking a '5' ball problem

ii)

tree diagram for ball problem

The probability(P) of getting a ‘5’ ball with 3 tries is the sum of each of the probabilities for each successful attempt.

The sum sign ‘+’ signifies an ‘OR’ decision.

The multiplication sign ‘x’ signifies the AND condition.

P = (q x q x p) + (q x p q) + (x q x q)

So the probability PL of the last ball drawn being a ‘5’ is given by:

PL = (q x q x p) = (9/10) (9/10) (1/10) = 81/1000 = 0.081

cumulative probability tables – the case of p<0.5

 

Cumulative probability tables – case of p<0.5

These give the tabulated value of P(Xx) . This means that the probability displayed is less than or equal to an observed value of x.
The random variable X is distributed Binomially, where there are n trials and probability of success p .

Before going into any detail about using the tables, we must first look at their structure.
There are a number of table designs, but they more or less contain the same data. It is just a matter of emphasis.

The tables we are using were issued by the Edexcel Examining Board(2009).

A binomial distribution ~ B(n, p) has values of p (across the top)and n (down the left side) in the following ranges:

p
0.05, 0.10, 0.15, 0.20, 0.25, 0.30, 0.35, 0.40, 0.45, 0.50
n
5, 6, 7, 8, 9, 10, 12, 15, 20, 25, 30, 40, 50

You can download a PDF copy of these tables, other tables and information on equations for A-level mathematics from the link below.

Mathematical Formulae Statistical Tables

(to download right click – “save target as” )

Case    P(X<x)

This is a straight forward lookup of the table.

Say that the random variable X has values X = 0, 1, 2, 3, 4, 5

also that the probability of success p=0.35 and the number of trials n=6 .

We want to know the value of the probability that X is less than or equal to 3 . That is , the value of : P(X<3)

cumulative probability table #1

From the table:           P(X<3) = 0.8826

The case      P(X=x)

Using the values of n and p from before, let’s say that we want to know the value of the probability that X is equal to 3. That is, the value of : P(X=3)

To understand this we must break down the values P(X<3) and P(X<2) into their constituent probabilities.

The probability that the random variable X is less than 3 or equal to 3 means that it can have values of ‘3’ or ‘2’ or ‘1’ or‘0’.

This can be written as the sum of probabilities. Remember for probability work, the operator ‘+‘ means OR .

P(X<3) = P(X=3) + P(X=2) + P(X=1) + P(X=0)

By the same reasoning,

P(X<2) = P(X=2) + P(X=1) + P(X=0)

subtracting the second equation from the first,

P(X<3) – P(X<2) = P(X=3)

turning the equation around,

P(X=3) = P(X<3) – P(X<2)

If we now look up the values for x = 3 and x = 2 from the tables:

P(X=3) = 0.8826 – 0.647  = 0.2355

The case     P(X<x)

Using the values of n and p from before, let’s say that we want to know the value of the probability that X is less than 3. That is, the value of : P(X<3)

The probability that the random variable X is less than 3 means that it can have values of ‘2’ or ‘1’ or ‘0’.

This can be written as the sum of probabilities. Remember for probability work, the operator ‘+‘ means OR .

P(X<3) = P(X=2) + P(X=1) + P(X=0)

but,

P(X<2) = P(X=2) + P(X=1) + P(X=0)

therefore,

P(X<3) = P(X<2)

since   P(X<2) = 0.6471

P(X<3) = 0.6471

The case     P(X>x)

Using the values of n and p from before, let’s say that we want to know the value of the probability that X is greater than 3. That is, the value of : P(X>3)

The probability that the random variable X is greater than 3 means that it can have values of ‘4’ or ‘5’ *

* the random variable X has values X = 0, 1, 2, 3, 4, 5

This can be written as the sum of probabilities. Remember for probability work, the operator ‘+‘ means OR .

P(X>3) = P(X=4) + P(X=5)

but the sum of all the individual probabilities equals ‘1’ .

