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# Geometric Probability Distributions

## Calculations of an experiment, with or without technology, conducted until the first defined success occurs

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Geometric Probability Distributions

### Geometric Probability Distributions

#### Defining Geometric Probability Distributions

Like the Poisson and binomial distributions, a geometric probability distribution describes a discrete random variable. Recall that in the binomial experiments, we tossed a coin a fixed number of times and counted the number, X\begin{align*}X\end{align*}, of heads as successes.

A geometric distribution describes a situation in which we toss the coin until the first head (success) appears. We assume, as in the binomial experiments, that the tosses are independent of each other.

#### Characteristics of a Geometric Probability Distribution

• The experiment consists of a sequence of independent trials.
• Each trial results in one of two outcomes: success, S\begin{align*}S\end{align*}, or failure, F\begin{align*}F\end{align*}.
• The geometric random variable X\begin{align*}X\end{align*} is defined as the number of trials until the first S\begin{align*}S\end{align*} is observed.
• The probability p(x)\begin{align*}p(x)\end{align*} is the same for each trial.

Why would we wait until a success is observed? One example is in the world of business. A business owner may want to know the length of time a customer will wait for some type of service. Another example would be an employer who is interviewing potential candidates for a vacant position and wants to know how many interviews he/she has to conduct until the perfect candidate for the job is found. Finally, a police detective might want to know the probability of getting a lead in a crime case after 10 people are questioned.

#### Probability Distribution, Mean, and Variance of a Geometric Random Variable

The probability distribution, mean, and variance of a geometric random variable are given as follows:

p(x)μσ2=(1p)x1px=1,2,3,=1p=1pp2\begin{align*}p(x) &= (1-p)^{x-1} p \quad x=1, 2, 3, \ldots\\ \mu &= \frac{1}{p}\\ \sigma^2 &= \frac{1-p}{p^2}\end{align*}

where:

p=\begin{align*}p =\end{align*} probability of an S\begin{align*}S\end{align*} outcome

x=\begin{align*}x =\end{align*} the number of trials until the first S\begin{align*}S\end{align*} is observed

The figure below plots a few geometric probability distributions. Note how the probabilities start high and drop off, with lower p\begin{align*}p\end{align*} values producing a faster drop-off.

#### Finding the Mean, Standard Deviation, and Probability

A court is conducting a jury selection. Let X\begin{align*}X\end{align*} be the number of prospective jurors who will be examined until one is admitted as a juror for a trial. Suppose that X\begin{align*}X\end{align*} is a geometric random variable, and p\begin{align*}p\end{align*}, the probability of a juror being admitted, is 0.50.

a. Find the mean and the standard deviation of X\begin{align*}X\end{align*}.

The mean and the standard deviation can be calculated as follows:

μσ2σ=1p=10.5=2=1pp2=10.50.52=2=2=1.41\begin{align*}\mu &= \frac{1}{p}=\frac{1}{0.5}=2\\ \sigma^2 &= \frac{1-p}{p^2}=\frac{1-0.5}{0.5^2}=2\\ \sigma&=\sqrt{2}= 1.41\end{align*}

b. Find the probability that more than two prospective jurors must be examined before one is admitted to the jury.

To find the probability that more than two prospective jurors will be examined before one is selected, you could try to add the probabilities that the number of jurors to be examined before one is selected is 3, 4, 5, and so on, as follows:

p(x>2)=p(3)+p(4)+p(5)+\begin{align*}p(x > 2)=p(3)+p(4)+p(5)+ \ldots\end{align*}

However, this is an infinitely large sum, so it is best to use the Complement Rule as shown:

p(x>2)=1p(x2)=1[p(1)+p(2)]\begin{align*}p(x > 2) &= 1-p(x \le 2)\\ &= 1-[p(1)+p(2)]\end{align*}

In order to actually calculate the probability, we need to find p(1)\begin{align*}p(1)\end{align*} and p(2)\begin{align*}p(2)\end{align*}. This can be done by substituting the appropriate values into the formula:

p(1)p(2)=(10.5)11(0.5)=(0.5)0(0.5)=0.5=(10.5)21(0.5)=(0.5)1(0.5)=0.25\begin{align*}p(1) &= (1-0.5)^{1-1}(0.5)=(0.5)^0(0.5)=0.5\\ p(2) &= (1-0.5)^{2-1}(0.5) =(0.5)^1 (0.5)=0.25\end{align*}

Now we can go back to the Complement Rule and plug in the appropriate values for p(1)\begin{align*}p(1)\end{align*} and p(2)\begin{align*}p(2)\end{align*}:

p(x>2)=1p(x2)=1(0.5+0.25)=0.25\begin{align*}p(x > 2) &= 1-p(x \le 2)\\ &= 1 - (0.5 + 0.25)=0.25\end{align*}

This means that there is a 0.25 chance that more than two prospective jurors will be examined before one is admitted to the jury.

