Suppose you were conducting an experiment where you tossed a fair coin 75 times and recorded your results. What is the probability that tails will come up 25 of those times?

### Watch This

First watch this video to learn about the binompdf function.

CK-12 Foundation: Chapter4BinompdfFunctionA

Then watch this video to see some examples.

CK-12 Foundation: Chapter4BinompdfFunctionB

### Guidance

The **binompdf function** on the TI-84 calculator can be used to solve problems involving the probability of a precise number of successes out of a certain number of trials. For example, if you wanted to know the probability of 45 successes out of 100 trials, you could use the binompdf function. Just like when computing the probability manually with paper and pencil, you also have to know the probability of success for any particular trial in order to find the answer. The syntax for the binompdf function is binompdf\begin{align*}(n, p, a)\end{align*}, where \begin{align*}n\end{align*} is the number of trials, \begin{align*}p\end{align*} is the probability of success for any particular trial, and \begin{align*}a\end{align*} is the number of successes.

#### Example A

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?

There are 10 trials, so \begin{align*}n = 10\end{align*}.

A success is a person using a card. You are interested in 5 successes. Therefore, \begin{align*}a = 5\end{align*}.

The probability of a success is 40%, or 0.40, and, thus, \begin{align*}p = 0.40\end{align*}.

Therefore, the probability of a failure is \begin{align*}1 - 0.40\end{align*}, or 0.60. From this, you know that \begin{align*}q = 0.60\end{align*}.

\begin{align*}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\end{align*}

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

You could have also 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 A, you could have used the binompdf function on the TI-84 calculator. Binompdf simply stands for binomial probability.

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

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

Notice that you typed in binompdf\begin{align*}(n, p, a)\end{align*} to solve the problem.

#### Example B

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

There are 5 trials, so \begin{align*}n = 5\end{align*}.

A success is when a girl is born, and we are interested in 3 girls. Therefore, \begin{align*}a = 3\end{align*}.

The probability of a success is 50%, or 0.50, and thus, \begin{align*}p = 0.50\end{align*}.

Therefore, the probability of a failure is \begin{align*}1 - 0.50\end{align*}, or 0.50. From this, you know that \begin{align*}q = 0.50\end{align*}.

\begin{align*}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\end{align*}

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 C

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

There are 50 trials, so \begin{align*}n = 50\end{align*}.

A success is getting a head, and we are interested in *exactly* 30 heads. Therefore, \begin{align*}a = 30\end{align*}.

The probability of a success is 50%, or 0.50, and, thus, \begin{align*}p = 0.50\end{align*}.

Therefore, the probability of a failure is \begin{align*}1 - 0.50\end{align*}, or 0.50. From this, you know that \begin{align*}q = 0.50\end{align*}.

\begin{align*}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\end{align*}

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:

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### Guided Practice

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?

**Answer:**

There are 15 shipments, so \begin{align*}n = 15\end{align*}.

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

Therefore, \begin{align*}a = 10\end{align*}.

The probability of a success is 90%, or 0.90, and, thus, \begin{align*}p = 0.90\end{align*}.

Therefore, the probability of a failure is \begin{align*}1 - 0.90\end{align*}, or 0.10. From this, you know that \begin{align*}q = 0.10\end{align*}.

\begin{align*}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\end{align*}

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

Using technology to check, you get the following:

### Explore More

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