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# The Binomial Distribution - PowerPoint PPT Presentation

The Binomial Distribution. In Statistics we often talk about trials. We mean an experiment, an investigation or the selection of a sample. e.g. We roll a die. There are 6 possible results ( outcomes ): 1 , 2 , 3 , 4 , 5 or 6.

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## PowerPoint Slideshow about ' The Binomial Distribution' - joel-foster

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Presentation Transcript

We mean an experiment, an investigation or the selection of a sample.

e.g. We roll a die.

There are 6 possible results ( outcomes ): 1, 2, 3, 4, 5 or 6.

However, if we are interested in getting a 6, we could say the trial has only 2 outcomes: a 6 or not a 6.

Lots of trials can be thought of as having 2 outcomes.

e.g. A seed is sown and the flower is either yellow or not yellow.

e.g. A computer chip is taken off a production line and it either works or it doesn’t.

The 2 possible outcomes of these trials are called success and failure.

Probabilty of success = p and probability failure = q.

Suppose that we repeat a trial several times and the probability of success doesn’t change from one trial to the next.

Suppose also that each result has no effect on the result of the other trials.

The trials are independent.

With these conditions all satisfied, we can use the binomial model to estimate the probability of success and to estimate the mean and variance.

What can you say about q + p ?

ANS:q + p = 1 since no other outcomes are possible.

• The Binomial distribution can be used to model a situation if all of the following conditions are met:

• A trial has 2 possible outcomes, success and failure.

• The trial is repeatedntimes.

• The probability of success in one trial is p and p is constant for all the trials.

• The trials are independent.

• n and p are called the parameters of the distribution.

• There are 2 outcomes to each trial. ( Success is getting a 4 and failure is not getting a 4 ).

• There is a constant probability ofsuccess ( getting a 4 ), so for every trial.

• The trials are independent.

e.g. We roll a fair die 4 times and we count the number of fours.

This experiment satisfies the conditions for the binomial model.

