The Binomial DistributionFor the very common case of

“Either-Or” experiments with only two possible outcomes

To accompany Hawkes lesson 5.2Original content by D.R.S.

Recognize Binomial Situations

• Only two possible outcomes in each trial.– Probability for one of the outcomes.– Probability for the other outcome.

• Some definite number of trials, .– They’re independent trials. don’t change.

• We’re interested in , probability of a certain count of how many times event happens in those trials.

A special kind of probability distribution

• It’s the familiar probability distribution• But only two rows for the two outcomes.• Note that – Because probabilities must always sum to 1.00000– And this leads to .

Outcomes Probabilities

One of the events

The other event

Total Exactly 1

The Binomial Probability Formula

• Question: If we have trials, what is the probability of occurrences of the “success” event (the one with probability )

• Answer:

Practice with the Formula

• Experiment: Roll two dice• Event of interest: “I rolled a 7 or an 11”• Probability of success: (from • Probability of failure: • Number of trials

Practice with the Formula

• Find P(2) successes in the seven/eleven game

• How about 3 successes?

Compute them all

Summary of the 7-11 experiment

X successes P(X successes)

0 times (no sevens or elevens)

1 time

2 times

3 times

4 times

5 times (all sevens and elevens)

Total (must equal 1.000000 !!)

Sometimes you add probabilities

• Probability of at least three wins in five trials– P(X≥3) = P(X=3) + P(X=4) + P(X=5) add them up!

• Probability of more than three wins– P(X>3) = P(X=4) + P(X=5)

• Probability of at most three wins – P(X≤3) = P(X=0) + P(X=1) + P(X=2) + P(X=3)

• Probability of fewer than three wins– P(X<3) = P(X=0) + P(X=1) + P(X=2)

Use the Complement to save time

• Example: • “Probability of at least 3 wins”• P(X≥3) =

P(X=3) + P(X=4) + … + P(X=49) + P(X=50)• This means 48 calculations and sum results.• EASIER: The complement is “fewer than 3”• Take 1 – [ P(X=0) + P(X=1) + P(X=2) ]

TI-84 Computations

• binompdf(n, p, X) = probability of X successes in n trials

• Recompute the table and make sure we get the same results as the by-hand calculations.

• The “pdf” in “binompdf” stands for “probability distribution function”

TI-84 Computations

• binomcdf(n, p, x) = P(X=0) + P(X=1) + … P(X=x) successes in n trials

• binomcdf(n, p, x) does lots of little binompdf() for you for x = 0, x = 1, etc. up to the x you told it, and it adds up the results

• The “cdf” in “binomcdf” stands for “cumulative distribution function”

Try and verify binomcdf(n,p,x)

X successes P(X successes)Using binompdf

P(0 thru X) successesUsing binomcdf

0 times (no sevens or elevens)1 time

2 times

3 times

4 times

5 times (all sevens and elevens)Total (must equal 1.000000 !!)

binomcdf() and complements

• Sevens or elevens, n = 50 trials again• P(no more than 10 successes)– binomcdf(50, 8/36, 10)

• P(fewer than 10 successes)– binomcdf(50, 8/36, 9)

• P(more than 10 successes) – use complement!– 1 minus binomcdf(50, 8/36, 10)

• P(at least 10 successes) – use complement!– 1 minus binomcdf(50, 8/36, 9)

Mean, Variance, and Standard Deviation

• We had formulas and methods for probability distributions in general.

• The special case of the Binomial Probability Distribution has special shortcut formulas– Mean = – Variance = – Standard deviation =

Mean, Variance, and Standard Deviation

• Compute these for the seven-eleven experiment with n = 5 trials– Mean = – Variance = – Standard deviation =

Mean, Variance, and Standard Deviation

• Compute these for the seven-eleven experiment with n = 50 trials– Mean = – Variance = – Standard deviation =

Mean, Variance, and Standard Deviation

• Compute these for the seven-eleven experiment with n = 100 trials– Mean = and Standard deviation =

• “Expected value” – in 100 tosses of two dice, how many seven-elevens are expected?– Remember, the mean of a probability distribution

is also called the “expected value”

Standard Deviation

• What happens to the standard deviation in the seven-eleven experiment as the number of trials, n, increases?

trials Standard deviation

5

50

100

Advanced TI-84 Exercise

• Y1=binompdf(20,8/36,X)

• seq(X,X,0,20) STO> L1

• seq(Y1(X),X,0,20) STO> L2

• STAT PLOT for these two lists, histogram• WINDOW

Xmin=0, Xmax=20, Ymin=-0.1,Ymax=0.6• ZOOM 9:ZoomStat