### Empirical Probability

A **probability simulation** is an experiment designed to determine a probability from many trials. By looking at the number of favorable outcomes and dividing by the number of trials, we can get an estimate of the true probability. The more trials we can do, the better our estimate of the true probability will be, but given that we can’t do infinitely many trials, the result we get will *always* be just an estimate of the true probability. Often we perform probability simulations because we can’t determine theoretical probability from looking at the sample space.

#### Real-World Application: TV Repair

A cable TV company sends out a repair technician to replace faulty receiver boxes. The company has boxes made by **Panasonic** and **Scientific Atlanta**, both in equal quantities. The technician carries 3 of each type of box in his van, and always replaces a box with one of the same brand. If the technician visits 4 homes before returning back to the depot, determine the probability that he will not have enough of one box type to make all the needed replacements.

This is a situation that we can model far more easily than we could conduct a real-life experiment. Since there are equal numbers of both boxes, we need to set up a model with a probability of \begin{align*}\frac{1}{2}\end{align*} for each element, like a coin toss. Visiting four houses where each house has an equal chance of needing one type of box or the other is like flipping a coin four times. Getting four heads or four tails is like needing four of one type of box, which is the only situation where the technician would not have enough of one type.

So let’s suppose we flip four coins 50 times and record the results, and they look like this:

Out of 50 trials, it’s as if the technician required four of the same type of box 6 times. So we can say that the probability that the technician will run out of one box type is approximately \begin{align*}\frac{3}{25}\end{align*} or **0.12**.

#### Solving Using Theoretical Probability

The problem in the first example can also be solved by finding the theoretical probability. Let's see how to do that.

Notice that instead of actually flipping the coins a lot of times, we could have used our earlier knowledge of theoretical probability and the sample space for four coin flips:

Here, we can see that we would expect the technician to run out of one type of box about 2 times out of 16, so the probability is about \begin{align*}\frac{1}{8}\end{align*} or **0.125**. But if we hadn’t known for sure what the odds of getting heads or tails on each flip were, we wouldn’t have been able to calculate the odds of getting four heads or tails this way, so we would have had to find out by experiment instead.

We also can check probability calculations like this against actual experimental data to see for ourselves whether something happens as often as we’d expect it to. If it doesn’t, something might be going on that we need to investigate.

**Finding the Experimental Probability of an Event**

In the previous examples, we saw how theoretical probability can be approximated by doing an experiment. We used a results table to approximate the probability of a certain **event** occurring (the technician running out of either type of box). We can approximate the probability of an event by using:

\begin{align*}P(E) \approx \frac{\text{number of matching events}}{\text{total number of trials}}\end{align*}

Randomness in the results will mean that we always get an approximation of the true probability, but the more trials we do, the more accurately our **experimental probability** will match the theoretical probability.

Nadia and Peter are playing dice, but Peter keeps winning and Nadia suspects him of cheating. She is suspicious about the number of times Peter rolls a six, and so she conducts the following experiment: She rolls the suspect die 100 times, writing down the result each time. The results are:

\begin{align*}& 4, 1, 4, 5, 3, 6, 2, 5, 1, 6, 2, 6, 4, 5, 1, 6, 4, 3, 6, 3, 2, 1, 1, 3, 4,\\ & 5, 5, 2, 3, 1, 1, 2, 3, 1, 2, 2, 1, 6, 6, 3, 4, 6, 3, 6, 6, 2, 2, 3, 4, 6,\\ & 1, 6, 6, 2, 6, 4, 3, 3, 2, 5, 3, 3, 2, 6, 6, 6, 6, 6, 1, 4, 1, 2, 6, 6, 6,\\ & 3, 6, 4, 5, 6, 3, 5, 4, 6, 6, 4, 6, 6, 6, 6, 6, 2, 6, 6, 1, 1, 5, 1, 4, 6.\end{align*}

Organize the data in a table and determine if 6 is more likely to come up then the other numbers.