X = No. of 4`s obtained

4C0 = 1 no. of ways of obtaining no 4`s

4C1 = 4 no. of ways of obtaining one 4

4C2 = 6 no. of ways of obtaining two 4`s

4C3 = 4 no. of ways of obtaining three 4`s

4C4 = 1 no. of ways of obtaining four 4`s heads

4444

4

4

4

4

4

4

4

4

4

4

4

4

4

4

For 4 trials throwing a die

X = No. of 4`s obtained

(q + p)4 =

1

q4

+ 4

q3p1

+ 6

q2p2

+ 4

q1p3

+ 1

p4

(q + p)4 =

4C0

4C0

q4

q4

+4C1

+4C1

q3p1

q3p1

+4C2

+4C2

q2p2

q2p2

+4C3

+4C3

q1p3

q1p3

+4C4

+4C4

p4

p4

P = probability of success q = probability of failure

(q + p)4 =

4C0

q4

+4C1

q3p1

+4C2

q2p2

+4C3

q1p3

+4C4

p4

This means get 4 failures

Remember these no.s are also Pascals Triange line 4

This means get 3 failures and 1 success

This means get 1 failure and 3 successes

As q + p = 1

=1

q4

+4C1

q3p1

+4C2

q2p2

+4C3

q1p3

+4C4

p4

For 4 trials throwing a die

X = No. of 4`s obtained

P(X=x)=

Where x = 0,1,2,3,4

So the P(X=4) = 4C4q0p4

= 1 (  )0(  )4 = 0.00077

So the P(X=3) = 4C3q1p3

= 4 (  )1(  )3 = 0.0154

So the P(X=2) = 4C2q2p2

= 6 (  )2(  )2 = 0.1157

q4

+4C1

q3p1

+4C2

q2p2

+4C3

q1p3

+4C4

p4

For 4 trials throwing a die

X = No. of 4`s obtained

P(X=x)=

Where x = 0,1,2,3,4

So the P(X=1) = 4C1q3p1

= 4 (  )3(  )1 = 0.3858

So the P(X=0) = 4C0q4p0

= 1 (  )4(  )0 = 0.4823

Sum of probabilities=0.00077+0.0154+0.1157+0.3858+0.4823

=1

4

4

4

4

4

4

4

4

4

4

4

4

4

4

For 4 trials throwing a die

X = No. of 4`s obtained

P(4444) = 4C4()0 ()4

P(444 ) = 4C3()1 ()3

4

4

4

4

4

4

4

4

4

4

4

4

4

4

4

For 4 trials throwing a dice

X = No. of 4`s obtained

P(4444) = 4C4()0 ()4

P(444 ) = 4C3()1 ()3

P(44 ) = 4C2()2()2

4

4

4

4

4

4

4

4

4

4

4

4

4

4

4

For 4 trials throwing a dice

X = No. of 4`s obtained

P(4444) = 4C4()0 ()4

P(444 ) = 4C3()1 ()3

P(44 ) = 4C2()2 ()2

P(4 ) = 4C1()3 ()1

4

4

4

4

4

4

4

4

4

4

4

4

4

4

4

For 4 trials throwing a dice

X = No. of 4`s obtained

P(4444) = 4C4()0 ()4

P(444 ) = 4C3()1 ()3

P(44 ) = 4C2()2 ()2

P(4 ) = 4C1()3 ()1

P( ) = 4C0()4 ()0

4

4

4

4

4

4

4

4

4

4

4

4

4

4

4

For 4 trials throwing a dice

X = No. of 4`s obtained

P(4444) = 4C4()0 ()4

= 6 (  )2(  )2 = 0.1157

How does this compare with the experimental value

Experimental demo

N = 10000 trials

P(X=2) = 0.12

p=  = 0.16666

In expt p = 0.17

Close – why not closer?

A probability distribution gives the probabilities for all possible values of a variable.

• Can you use the formula we developed for rolling a die 4 times to work for a die thrown

• 3 times X = no. of 6`s

• 6 times X = no. of 1`s

nCx =

So P(X=x) =

For 3 trials throwing a dice

X = No. of 6`s obtained

q1p2

q2p1

q0p3

+3C1

+3C2

+3C3

q3p0

(q + p)3=

3C0

P(X=x) = nCx qn–x px

P(X=x) = 3Cx ()3–x ()x

For 6 trials throwing a dice

X = No. of 1`s obtained

6C0

q6p0

+6C1

q5p1

+6C2

q4p2

+6C3

q3p3

(q + p)6=

+6C4

q2p4

+6C5

q1p5

+6C6

q0p6

P(X=x) = nCx qn–xpx

P(X=x) = 6Cx ()6–x ()x

Using the notation in the formula book

If X is the random variable “ the number of sixes when a die is rolled 5 times ” then X has a binomial distribution and

are equal

If X is the random variable “ the number of sixes when a die is rolled 5 times ” then X has a binomial distribution and

and this . . .

is the sum of these

( 4 d.p. )

Tip: It saves some fiddling on the calculator if you remember that

It’s useful to remember that

We can simplify the expression using a calculator:

We can find the probabilities of getting 0, 1, 2, 4 and 5 sixes in the same way.

Can you find the probabilities thatX = 0 and X = 4andX = 5 ?

( Give the answers correct to 4 d.p. )

Since the sum of the probabilities is 1, I added the others and subtracted from 1.

The probabilities are:

Tip: If you have answers listed like this you need not write them out in a table.

The probability isn’t exactly zero so we need to show the 4 noughts to give the answer correct to 4 d.p.

In general, if X is a random variable with a binomial distribution, then we write

where n is the number of trials and

p is the probability of success in one trial.

The probabilities of0, 1, 2, 3, . . . n successes are given by

where x =0, 1, 2, 3, . . . n and q = 1 - p

e.g.1 If find the probability that X equals 0 or 1 giving the answer correct to 3 d.p.

In order to find this probability we have to add 2 results.

If we had used 3 d.p. for the individual probabilities we would have got for the answer, which is incorrect.

To be sure of the accuracy of the answer, we must use 4 decimal places in the individual calculations.

Solution:

When adding numbers, always use 1 more d.p. than you need in the answer OR store each individual number in your calculator’s memories.

(b)

(c)

(a)

(a)

Don’t forget that the binomial always has X = 0 as one possibility.

(b)

< 2

= 0 or 1

Solution:

(b)

(c)

(a)

c)

Can you see the quick way of doing this?

We found this in part (b)

ANS: Subtract the probabilities that we don’t want from 1.

Solution:

(b)

(c)

(a)

(a)

(b)

(c)

Exercise

Solution: