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Chapter 17:The binomial model of probability

Part 2

AP Statistics

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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.

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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:

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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…….

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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

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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.

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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.

1. L L L R R2. L L R L R3. L L R R L4. L R L L R5. L R L R L6. L R R L L7. R L L L R8. R L R L L9. R L L R L10. R R L L L

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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:

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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.

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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:

5-k k (5 k) q5-n pn (5 k)q5-npn Lefties0 01 12 23 34 45 5

Totals:

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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:

5-k k (5 k) q5-n pn (5 k)q5-npn Lefties5-0=5 0 5!/5!0! 0.87^5 0.13^0 05-1=4 1 5!/1!4! 0.87^4 0.13^1 1

2 5!/2!3! 23 34 45 5

Totals:

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

5-k k (5 k) q5-n pn (5 k)q5-npn Lefties5-0=5 0 5!/5!0! 0.87^5 0.13^0 05-1=4 1 5!/1!4! 0.87^4 0.13^1 15-2=3 2 5!/2!3! 0.87^3 0.13^2 25-3=2 3 5!/3!2! 0.87^2 0.13^3 35-4=1 4 5!/4!1! 0.87^1 0.13^4 45-5=0 5 5!/5!0! 0.87^0 0.13^5 5

Totals:

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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:

5-k k (5 k) q5-k pk (5 k)q5-kpk Lefties0 01 12 23 34 45 5

Totals:

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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 .

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

5-k k (5 k) q5-k pk (5 k)q5-kpk Lefties5 0 1 0.4984209 1.0000000 0.4984209 04 1 5 0.5728976 0.1300000 0.3723834 13 2 10 0.6585030 0.0169000 0.1112870 22 3 10 0.7569000 0.0021970 0.0166291 31 4 5 0.8700000 0.0002856 0.0012424 40 5 1 1.0000000 3.71E-05 3.713E-05 5

Totals: 32 1

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The binomial model/Exercise 13(d):Important points

1. (5 k)q5-kpk is the probability that you will get k events in 5 trials.

2. The column headed by (5 k)q5-kpk adds up to 1, which tells us that this is the TOTAL probability model.

3. Note that we start with k=0….we DO have the probability that we will get zero lefties.

4. If you have k trials, you will have k+1 entries in your table.

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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:

5-k k (5 k) q5-k pk (5 k)q5-kpk Lefties5 0 1 0.4984209 1.0000000 0.4984209 04 1 5 0.5728976 0.1300000 0.3723834 13 2 10 0.6585030 0.0169000 0.1112870 22 3 10 0.7569000 0.0021970 0.0166291 31 4 5 0.8700000 0.0002856 0.0012424 40 5 1 1.0000000 3.71E-05 3.713E-05 5

Totals: 32 1

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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:

5-k k (5 k) q5-k pk (5 k)q5-kpk Lefties5 0 1 0.4984209 1.0000000 0.4984209 04 1 5 0.5728976 0.1300000 0.3723834 13 2 10 0.6585030 0.0169000 0.1112870 22 3 10 0.7569000 0.0021970 0.0166291 31 4 5 0.8700000 0.0002856 0.0012424 40 5 1 1.0000000 3.71E-05 3.713E-05 5

Totals: 32 1

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The binomial model:Answers to Exercises 13(e) and 13(f)

• Exercise 13(e): 0.982–Does this answer make sense?

• Exercise 13(f): 0.0179–Does THIS answer make sense? 5-k k (5 k) q5-k pk (5 k)q5-kpk Lefties5 0 1 0.4984209 1.0000000 0.4984209 04 1 5 0.5728976 0.1300000 0.3723834 13 2 10 0.6585030 0.0169000 0.1112870 22 3 10 0.7569000 0.0021970 0.0166291 31 4 5 0.8700000 0.0002856 0.0012424 40 5 1 1.0000000 3.71E-05 3.713E-05 5

Totals: 32 1

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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.

6-k k (6 k) q6-k pk (6 n)q6-kpk bull's-eyes1 0 02 1 13 2 24 3 35 4 46 5 57 6 6

TOTAL:

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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.

6-k k (6 k) q6-k pk (6 n)q6-kpk bull's-eyes1 0 1 0.000064 1 0.000064 02 1 6 0.00032 0.8 0.001536 13 2 15 0.0016 0.64 0.015360 24 3 20 0.008 0.512 0.081920 35 4 15 0.04 0.4096 0.245760 46 5 6 0.2 0.32768 0.393216 57 6 1 1 0.262144 0.262144 6

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The binomial model/Exercise 14:Answers

d) 0.246 (yellow shading)e) 0.0179 (adding numbers circled in redf) 0.345 (adding numbers circled in green)

6-k k (6 k) q6-k pk (6 n)q6-kpk bull's-eyes6 0 1 0.000064 1 0.000064 05 1 6 0.00032 0.8 0.001536 14 2 15 0.0016 0.64 0.015360 23 3 20 0.008 0.512 0.081920 32 4 15 0.04 0.4096 0.245760 41 5 6 0.2 0.32768 0.393216 50 6 1 1 0.262144 0.262144 6

64 1

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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

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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:a. All 6 serves go inb. Exactly 4 serves go inc. At least 4 serves go ind. No more than 4 serves go in

• Spend 5-10 minutes calculating (a) through (c), then advance to the next slide for the analysis.

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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

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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 0.0216 to get 0.324

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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)

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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=6×0.0504 (from slide 27)– P(6)=0.118– TOT: 0.743

• Whenever you have multiple outcomes, calculate and add them together!!

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The binomial model:Homework

• Chapter 17, Exercises 15, 16, 17 18 and 20.