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# 4.2: Binomial Distributions

Difficulty Level: At Grade Created by: CK-12

In the last chapter, you found that binomial experiments are ones that involve only 2 choices. Each observation from the experiment, therefore, falls into the category of a success or a failure. For example, if you tossed a coin to see if a 6 appears, it would be a binomial experiment. A successful event is the 6 appearing. Every other roll (1, 2, 3, 4, or 5) would be a failure. Asking your classmates if they watched American Idol last evening is also an example of a binomial experiment. There are only 2 choices (yes and no), and you can deem a yes answer to be a success and a no answer to be a failure. Coin tossing is another example of a binomial experiment, as you saw in Chapter 3. There are only 2 possible outcomes (heads and tails), and you can say that heads are successes if you are looking to count how many heads can be obtained. Tails would then be failures. You should also note that the observations are independent of each other. In other words, whether or not Alisha watched American Idol does not affect whether or not Jack watched the show. In fact, knowing that Alisha watched American Idol does not tell you anything about any of your other classmates. Notice, then, that the probability for success for each trial is the same.

The distribution of the observations in a binomial experiment is known as a binomial distribution. For binomial experiments, there is also a fixed number of trials. As the number of trials increases, the binomial distribution becomes closer to a normal distribution. You should also remember that for normal distributions, the random variable is continuous, and for binomial distributions, the random variable is discrete. This is because binomial experiments have 2 outcomes (successes and failures), and the counts of both are discrete.

Of course, as the sample size increases, the accuracy of a binomial distribution also increases. Let’s look at an example.

Example 4

Keith took a poll of the students in his school to see if they agreed with the new “no cell phones” policy. He found the following results:

Age Number Responding No
14 56
15 65
16 90
17 95
18 60

Is his data normally distributed?

Solution:

Keith plotted the data and found that it was distributed as follows:

Clearly, Keith’s data is not normally distributed.

Example 5

Jake sent out the same survey as Keith, except he sent it to every junior high school and high school in the district. These schools had students from grade 6 (age 12) to grade 12 (up to age 20). He found the following data:

Age Number Responding No
12 274
13 261
14 259
15 289
16 233
17 225
18 253
19 245
20 216

Is his data normally distributed? What can Jake conclude from his data?

Solution:

Jake plotted his data and found that the data was distributed as follows:

Because Jake expanded the survey to include more people (and a wider age range), his data turned out to fit more closely to a normal distribution.

Notice that in Jake’s sample, the number of people responding no on the survey was 2,225. This number has a huge advantage over the number of students who responded no on Keith's survey, which was 366. You can make much more accurate conclusions with Jake's data. For example, Jake could say that the biggest opposition to the new “no cell phones” policy came from the middle or junior high schools, where students were 12–15 years old. This is not the same conclusion Keith would have made based on his small sample.

In Chapter 3, you did a little work with the formula used to calculate probability for binomial experiments. Here is the general formula for finding the probability of a binomial experiment from Chapter 3.

The probability of getting $X$ successes in $n$ trials is given by:

$P(X = a) = {_n}C_a \times p^a \times q^{(n - a)}$

where:

$a$ is the number of successes from the trials.

$p$ is the probability of success.

$q$ is the probability of failure.

Let’s look at a simple example of a binomial probability distribution just to recap what you learned in Chapter 3.

Example 6

A coin is tossed 3 times. Find the probability of getting exactly 2 tails.

Solution:

There are 3 trials, so $n = 3$.

A success is getting a tail. We are interested in exactly 2 successes. Therefore, $a = 2$.

The probability of a success is $\frac{1}{2}$, and, thus, $p = \frac{1}{2}$.

Therefore, the probability of a failure is $1 - \frac{1}{2}$, or $\frac{1}{2}$. From this, you know that $q = \frac{1}{2}$.

$P(X = a) & = {_n}C_a \times p^a \times q^{(n - a)}\\P(2 \ \text{tails}) & = {_3}C_2 \times p^2 \times q^1\\P(2 \ \text{tails}) & = {_3}C_2 \times \left ( \frac{1}{2} \right )^2 \times \left ( \frac{1}{2} \right )^1\\P(2 \ \text{tails}) & = 3 \times \frac{1}{4} \times \frac{1}{2} \\P(2 \ \text{tails}) & = \frac{3}{8}$

Therefore, the probability of seeing exactly 2 tails in 3 tosses is $\frac{3}{8}$, or 37.5%.

