STATISTICS - Section 2

 

Binomial Distribution 2

 

 

Combinations nCr

The Binomial 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 n items'

 

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

 

It must be emphasized that 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

 

 

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The Binomial Equation

 

 

 

 

where the probability of failure,

 

 

Since, 

 

and

 

 

 

the Binomial equation 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)

 

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 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

 

 

 

 

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