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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(n, p)
The full notation describing a Binomial distribution is:
where,
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,
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:
*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?
Using a calculator,
52 SHIFT nCr 3 = 22100
The Binomial Equation
where the probability of failure,
Since,
and
the Binomial equation (i is sometimes expressed as:
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
p probability of single successful trial (ie a ‘5’ ball) = 1/10
ii)
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 x q) + (p 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(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 .
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 X ~ B(n, p) has values of p (across the top)and n (down the left side) in the following ranges:
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)
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 + pY = 1
In the same way as X, Y 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 X (success) and Y (failure) looks like this:
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:
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
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).
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).
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).
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).
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:
X ~ 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,
X ~ 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 q .
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,
X ~ 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