Example 7

A local food chain has determined that 40% of the people who shop in the store use an incentive card, such as air miles. If 10 people walk into the store, what is the probability that half of them will be using an incentive card?

Solution:

There are 10 trials, so $n = 10$.

A success is a person using a card. You are interested in 5 successes. Therefore, $a = 5$.

The probability of a success is 40%, or 0.40, and, thus, $p = 0.40$.

Therefore, the probability of a failure is $1 - 0.40$, or 0.60. From this, you know that $q = 0.60$.

$P(X = a) & = {_n}C_a \times p^a \times q^{(n - a)}\\P(5 \ \text{people using a card}) & = {_{10}}C_5 \times p^5 \times q^5\\P(5 \ \text{people using a card}) & = {_{10}}C_5 \times (0.40)^5 \times (0.60)^5\\P(5 \ \text{people using a card}) & = 252 \times 0.01024 \times 0.07776\\P(5 \ \text{people using a card}) & = 0.201$

Therefore, the probability of seeing 5 people using a card in a random set of 10 people is 20.1%.

Technology Note

You could have used technology to solve this problem, rather than pencil and paper calculations. However, with technology, it is often very helpful to check our answers using pencil and paper as well. With Example 7, you could have used the binompdf function on the TI-84 calculator.

The key sequence for using the binompdf function is as follows:

If you used the data from Example 7, you would find the following:

Notice that you typed in binompdf$(n, p, a)$. What if you changed the problem to include the phrase at most? In reality, probability problems are normally ones that use at least, more than, less than, or at most. You do not always have a probability problem using the word exactly. Take a look at Example 8, a problem similar to Example 7.

Example 8

A local food chain has determined that 40% of the people who shop in the store use an incentive card, such as air miles. If 10 people walk into the store, what is the probability that at most half of these will be using an incentive card?

Solution:

There are 10 trials, so $n = 10$.

A success is using a card. We are interested in at most 5 people using a card. That is, we are interested in 0, 1, 2, 3, 4, or 5 people using a card. Therefore, $a = 5, 4, 3, 2, 1,$ and $0$.

The probability of a success is 40%, or 0.40, and, thus, $p = 0.40$.

Therefore, the probability of a failure is $1 - 0.40$, or 0.60. From this, you know that $q = 0.60$.

$P(X = a) & = {_n}C_a \times p^a \times q^{(n - a)}\\P(5 \ \text{people using a card}) & = {_{10}}C_5 \times p^5 \times q^5\\P(5 \ \text{people using a card}) & = {_{10}}C_5 \times (0.40)^5 \times (0.60)^5\\P(5 \ \text{people using a card}) & = 252 \times 0.01024 \times 0.07776\\P(5 \ \text{people using a card}) & = 0.201$

$P(4 \ \text{people using a card}) & = {_{10}}C_4 \times p^4 \times q^6\\P(4 \ \text{people using a card}) & = {_{10}}C_4 \times (0.40)^4 \times (0.60)^6\\P(4 \ \text{people using a card}) & = 210 \times 0.0256 \times 0.04666\\P(4 \ \text{people using a card}) & = 0.251$

$P(3 \ \text{people using a card}) & = {_{10}}C_3 \times p^3 \times q^7\\P(3 \ \text{people using a card}) & = {_{10}}C_3 \times (0.40)^3 \times (0.60)^7\\P(3 \ \text{people using a card}) & = 120 \times 0.064 \times 0.02799\\P(3 \ \text{people using a card}) & = 0.215$

$P(2 \ \text{people using a card}) & = {_{10}}C_2 \times p^2 \times q^8\\P(2 \ \text{people using a card}) & = {_{10}}C_2 \times (0.40)^2 \times (0.60)^8\\P(2 \ \text{people using a card}) & = 45 \times 0.16 \times 0.01680\\P(2 \ \text{people using a card}) & = 0.121$

$P(1 \ \text{person using a card}) & = {_{10}}C_1 \times p^1 \times q^9\\P(1 \ \text{person using a card}) & = {_{10}}C_1\times (0.40)^1 \times (0.60)^9\\P(1 \ \text{person using a card}) & = 10 \times 0.40 \times 0.01008\\P(1 \ \text{person using a card}) & = 0.0403$

$P(0 \ \text{people using a card}) & = {_{10}}C_0 \times p^0 \times q^{10}\\P(0 \ \text{people using a card}) & = {_{10}}C_0 \times (0.40)^0 \times (0.60)^{10}\\P(0 \ \text{people using a card}) & = 1 \times 1 \times 0.00605\\P(0 \ \text{people using a card}) & = 0.00605$

The total probability for this example is calculated as follows:

$P(X \le 5) & = 0.201 + 0.251 + 0.215 + 0.121 + 0.0403 + 0.0605\\P(X \le 5) & = 0.834$

Therefore, the probability of seeing at most 5 people using a card in a random set of 10 people is 82.8%.

