Part 10: The Binomial Distribution

Transcription

1 Part 10: The Binomial Distribution The binomial distribution is an important example of a probability distribution for a discrete random variable. It has wide ranging applications. One readily available application is quality control in manufacturing production and supply. However, the development of its formula can be a little bit hard to understand. Therefore, at each step of the general development below, we will illustrate with the corresponding steps in a concrete example. Hopefully this makes the general theory more understandable. Suppose we have a probability experiment with only two outcomes. One of these we ll call success and assign it probability p. The other we will call failure, which therefore has probability q = 1 p. For example: Suppose the experiment consists of tossing a fair die. Let s say we win if we roll a 5 or a 6 and lose otherwise. So, in this case p = 1/3 and q = 2/3. Now we form a new experiment that consists of performing the first experiment n times. Each repetition is called a trial (or a Bernoulli trial) and we require that the trials be independent. For example: Let s toss the die 10 times. (Certainly the die outcomes are independent events.) What is the probability of attaining success on the first r trials followed by failures on the remaining n r trials? Clearly, the number of successes plus the number of failures equals the number of n trials: r + (n-r) = n. For example: What is the probability of winning on the first 3 tosses of the die and losing on the remaining 7? That is, what s the value of the stated probability as follows: P(win and win and win and lose and lose and lose and lose and lose and lose and lose)? Since these events are independent, we can easily use the Multiplication Rule for independent events to deduce that this is (1/3) 3 (2/3) 7. Following the thinking of our example, we can now see that in the general case, the probability of r successes followed by n r failures is p r q (n r). Now, would this computation be any different if the r successes occurred in trials other than simply the first ones? That is, what if they were in the middle or the end or sprinkled throughout the n trials? For example: P(lose and win and lose and lose and lose and win and lose and win and lose and lose) For example: What is the probability of 3 wins and 7 losses in some other particular order? Perhaps by writing down some random looking arrangement of 3 wins and 7 losses, you can see that the probability of this sequence is still calculated using the Multiplication Rule and by the end we will again have multiplied by (1/3) 3 times and (2/3) 7 times. Since multiplication is commutative, of course, the result is again (1/3) 3 (2/3) 7. Again following the thinking of our example, we see that the probability of r successes and n r failures in some particular order is still p r q (n r) no matter what that order is. Now, as a final step, what is the probability of r successes and n r failures? Here, we re not asking for the probability that the r successes occur in any particular order, but the probability that they occur at all. So, this event includes all possible positions of the r successes among the n trials. For example: What is the probability of 3 wins amongst our 10 trials? The 3 wins could occur at the beginning or in another 3 positions or in yet another 3 positions or... etc. All those or occurrences tell us we need to use the Addition Rule. And since two different placements of the 3 wins certainly 1

2 can t simultaneously occur, we don t need to worry about subtracting anything off. In other words, in P(A or B) = P(A) + P(B) P(A and B), the P(A and B) = 0. Instead, we can simply add them all up. Even better, we know that each such positioning of the 3 wins has a probability of (1/3) 3 (2/3) 7 and so we simply need to count the number of ways the 3 wins can be placed among the 10 trials and multiply (1/3) 3 (2/3) 7 by that number. A few moments thought should reveal that there are exactly C(10,3) or 10C3 ways to place the 3 wins among the 10 trials. Thus, the probability of 3 wins (in any order whatsoever) is C(10,3)(1/3) 3 (2/3) 7. That is right, we can think of the 3 wins as placing 3 people in 10 chairs where the order does not matter. We are talking about a combination of 3 wins and 7 losses, i.e., 10C3. We have finally arrived at our general result: Suppose we have n (Bernoulli) trials and let X be the random variable giving the number of successes in the n trials. Then, the probability of r successes, denoted P(X = r), is given by C(n,r)p r q (n r). This probability distribution is known as the Binomial Distribution. Please note that the terms success and failure are traditional but completely arbitrary; one s success can be another s failure, one can succeed in falling down. Also, you shouldn t expect a homework or exam problem to say By the way, this variable has a binomial distribution. Rather, you will need to recognize them for yourself, and so you should look for the sort of situation described above: repeated and identical steps, each one having only two possibilities, and a question involving the probability of a certain number of those steps having one of the two outcomes. Example 1: Recall our example for the random variable being the number of boys in 3 births. Let us think of having a boy as a success and not(a boy) as a failure. Then, we have p=.5 and 1-p=q =.5. Clearly, this satisfies the criteria for a binomial distribution and we can compute the probability of r successes out of n trials. Let us confirm our previous results from Part 9 Example 8 using the binomial distribution for the random variable X= C(3,r).5 r.5 (3 r) for r = 0,1,2,3. X boys Prob 2

3 Example 2: Suppose a jar contains 3 red marbles and 5 blue marbles. A trial consists of randomly choosing one marble and then replacing it. (a) Find the probability distribution for X, the number of red marbles selected, if 2 trials are performed. Use the Binomial Distribution. (b) Repeat part (a) for 3 trials, and find P(X 2). (c) Find P(X 2) if 10 trials are performed. Let us draw a tree diagram to illustrate the scenario in part b) For a binomial variable X, some fairly unpleasant calculations reveal that the expected value or the mean is µ = np and the standard deviation is σ = npq. The first of these shouldn t be surprising. For example, it tells us that when tossing a coin 10 times, we should expect 5 heads. Be sure to make use of these two formulas when calculating µ and σ directly since using the formulas given in Part 9 would be quite time consuming for the binomial variables encountered in the homework and exams. On the other hand, don t forget that these formulas only apply to binomial variables. For any other sort of variable we will need to use the formulas given in Part 9 to find µ and σ (that is, except for some variables we have yet to learn about). You can find most of the formulas on your formula cards or handout. Write them in if they are not on your card or handout. 3

4 Example 3: On an exam consisting of 25 multiple choice questions, each having 4 possible answers, Mr. Rong decides to guess on every question. What is the probability that Mr. Rong gets exactly 7 correct answers? What is the probability that that Mr. Rong gets at the least 3 correct answers? How many questions should Mr. X expect to answer correctly? What is the standard deviation for his number of correct answers? Example 4: Where do teenagers get their spending money? A recent survey of households in California found that 40% of teenagers receive their spending money from part-time jobs. In a group of 5 teenagers going to the movies a) what is the probability that all 5 of them have part-time jobs to pay for the movie? b) what is the probability that at least 4 of them will have part-time jobs? 4

5 Example 5: A survey of Bay Area households showed that 20% of those with televisions watch NFL football at some point during any given week during the regular season. Advertisers are making decision on where to spend their dollars online advertising, television, or radio. They conduct their own preliminary survey by contacting 15 homes in the Bay Area with televisions. What is the probability that None of the 15 homes are watching football that week? At least one home is watching football that week? At most one home is watching football that week? If at most one home is watching NFL football that week, is this unusual?? 5