### The Binomcdf Function

The **binomcdf function** on the TI-84 calculator can be used to solve problems involving the probability of less than or equal to a number of successes out of a certain number of trials. For example, if you wanted to know the probability of less than or equal to 45 successes out of 100 trials, you could use the binomcdf 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 binomcdf function is binomcdf\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.

#### Real-World Application: Rewards Cards

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?

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

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, \begin{align*}a = 5, 4, 3, 2, 1,\end{align*} and .

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*}

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

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

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

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

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

The total probability for this example is calculated as follows:

\begin{align*}P(X \le 5) & = 0.201 + 0.251 + 0.215 + 0.121 + 0.0403 + 0.00605\\ P(X \le 5) & = 0.834\end{align*}

Therefore, the probability of seeing at most 5 people using a card in a random set of 10 people is 83.4%.

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 the problem above. Binomcdf stands for binomial cumulative probability.

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

If you used the data from the problem above, 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 the problem above.

#### Calculating Probability

1. Karen and Danny want to have 5 children after they get married. What is the probability that they will have *less than* 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 *less than* 3 girls. Therefore, \begin{align*}a = 2, 1,\end{align*} and .

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(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\end{align*}

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

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

The total probability for this example is calculated as follows:

\begin{align*}P(X < 3) & = 0.3125 + 0.1563 + 0.3125\\ P(X < 3) & = 0.500\end{align*}

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.

2. A fair coin is tossed 50 times. What is the probability that you will get heads in *at most* 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 *at most* 30 heads. Therefore, \begin{align*}a = 30, 29, 28, 27, 26, 25,\end{align*} \begin{align*}24, 23, 22, 21, 20, 19, 18, 17, 16, 15, 14, 13, 12, 11, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1\end{align*}, and, .

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*}.

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

#### Example 1

A fair coin is tossed 50 times. What is the probability that you will get heads in *at least* 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 *at least* 30 heads. Therefore, \begin{align*}a = 50, 49, 48, 47, 46, 45,\end{align*} \begin{align*}44, 43, 42, 41, 40, 39, 38, 37, 36, 35, 34, 33, 32, 31,\end{align*} and \begin{align*}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*}.

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 \begin{align*}-\end{align*} binomcdf \begin{align*}(n, p, a - 1)\end{align*}. Since \begin{align*}a = 30\end{align*}, at least \begin{align*}a\end{align*} 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%.

### Review

- It is determined that because of a particular genetic trend in a family, the probability of having a boy is 60%. Janet and David decide to have 4 children. What is the probability that Janet and David will have at least 2 boys?
- For question 1, what is the probability that Janet and David will have at most 2 boys?
- Suppose the probability of rain for each of the next 4 days is 40%. What is the probability that it rains on 2 or more of the days?
- If a coin is flipped 35 times, what is the probability of getting 15 or fewer heads?
- What is the probability of getting at least 28 tails when flipping a coin 40 times?
- Suppose the probability of getting selected for jury duty in any particular year is 9%. What is the probability of getting selected for jury duty in at most 2 of the next 5 years?
- If you randomly choose a card from a deck of cards 24 times and return the card to the deck each time, what is the probability of getting fewer than 7 clubs?
- Brady is playing bingo, and he calculated the probability of winning any particular game to be 4%. If Brady plays 13 games of bingo, what is the probability that he wins at least twice?
- The probability of getting an even number when rolling a fair die is 50%. If a fair die is rolled 200 times, what is the probability of getting more than 106 even numbers?
- The probability of getting stopped by a red light at a particular intersection is 16%. If you randomly pass through the intersection 22 times, what is the probability that you get stopped by a red light 3 or fewer times?

### Review (Answers)

To view the Review answers, open this PDF file and look for section 4.4.