Section 2 : The Binomial Distribution(2)
 

[combinations nCr][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:

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

 

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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

p 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 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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