A Discrete Probability Distribution: The Binomial Distribution

1. A Discrete Probability Distribution:The Binomial Distribution MSIT 3000 Lecture 06

2. Objectives: • Recognize situations in which it is appropriate to use a binomial distribution. • Be able to use the binomial distribution to calculate probabilities. • Be able to calculate mean, variance and standard deviation for a binomially distributed random variable. References: Text: 4.3

3. The Binomial Distribution • Motivating Example • When we considered the example with three coins and x was the number of heads, a question is often raised: • How do you calculate probabilities if you have 100 coin flips instead of just 3? • You recognize this as a “binomial” problem & use the binomial probability distribution.

4. Characteristics of a Binomial Random Variable: • The experiment consist of n identical trials (a.k.a. Bernoulli trials). • There are only two possible outcomes on each trial, S (success) & F (failure). • P(S) = p is constant. P(F) = q = 1-p. • The trials are independent. • The binomial random variable x is the number of successes (S) in n trials.

5. Example: 3 heads in10 coin flips • The number of flips: n=10. • 3 successes: x=3 • Example outcome: SFFSFFFFSF • Probability(SFFSFFFFSF) = pqqpqqqqpq=p3q7= pxqn-x • Questions: • How many sample points are possible that have x successes out of n? • What is the probability of each of these sample points?

6. Answer to Q1: n choose x • The number of ways to organize x successes out of N trials is written: • Recall the definition of the factorial: y! = y*(y-1)*...*1 • Then:

7. Answer to Q2 • Each sample point has the same probability of occurring because each trial is independent, and in a product, the order of the factors does not matter. • E.g. Probability(SFFSFFFFSF) = pqqpqqqqpq=p3q7= pxqn-x

8. Put the pieces together: • There are ways of organizing 3 successes out of 10 trials, and • each sample point has probability: • Therefore the probability of getting 3 heads out of ten tries is:

9. Very Small example: Number of heads out of three coins

10. How many ways can we choose 2 heads? • Count from chart: 3 • Using the formula: • 3*2/2 = 3 • 3!/2!(3-2)! = 3*2*1/2*1(1) = 6/2 = 3

11. And what is the probability of seeing some particular order of exactly two heads & one tail? • .5*.5*(1-.5) = ppq = 0.125 = 1/8

12. Putting the pieces together: • The probability of getting 2 heads out of 3 flips= • P(x=2) = 3(1/8)=3/8 • Note that this is exactly what we got from counting. • What is the probability of getting exactly 5 heads out of 10 flips? • (10!/5!(10-5)!)*[(.5)5(1-.5)5]=252*(0.5)10 0.2461

13. Parameters of binomially distributed random variables: • Mean: •  = np • Variance • 2 = npq • Standard deviation •  = (npq)

14. Cumulative binomial tables: • What is the probability that we got less than 8 heads out of 10 coins? • We can either calculate all the probabilities (for x=0, for x=1, etc…) and add them up, • Or we can look up the answer in a table (found in the back of the text).

15. Conclusion: • Objectives addressed: • Recognize situations in which it is appropriate to use a binomial distribution. • Be able to use the binomial distribution to calculate probabilities. • Be able to calculate mean, variance and standard deviation for a binomially distributed random variable. • Text problems: (4.27d, 4.28b, 4.29f), 4.31, 4.35, (4.36) • Exam 1A: 25