1. Unit 18 Section 18C The Binomial Distribution

2. Example 1: If a coin is tossed 3 times, what is the probability of obtaining exactly 2 heads Solution: We can solve this problem using a tree diagram

3. H=0.5 T=0.5 H=0.5 T=0.5 HHH HTH HTT THH THT TTH TTT HHT There are three branches with exactly 2 heads

4. P(exactly 2 heads) = (0.5)(0.5)(0.5) + (0.5)(0.5)(0.5) + (0.5)(0.5)(0.5) = 3 (0.5)(0.5)(0.5) = 3(0.125) = 0.375 We can calculate this probability using a Binomial Expansion

5. If we examine the term And let h= P(head) = 0.5, t=P(tail)= 0.5, we have Represents the number of branches with exactly 2 heads

6. The Binomial Model used above applies to situations that are equivalent to drawing from a hat with replacement. These are called BERNOULI TRIALS Bernouli Trials have the following characteristics: Each Trial has exactly 2 outcomes, success or failure Each trial is independent The probability of each outcome is the same for each trial of the experiment.

7. The probabilities for various events determined by a Bernouli experiment can be found easily by using the binomial expansion. Where p = probability of success on any trial, q = probability of failure, and p + q = 1 represents the probability of having exactly r successes in n trials if the probability of success is p.

8. Example 2: Determine the probability of having at least 2 girls in a family of 4 children if P(boy)=P(girl) = 0.5 The probability of having at least 2 girls is the sum of the probabilities of having 2, 3, or 4 girls An alternative approach considers the complementary event. The probability of having at least 2 girls is 1 – P(0 or 1 girl).

9. A sequence of Bernouli trials has a binomial distribution. If the random variable X represents the number of successes in n trials, then Here p = the probability of success on any single trial of the experiment and p + q = 1 or q = 1 –p, hence we have We can also express the binomial distribution in compact form, written as X~B(n, p), read as X is distributed binomially with parameters n and p, where n is the number of trials and p = P(success)

10. Example 3: Write a probability function for the experiment of tossing a coin 4 times where the random variable X represents the number of heads. Let t represent the probability of a tail and h the probability of a head. Select h= 0, 1, 2, 3, 4 and calculate

12. So we have, X01234 P(X=x)0.06250.250.3750.250.0625

13. Example 4: If X~B (7, 0.8), find P(X = 5) Solution: X~B (7, 0.8) means that P(X = 5) is the probability of success 5 times in 7 trials where each trial has a 0.8 chance of success and a 0.2 chance of failure. We can use the TI- 83 Binompdf(7, 0.8, 5)

14. Example 5: Bailey has decided that he is not going to study for the final exam in math but will instead guess at every question. If the exam consists of 50 multiple choice questions, each with 5 possible answers, what is the probability that Bailey passes the exam. Let X represent the number of correct answers, therefore we have X~B (50, 0.2), and we want to find P(X ≥25). P(X ≥25) = P(X = 25) + P(X = 26) + …+ P(X = 50) This could be extremely time consuming. By using the complementary event, the calculator can help us. We will find

15. P(X ≥25) = 1 – P(X ≤24) P(X ≥25) = 1 – binomcdf(50,0.2,24) P(X ≥25) = 1 – 0.9999979051 = 0.00000209485 Bailey had better study. P(X ≥25) = 1 – [P(X = 0) + P(X = 1) + … + P(X = 24) Using the TI-83

16. Finally, it can be shown that for the binoimial distribution X ~ B (n, p), we have The expected value of X is µ = E(X) = np The mode of X is the value of x which has the largest probability The variance of X is  2 = Var(X) = npq = np( 1 – p) The standard deviation of X is Sd(X) =

17. HOMEWORK Page 559 # 1- 14