Technology Note

You can see now that the use of the TI-84 calculator can save a great deal of time when solving problems involving the phrases at least, more than, less than, or at most. This is due to the fact that the calculations become much more cumbersome. You could have used the binomcdf function on the TI-84 calculator to solve Example 8. Binomcdf stands for binomial cumulative probability. Binompdf simply stands for binomial probability.

The key sequence for using the binompdf function is as follows:

If you used the data from Example 8, you would find the following:

You can see how using the binomcdf function is a lot easier than actually calculating 6 probabilities and adding them up. If you were to round 0.8337613824 to 3 decimal places, you would get 0.834, which is our calculated value found in Example 8.

Example 9

Karen and Danny want to have 5 children after they get married. What is the probability that they will have exactly 3 girls?

Solution:

There are 5 trials, so $n = 5$.

A success is when a girl is born, and we are interested in 3 girls. Therefore, $a = 3$.

The probability of a success is 50%, or 0.50, and thus, $p = 0.50$.

Therefore, the probability of a failure is $1 - 0.50$, or 0.50. From this, you know that $q = 0.50$.

$P(X = a) & = {_n}C_a \times p^a \times q^{(n - a)}\\P(3 \ \text{girls}) & = {_5}C_3 \times p^3 \times q^2\\P(3 \ \text{girls}) & = {_5}C_3 \times (0.50)^3 \times (0.50)^2\\P(3 \ \text{girls}) & = 10 \times 0.125 \times 0.25\\P(3 \ \text{girls}) & = 0.3125$

Therefore, the probability of having exactly 3 girls from the 5 children is 31.25%.

When using technology, you will select the binompdf function, because you are looking for the probability of exactly 3 girls from the 5 children.

Using the TI-84 calculator gave us the same result as our calculation (and was a great deal quicker).

Example 10

Karen and Danny want to have 5 children after they get married. What is the probability that they will have less than 3 girls?

Solution:

There are 5 trials, so $n = 5$.

A success is when a girl is born, and we are interested in less than 3 girls. Therefore, $a = 2, 1,$ and $0$.

The probability of a success is 50%, or 0.50, and, thus, $p = 0.50$.

Therefore, the probability of a failure is $1 - 0.50$, or 0.50. From this, you know that $q = 0.50$.

$P(X = a) & = {_n}C_a \times p^a \times q^{(n - a)}\\P(2 \ \text{girls}) & = {_5}C_2 \times p^2 \times q^3\\P(2 \ \text{girls}) & = {_5}C_2 \times (0.50)^2 \times (0.50)^3\\P(2 \ \text{girls}) & = 10 \times 0.25 \times 0.125\\P(2 \ \text{girls}) & = 0.3125$

$P(1 \ \text{girl}) & = {_5}C_1 \times p^1 \times q^4\\P(1 \ \text{girl}) & = {_5}C_1 \times (0.50)^1 \times (0.50)^4\\P(1 \ \text{girl}) & = 5 \times 0.50 \times 0.0625\\P(1 \ \text{girl}) & = 0.1563$ $P(0 \ \text{girls}) & = {_5}C_0 \times p^0 \times q^5\\P(0 \ \text{girls}) & = {_5}C_0 \times (0.50)^0 \times (0.50)^5\\P(0 \ \text{girls}) & = 1 \times 1 \times 0.03125\\P(0 \ \text{girls}) & = 0.03125$

The total probability for this example is calculated as follows:

$P(X < 3) & = 0.3125 + 0.1563 + 0.3125\\P(X < 3) & = 0.500$

Therefore, the probability of having less than 3 girls in 5 children is 50.0%.

When using technology, you will select the binomcdf function, because you are looking for the probability of less than 3 girls from the 5 children.

