# Chapter 17: The binomial model of probability Part 2

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Chapter 17: The binomial model of probability Part 2
AP Statistics

The binomial model: jumping in with both feet
There’s a close connection between tree diagrams, combinations, expanding binomial expressions, and the binomial model I’ve debated about giving you a lot of background on it to show you the connections FIRST Instead, I think it will be more effective to show you how to solve the problems, and then show you the connections, which then may be more meaningful.

The binomial model: formulae
So here they are, the three formulae you need to recognize and use. Let’s do the easiest two first, mean and standard deviation Check out the Math Box on p. 392 if you want to see the derivations Mean: Standard deviation:

The binomial model: probability of exactly k successes in n trials
This is the key equation to recognize (memorizing it is hard, at least at first) Remember that p=probability of success, q=probability of failure, and the binomial coefficient is the number of combinations that satisfy k successes in n trials Here it is……wait for it…….

The binomial model: how to dissect the binomial formula
It’s the number of combinations you can get from taking k samples from a population of n That, in turn, is equal to These are the calculations we’ll be working on today

The binomial model/Exercise 13(d): Starting with Exercise 13(d)
We were not able to do Exercises 13(d)-(f) or 14(d)-(f) because we needed to apply the binomial model. So let’s start with Exercise 13(d), which asks for what percentage of the populations has exactly 3 lefties. Before going on to the next slide, work for 2-3 minutes and see how many different combinations you can make using three lefties and two righties.

The binomial model/Exercise 13(d): combinations of 3 lefties, 2 righties
Here’s the distribution on the right. It took me about 5 minutes to derive and check it; YMMV. But even if you do it faster than I did, what about 3 out 10? Or 3 out of 15? Or even 3 out of 8? It will take way too long. We need a better (faster) method. L L L R R L L R L R L L R R L L R L L R L R L R L L R R L L R L L L R R L R L L R L L R L R R L L L

The binomial model/Exercise 13(d): Using the binomial coefficient
The binomial coefficient will save us from the tedium of finding such distributions For n=5 and k = 3 (the conditions of our problem), we can calculate the number of combinations directly:

The binomial model/Exercise 13(d): Applying the binomial coefficient to the probabilities
We have the binomial coefficient, which is 10 for this problem. Now, how to apply the q and p probabilities? The best thing to do is to create a table. We’re going to spend some time doing this, and it should help to make everything clear.

The binomial model/Exercise 13(d): Making a probability table
Because the calculations are complicated, we’re going to make a couple of tables to help make things clear. Here’s the form of the tables to copy for this part of the exercise. Make two of them:

The binomial model/Exercise 13(d): Filling in the 1st probability table with formulae
Note that the table k=0 through k=5. We’ll discuss why in a bit as we fill it in. Now, let’s fill in the formulas before you do the calculations. You’ll do the calculations in the 2nd table. Fill in the blanks as indicated:

The binomial model/Exercise 13(d): What the 1st table should look like when you’re done

The binomial model/Exercise 13(d): Starting the 2nd table
Your second table will contain the calculations using the formulas in the 1st table. As you start, it should look like this:

The binomial model/Exercise 13(d): Completing the 2nd table
Using the formulas in Table 1, do the calculations. Be sure to fill in the calculations in the second column from the right that I’ve grayed out. By the end, you should have a table of all the values possible where there are 0 lefties, 1 lefty, …., 4 lefties, and 5 lefties. It will probably take about 10 minutes to complete this. Be sure to complete the totals for the columns headed by (5 n) and by (5 k)q5-npn .

The binomial model/Exercise 13(d): What the second table should like and its meaning

The binomial model/Exercise 13(d): Important points
(5 k)q5-kpk is the probability that you will get k events in 5 trials. The column headed by (5 k)q5-kpk adds up to 1, which tells us that this is the TOTAL probability model. Note that we start with k=0….we DO have the probability that we will get zero lefties. If you have k trials, you will have k+1 entries in your table.

The binomial model/Exercise 13 (e): Using the table to determine summed probabilities
13(e): at least 3 lefties in the group. That means there could be 3….or 4…..or 5. Find the total by adding the areas marked:

The binomial model/Exercise 13 (f): Using the table to determine summed probabilities (2)
13(d): no more than 3 lefties in the group. That means there could be 3, 2, 1, or 0 Find the total by adding the areas marked:

The binomial model: Answers to Exercises 13(e) and 13(f)
Exercise 13(e): 0.982 Does this answer make sense? Exercise 13(f): Does THIS answer make sense?

The binomial model/Exercise 14: Preparing the chart
Same as 13, except we now have 6 arrows, not 5 people Do the tables again, starting with a blank table like this: Go to the next slide when done.

The binomial model/Exercise 14: The complete chart
Here’s what you should have come up with. The answers are similar to what you did for 13. Use your chart to calculate them. Answers on next slide.

The binomial model: Review of how to calculate combinations
Remember the formula: k is the number of events you’re looking for in the population n Be sure you put the k with the p, which, as you’ve seen, may change from problem to problem. Don’t forget to calculate all three parts of the equation and do the multiplication

The binomial model: Practice: Exercise 19
Similar to what we’ve done already Tennis player has successful first serve 70% of the time. Assuming independence and 6 serves (n=6), calculate the following probabilities: All 6 serves go in Exactly 4 serves go in At least 4 serves go in No more than 4 serves go in Spend 5-10 minutes calculating (a) through (c), then advance to the next slide for the analysis.

The binomial model: Practice: Exercise 19(a)
All six serves go in. n=6, k=number of good serves=6, p=0.7, so q=0.3 Calculate combinations first: Next, pkqn-k=(0.7)6(0.3)6-6=(0.7)6(0.3)0=(0.7)6=0.118

The binomial model: Exercise 19(b)
Exactly 4 serves go in. Similar to (a), but with different numbers: n still 6, but k=4 Binomial coefficient: Next, pkqn-k=(0.7)4(0.3)6-4 = (0.7)4(0.3)2 =(0.24)(0.09)=0.0216 Multiply 15 times to get 0.324

The binomial model: Exercise 19(c)
At least 4 serves in means she got 4, 5 or 6 serves in. If we know probabilities, we can add them. Already know 6 from (a) and 4 from (b) Calculate probabilities of 5: Calculate pkqn-k (continued on next slide)

The binomial model: Putting all of 19(c) together
Remember that the total probability of at least 4 serves going in is P(4)+P(5)+P(6) From previous slides, we have P(4)= 0.324 P(5)=0.301=6× (from slide 27) P(6)=0.118 TOT: Whenever you have multiple outcomes, calculate and add them together!!

The binomial model: Homework
Chapter 17, Exercises 15, 16, and 20.