P(X=0) + P(X=1) + P(X=2) + P(X=3) + P(X=4) + P(X=5) = 1

rearranging, making P(X=4) + P(X=5) the subject,

P(X=4) + P(X=5) = 1 – [ P(X=0) + P(X=1) + P(X=2) + P(X=3)]

but,

P(X<3) = P(X=0) + P(X=1) + P(X=2) + P(X=3)

therefore,

P(X=4) + P(X=5) = 1 – P(X<3)

hence,

P(X>3) = 1 – P(X<3)

from the table,

P(X<3) = 0.8826

therefore,

P(X>3) = 1 – 0.8826 = 0.1174

The case     P(X>x)

Using the values of n and p from before, let’s say that we want to know the value of the probability that X is greater than or equal to 3. That is, the value of : P(X>3)

The probability that the random variable X is greater than or equal to 3 means that it can have values of ‘3’ or ‘4’ or ‘5’ *

* the random variable X has values X = 0, 1, 2, 3, 4, 5

This can be written as the sum of probabilities. Remember for probability work, the operator ‘+‘ means OR .

P(X>3) = P(X=3) + P(X=4) + P(X=5)

but the sum of all the individual probabilities equals ‘1’ .

P(X=0) + P(X=1) + P(X=2) + P(X=3) + P(X=4) + P(X=5) = 1

rearranging, making P(X=3) + P(X=4) + P(X=5) the subject,

P(X=3) + P(X=4) + P(X=5) = 1 – [P(X=0) + P(X=1) + P(X=2)]

but,

P(X>3) = P(X=3) + P(X=4) + P(X=5)

and,

P(X<2) = P(X=0) + P(X=1) + P(X=2)

it follows that ,

P(X>3) = 1 – P(X<2)

putting in values,

P(X>3) = 1 – 0.6471 = 0.3529

Cumulative probability tables – case of p>0.5

These give the tabulated value of P(X< x) . This means that the probability displayed is less than or equal to an observed value of x.
The random variable X is distributed Binomially, where there are n trials and probability of success p .

Working out values of random variable probabilty P(X) for the case of p>0.5 is complicated by the fact that values of p only go up to 0.5 .

The way around this problem is to consider another random variable Y , representing failure.

So we have:                      pX + p= 1

In the same way as XY is distributed binomially:

Y ~ B(n,1 – pY)

Say that the random variable X has values X = 0, 1, 2, 3, 4, 5

A table of values for (success) and Y (failure) looks like this:

X
0
1
2
3
4
5
Y
5
4
3
2
1
0

The method is to use the table to produce an expression in Y that will use values of p<0.5 .

The case      P(X<x)

Say we wish to find the value of P(X<4) for :

X = 0, 1, 2, 3, 4, 5      pX=0.85*       n=6

*tables only go up to 0.5

X (success) and Y (failure) are related so:

X
0
1
2
3
4
5
Y
5
4
3
2
1
0

The sum of successes and failures for each outcome must always be the same (ie 5).

The probability for Y becomes py=0.15*     (pX + pY= 1)

* py <0.5 and therefore on the table

From the table,

P(X=0) + P(X=1) + P(X=2) + P(X=3) + P(X=4)

P(Y=5) + P(Y=4) + P(Y=3) + P(Y=2) + P(Y=1)

this can be written:

P(X<4) = P(Y>1)

since,

P(Y=0) + P(Y=1) + P(Y=2) + P(Y=3) + P(Y=4) + P(Y=5) = 1

P(Y=1) + P(Y=2) + P(Y=3) + P(Y=4) + P(Y=5) = 1 – P(Y=0)

in other words,

P(Y>1) = 1 – P(Y<0)

hence the original inequality can be rewritten :

P(X<4) = 1 – P(Y<0)

Using the tables to find the value of P(Y<0) for n=6 pY=0.15 ,Y=0 :

P(Y<0) = 0.3771

binomial distribution - cumulative probability table #2

hence,

P(X<4) = 1 – 0.3771 = 0.6229

The case      P(X=x)

Consider the binomial distribution for success,

X ~ B(n, pX)

X = 0, 1, 2, 3, 4, 5      pX=0.85       n=6

also the binomial distribution for failure,

Y ~ B(n, pY)

Y= 0, 1, 2, 3, 4, 5      pY=0.15       n=6

Say we want to find P(X=4).