Example 11

A fair coin is tossed 50 times. What is the probability that you will get heads in 30 of these tosses?

Solution:

There are 50 trials, so $n = 50$.

A success is getting a head, and we are interested in exactly 30 heads. Therefore, $a = 30$.

The probability of a success is 50%, or 0.50, and, thus, $p = 0.50$.

Therefore, the probability of a failure is $1 - 0.50$, or 0.50. From this, you know that $q = 0.50$.

$P(X = a) & = {_n}C_a \times p^a \times q^{(n - a)}\\P(30 \ \text{heads}) & = {_{50}}C_{30} \times p^{30} \times q^{20}\\P(30 \ \text{heads}) & = {_{50}}C_{30} \times (0.50)^{30} \times (0.50)^{20}\\P(30 \ \text{heads}) & = (4.713 \times 10^{13}) \times (9.313 \times 10^{-10}) \times (9.537 \times 10^{-7})\\P(30 \ \text{heads}) & = 0.0419$

Therefore, the probability of getting exactly 30 heads from 50 tosses of a fair coin is 4.2%.

Using technology to check, you get the following:

Example 12

A fair coin is tossed 50 times. What is the probability that you will get heads in at most 30 of these tosses?

Solution:

There are 50 trials, so $n = 50$.

A success is getting a head, and we are interested in at most 30 heads. Therefore, $a = 30, 29, 28, 27, 26, 25,$ $24, 23, 22, 21, 20, 19, 18, 17, 16, 15, 14, 13, 12, 11, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1$, and, $0$.

The probability of a success is 50%, or 0.50, and, thus, $p = 0.50$.

Therefore, the probability of a failure is $1 - 0.50$, or 0.50. From this, you know that $q = 0.50$.

Obviously, you will be using technology to solve this problem, as it would take us a long time to calculate all of the individual probabilities. The binomcdf function can be used as follows:

Therefore, the probability of having at most 30 heads from 50 tosses of a fair coin is 94.1%.

Example 13

A fair coin is tossed 50 times. What is the probability that you will get heads in at least 30 of these tosses?

Solution:

There are 50 trials, so $n = 50$.

A success is getting a head, and we are interested in at least 30 heads. Therefore, $a = 50, 49, 48, 47, 46, 45,$ $44, 43, 42, 41, 40, 39, 38, 37, 36, 35, 34, 33, 32, 31,$ and $30$.

The probability of a success is 50%, or 0.50, and, thus, $p = 0.50$.

Therefore, the probability of a failure is $1 - 0.50$, or 0.50. From this, you know that $q = 0.50$.

Again, you will obviously be using technology to solve this problem, as it would take us a long time to calculate all of the individual probabilities. The binomcdf function can be used as follows:

Notice that when you use the phrase at least, you used the numbers 50, 0.5, 29. In other words, you would type in 1 $-$ binomcdf $(n, p, a - 1)$. Since $a = 30$, at least $a$ would be anything greater than 29. Therefore, the probability of having at least 30 heads from 50 tosses of a fair coin is 10.1%.

Example 14

You have a summer job at a jelly bean factory as a quality control clerk. Your job is to ensure that the jelly beans coming through the line are the right size and shape. The probability that any particular shipment of jelly beans passes inspection is 90%. A normal day at the jelly bean factory means 15 shipments are produced. What is the probability that exactly 10 will pass inspection?

Solution:

There are 15 shipments, so $n = 15$.

A success is a shipment passing inspection, and we are interested in exactly 10 passing inspection.

Therefore, $a = 10$.

The probability of a success is 90%, or 0.90, and, thus, $p = 0.90$.

Therefore, the probability of a failure is $1 - 0.90$, or 0.10. From this, you know that $q = 0.10$.

$P(X = a) & = {_n}C_a \times p^a \times q^{(n - a)}\\P(10 \ \text{shipments passing}) & = {_{15}}C_{10} \times p^{10} \times q^5\\P(10 \ \text{shipments passing}) & = {_{15}}C_{10} \times (0.90)^{10} \times (0.10)^5\\P(10 \ \text{shipments passing}) & = 3003 \times 0.3487 \times (1.00 \times 10^{-5})\\P(10 \ \text{shipments passing}) & = 0.0105$

Therefore, the probability that exactly 10 of the 15 shipments will pass inspection is 1.05%.

Feb 23, 2012

Jan 27, 2015