Here’s what we get if we tally up all the results in a table:

Number |
Tally |
Total |
P(number) |
---|---|---|---|

1 | \begin{align*}\bcancel{||||} \ \bcancel{||||} \ \bcancel{||||}\end{align*} | 15 | \begin{align*}P(1)= \frac{15}{100} \approx 0.15\end{align*} |

2 | \begin{align*}\bcancel{||||} \ \bcancel{||||} \ ||||\end{align*} | 14 | \begin{align*}P(2)= \frac{14}{100} \approx 0.14\end{align*} |

3 | \begin{align*}\bcancel{||||} \ \bcancel{||||} \ \bcancel{||||}\end{align*} | 15 | \begin{align*}P(3)= \frac{15}{100} \approx 0.15\end{align*} |

4 | \begin{align*}\bcancel{||||} \ \bcancel{||||} \ |||\end{align*} | 13 | \begin{align*}P(4)= \frac{13}{100} \approx 0.13\end{align*} |

5 | \begin{align*}\bcancel{||||} \ ||||\end{align*} | 9 | \begin{align*}P(5)= \frac{9}{100} \approx 0.09\end{align*} |

6 | \begin{align*}\bcancel{||||} \ \bcancel{||||} \ \bcancel{||||} \ \bcancel{||||} \ \bcancel{||||} \ \bcancel{||||} \ ||||\end{align*} | 34 | \begin{align*}P(6)= \frac{34}{100} \approx 0.34\end{align*} |

It’s clear looking at the table that something strange is going on with the die in question – 6 occurs approximately twice as often as the other numbers, so we could *reasonably* assume that the die is weighted unfairly. However, we still can’t be 100% certain that the results we are seeing are not just due to chance. We must therefore talk only in terms of **likelihood**, and not **certainty**.

### Example

#### Example 1

Juan suspects that his lucky coin is actually weighted so that he gets heads more often than tails in a coin toss. The reason he is suspicious is because he seems to get several heads in a row when he tosses it. He conducts a probability simulation and gets the following results:

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

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

What is the experimental probability of getting heads with Juan's coin in this case?

First, find the number of total tosses, the number of heads, and the number of tails:

40 tosses

23 heads

17 tails

This means that the experimental probability of getting heads with Juan's coin is:

\begin{align*}P(E) \approx \frac{\text{number of matching events}}{\text{total number of trials}}=\frac{26}{40}= 0.65\end{align*}

For the coin to be fair, we would expect heads around 50% of the time. 65% is higher than 50%, but this difference may be due to random chance. Further experimentation would help to get a more accurate answer.

### Review

- Peter and Andrew each visit the hardware store in the high street every week. The store is open 6 days a week (it is closed on Sundays) and Peter and Andrew visit the store on random days when it is open.
- Use a pair of dice to simulate what day Andrew and Peter each visit the store, and determine experimentally the probability that they both visit the store on the same day.
- What would you expect the theoretical probability to be?

- Find experimentally both the probability and odds for the next car passing a stoplight being red if the previous 25 car colors were: red, blue, white, blue, silver, red, black, black, white, red, green, red, black, blue, white, red, silver, white, red, black, white, blue, silver, red, black.

For 3-13, determine whether you could calculate the theoretical probability of the given event based on your knowledge of the possible outcomes, or whether you would have to do a test (or get more real-world information some other way) to find the experimental probability:

- Flipping a coin three times in a row and getting three heads.
- Pulling a nickel from your pocket when you know you have three nickels and five dimes in your pocket.
- Pulling a nickel from your pocket when you know you have ten coins in your pocket but can’t remember what they are.
- Guessing the right answer on a multiple-choice question.
- Guessing the right answer on a free-response question.
- Getting a perfect score on a twenty-question multiple-choice test.
- Getting a perfect score on a test that has ten multiple-choice questions and ten free-response questions.
- Guessing a randomly chosen high school student’s age correctly.
- Sharing a birthday with one of your three best friends.
- Getting a flat tire while driving home.
- Having your left front tire be the one that goes flat, whenever you
*do*get your next flat tire.

### Review (Answers)

To view the Review answers, open this PDF file and look for section 13.2.