Lesson 10.7

 Shown at the right are the first six rows of Pascal’s triangle.  It contains many different patterns that have been studied for centuries. ◦ The triangle begins with a 1, then there are two 1’s below it. ◦ Each successive row is filled in with numbers formed by adding the two numbers above it. For example, each of the 4’s is the sum of a 1 and a 3 in the previous row. ◦ Every row begins and ends with a 1.

In Lesson 10.6, you studied numbers of combinations. These numbers occur in the rows of Pascal’s triangle. For example, the numbers 1, 5, 10, 10, 5, and 1 in the sixth row are the values of 5 C r :

A group of five students regularly eats lunch together, but each day only three of them can show up.  Step 1 How many groups of three students could there be? Express your answer in the form n C r and as a numeral.  Step 2 If Leora is definitely at the table, how many other students are at the table? How many students are there to choose from? Find the number of combinations of students possible in this instance. Express your answer in the form n C r and as a numeral.

 Step 3 How many three-student combinations are there that don’t include Leora? Consider how many students there are to select from and how many are to be chosen. Express your answer in the form n C r and as a numeral.  Step 4 Repeat Steps 1–3 for groups of four of the five students.  Step 5 What patterns do you notice in your answers to Steps 1–3 for groups of three students and four students? Write a general rule that expresses n C r as a sum of other combination numbers.  Step 6 How does this rule relate to Pascal’s triangle?

 Pascal’s triangle can provide a shortcut for expanding binomials. For example, the expansion of (x +y) 3 is 1x 3 +3x 2 y+3xy 2 +1y 3. Note that the coefficients of this expansion are the numbers in the fourth row of Pascal’s triangle.

 Expand (H +T) 3. Relate the coefficients in the expansion to combinations.

 Suppose that a hatching yellow-bellied sapsucker has a 0.58 probability of surviving to adulthood. Assume the chance of survival for each egg in a nest is independent of the outcomes for the other eggs. Given a nest of six eggs, what is the probability that exactly four birds will survive to adulthood? You are given the probability of survival, P(S) 0.58, so you know the probability of nonsurvival, P(N) = = If you represent these values as s and n respectively, then the probability that the first four birds survive and the last two do not is s 4 n 2. But it may not be the first four that survive. The 4 surviving birds can be chosen from the 6 birds in 6 C 4 ways, represented by SSNSNS, SNNSSS, and so on. So the probability that exactly 4 birds survive is

Can you extend this method to find the probabilities of survival for different numbers of birds? In Example B, s + n is equal to 1, as is (s + n) 6. But if you expand the latter expression, you will find the probabilities of survival for 0, 1, 2, 3, 4, 5, and 6 birds.

You could show these probabilities with a histogram, as shown at right.

Suppose you want to know the probability that at least 4 birds survive or that at most 3 birds survive. The table and histogram can be extended to calculate probabilities such as these. How could you calculate the values in the table below? In Exercise 3 you will calculate the missing table values.

You can think of the 6 birds as a sample taken from a population of many birds. You know the probability of success in the population, and you want to calculate various probabilities of success in the sample. You saw in Example B that a survival rate of 4 birds was most likely. What if the situation is reversed? Suppose you don’t know what the probability of success in the population is, but you do know that on average 4 out of 6 birds survive to adulthood. This ratio would indicate a population success probability of 4/6, or 0.67; but you saw that this ratio is also highly likely with a population success probability of So more than one population success probability yields the highest probability that 4 birds will survive in a sample of 6.

 A random sample of 32 voters is taken from a large population. In the sample, 20 voters favor passing a proposal. Is it likely that the proposal will pass? You need to find out how likely it is that more than 50% of the voting population supports the proposal. Suppose f represents the number of survey respondents who favor the proposal. If exactly 50% of voters actually favor the proposal, then with a sample of 32 voters (n = 32), P( f=20) is 32 C 20 (0.5) 20( 0.5) 12 or So it’s not very likely that exactly 20 of the 32 survey respondents would favor the proposal if half of the voters overall did not. And the probability that f =20 would be even smaller if less than 50% of the voters actually favored the proposal.

 But the probability of any single event, like exactly 20 respondents in favor of the proposal, is small. So instead, consider the probability of this particular event or a more extreme one. Thus, you look at the probability P( f =20)  or the sum of all the probabilities that from 20 to 32 voters support the proposal. When x 0.5, this probability is equal to

 If the actual percentage of supporters were 50% or less, then there would be only a 10.8% or less probability that you would have had this particular survey result (or a more extreme one). Because this value is small, the actual proportion supporting the proposal is most likely to be greater than 50%. Thus, the proposal is likely to pass.