X
0
1
2
3
4
5
Y
5
4
3
2
1
0

From the table it follows that,

P(X=4) = P(Y=1)

P(Y=1) = P(Y<1) – P(Y<0)

P(X=4) = P(Y<1) – P(Y<0)

P(X=4) = 0.7765 – 0.3771 = 0.3994

The case     P(X<x)

Consider the binomial distribution for success,

X ~ B(n, pX)

X = 0, 1, 2, 3, 4, 5      pX=0.85       n=6

also the binomial distribution for failure,

Y ~ B(n, pY)

Y= 0, 1, 2, 3, 4, 5      pY=0.15       n=6

Say we want to find P(X<4).

X
0
1
2
3
4
5
Y
5
4
3
2
1
0

From the table it follows that,

P(X<4) = P(Y>1)

and

P(Y>1) = P(Y=2) + P(Y=3) + P(Y=4) + P(Y=5)

it follows that,

P(Y>1) = P(Y>2)

P(Y>2) = P(Y=2) + P(Y=3) + P(Y=4) + P(Y=5)

P(Y=0) + P(Y=1) + P(Y=2) + P(Y=3) + P(Y=4) + P(Y=5) = 1

P(Y=2) + P(Y=3) + P(Y=4) + P(Y=5) = 1 – [ P(Y=0) + P(Y=1)]

P(Y>2) = 1 – [ P(Y=0) + P(Y=1)]

P(Y>2) = 1 – P(Y<1)

P(X<4) = 1 – P(Y<1)

P(X<4) = 1 – 0.7765 = 0.2235

The case     P(X>x)

Consider the binomial distribution for success,

X ~ B(n, pX)

X = 0, 1, 2, 3, 4, 5      pX=0.85       n=6

also the binomial distribution for failure,

Y ~ B(n, pY)

Y = 0, 1, 2, 3, 4, 5      pY=0.15       n=6

Say we want to find P(X>4).

X
0
1
2
3
4
5
Y
5
4
3
2
1
0

P(X>4) = P(Y<1)

P(Y<1) = P(Y=0)

P(Y=0) = P(Y<0)

P(X>4) = P(Y<0)

reading P(Y<0) directly from the tables,

P(X>4) = 0.3771

The case     P(X>x)

Consider the binomial distribution for success,

X ~ B(n, pX)

X = 0, 1, 2, 3, 4, 5      pX=0.85       n=6

also the binomial distribution for failure,

Y ~ B(n, pY)

Y = 0, 1, 2, 3, 4, 5      pY=0.15       n=6

Say we want to find P(X>4).

X
0
1
2
3
4
5
Y
5
4
3
2
1
0

P(X>4) = P(Y<1)

reading P(Y<1) directly from the tables,

P(X>4) = 0.7765

The Mean

A random variable X distributed binomially, with trials n and constant probability of success p is described as:

B(n, p)

By definition the mean is described as,

mean = μ (mu) = E(X) = np

where E(X) is the expectation/expected value of X .

Example

The chance of getting a red sweet from a box of 40 coloured sweets is 1/10 . How many red sweets would you expect in each box?

Let the random variable of getting a red sweet be X .

Therefore,

B(40,1/10)

since the mean/expected value is given by:

μ = E(X) = np

μ = 40 x 1/10 = 4

answer: you would expect 4 red sweets in each box

Variance

For random variable X distributed binomially, with trials n and constant probability of success p ,

variance is defined as:

variance = σ2 (sigma squared) = Var(X) = np(1 – p)

Sometimes variance is written in terms of the probability of failure .

Since p + q = 1, then q = (1 – p). The equation for variance now becomes:

variance = npq

Example

A five sided spinner with numbers 1, 2, 3, 4, 5 on each sector is twirled 20 times and the number of ‘3’ s scored recorded each time.

i) How many times would you expect the ‘3’ to appear?
ii)What is the variance?

i) If X is the random variable distributed binomially,

B(20,1/5)

μ = E(X) = np

μ =20 x 1/5 = 4

answer: you would expect a ‘3‘ to be recorded 4 times

ii) since,

variance = npq

variance = 20 x 1/5 x 4/5 = 80/25 = 3.2

answer: the variance is 3.2

ASSIGNMENT : The Binomial Distribution Assignment MARKS : 30  DURATION : 1 week, 3 days

 

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