The mean and variance of a binomial random variable are(),()(1)E Xnp Var Xnpp==-

68 Binomial Example: A die is rolled three times and the random variable Xis defined to be the number of times the face 6 turns up in the three tosses. Find the probability that X=1, that is exactly one 6 appears on three rolls of the dice. We have X~Bin(3,1/6) since the probability of a 6 turning up is 1/6 and the dice rolls are independent. By the binomial formula 123153!25(1)0.351661!(31)!216P X ==== - This result is easily checked usually a binomial calculator such as the one found at The following calculates this probability in Stata: . display binomial(3,1,.1667)-binomial(3,0,.1667) .34726388

69 Binomial Example: A basketball player traditionally makes 85% of her free throws. Suppose she shoots 10 baskets and counts the number she makes. What is the probability that she makes less than 8 baskets. Define the random variable Xto be the number of baskets that she makes. Then we are looking for P(X<8). Note that since the binomial is a discrete random variable that only takes on integer values, (8)(7).P XP X<=≤Using Stata . display binomial(10,7,.85) .17980352 So (7)0.18P X≤=Binomial Example: Traditionally, about 70% of students in a particular economics course at Harvard are successful. Suppose 20 students are selected at random from all previous students in this course. What is the probability that more than 15 of them will have been successful in the course? Let's do a quick overview of the criteria for a binomial experiment to see if this fits. •A fixed number of trials - The students are our trials. •Each trial is independent of the others - Since they're randomly selected, we can assume they are independent of each other. •There are only two outcomes - Each student either was successful or was not successful. •The probability of each outcome remains constant from trial to trial. - Because the students were independent, we can assume this probability is constant. Define the random variable Xto be the number of baskets that she makes. Then we are looking for P(X>15). Note that since the binomial is a discrete random variable that only takes on integer values, (15)1(15).P XP X>=-≤Using Stata . display 1-binomial(20,15,.70) .23750778 So (15)0.24P X>=

70 Binomial Process For a sequence of ntrials, each with an outcome of either success or failure, each with a probability of pto succeed – the probability to get xsuccesses is equal to the Basic Counting Rule formula (above) times px(1-p)n-x. ()p,nxXP=()()xnxpp!xn!x!n---=1If an airline takes 20 reservations, and there is a 0.9 probability that each passenger will show up, then the probability that exactly16 passengers will show is: !!!41620(0.9)16(0.1)4= 0.08978 Binomial End Points All successes or all failures (0)(1)nP Xp==-()nP Xnp==Binomial Expected Value The expected value of a Binomial Process, given ntrials and pprobability.