#### Calculating the Probability of Success

Using the geometric distribution with a success probability of 0.4, calculate the probability of getting your first success on the third trial.

The distribution we are working with is a geometric distribution with a success probability of 0.4, or X\begin{align*}X\end{align*}~G(0.4)\begin{align*}G(0.4)\end{align*}. Using the probability formula:

P(X=3)=(10.4)31(0.4)=(0.6)2(0.4)=0.144\begin{align*}P(X=3)=(1-0.4)^{3-1}(0.4)=(0.6)^2(0.4)=0.144\end{align*}

The probability that the first success is on the third trial is 0.144.

### Technology Notes: Calculating Geometric Probabilities on the TI-83/84 Calculator

Press [2ND][DISTR] and scroll down to 'geometpdf('. Press [ENTER] to place 'geometpdf(' on your home screen. Type in values of p\begin{align*}p\end{align*} and x\begin{align*}x\end{align*} separated by a comma, with p\begin{align*}p\end{align*} being the probability of success and x\begin{align*}x\end{align*} being the number of trials before you see your first success. Press [ENTER]. The calculator will return the probability of having the first success on trial number x\begin{align*}x\end{align*}.

Use 'geometcdf(' for the probability of at most x\begin{align*}x\end{align*} trials before your first success.

Note: It is not necessary to close the parentheses.

#### Using Technology

A venture capitalist invests in start-up companies in Silicon Valley, California. Each start-up company either succeeds or fails. If a company fails the venture capitalist loses $3 million dollars; if the company succeeds the capitalist gains$8 million dollars. What is the probability that the investor will fail with the first 11 companies and succeed for the first time on his/her 12th investment? What assumption do you have to make to determine this probability?

Assume that a venture capitalist randomly chooses whether a venture will be successful or not. Then this is a geometric distribution with probability of success =.5\begin{align*}= .5\end{align*}. Use the TI calculator geompdf(.5,12)=.0002\begin{align*}(.5,12) = .0002\end{align*}. So there is a very small probability that the investor will not be successful until his 12th company.

### Example

#### Example 1

Matthew is a high school basketball player and a 75% free throw shooter. What is the probability that Matthew makes his first free throw on his fifth shot?

The distribution we are working with is a geometric distribution with a success probability of 0.75, or X\begin{align*}X\end{align*}~G(0.75)\begin{align*}G(0.75)\end{align*}. Using the probability formula:

P(X=5)=(10.75)51(0.75)=(0.25)4(0.75)=0.002930.003\begin{align*}P(X=5)&=(1-0.75)^{5-1}(0.75)\\ &=(0.25)^4(0.75)\\ &=0.00293\\ & \approx 0.003\end{align*}

The probability that the first success is on the third trial is approximately 0.003.

### Review

1. A prison reports that the number of escape attempts per month has a Poisson distribution with a mean value of 1.5.
1. Calculate the probability that exactly three escapes will be attempted during the next month.
2. Calculate the probability that exactly one escape will be attempted during the next month.
2. The mean number of patients entering an emergency room at a hospital is 2.5. If the number of available beds today is 4 beds for new patients, what is the probability that the hospital will not have enough beds to accommodate its new patients?
3. An oil company has determined that the probability of finding oil at a particular drilling operation is 0.20. What is the probability that it would drill four dry wells before finding oil at the fifth one? (Hint: This is an example of a geometric random variable.)
1. Suppose the probability of a high school senior working full-time when in college is .234 and suppose you randomly select one senior until you find one who expects to work full-time while in college. You are interested in the number of seniors you must ask.
1. In words, define the random variable.
2. What is the distribution of this random variable?
3. Construct the probability function for X\begin{align*}X\end{align*}. Stop at X=6\begin{align*}X = 6\end{align*}.
4. On average, how many seniors would you expect to have to ask until you found one who expects to work full-time when in college?
5. What is the probability that you will have to ask fewer than 4 high school seniors?
6. Construct a histogram for this distribution.
2. What are the four conditions for the geometric probability setting?
3. Explain the difference between the binomial probability distribution and the geometric probability distribution.
4. If X\begin{align*}X\end{align*} has a geometric distribution with probability of success p\begin{align*}p\end{align*}, what does (1p)(n1)\begin{align*}(1-p)^{(n-1)}\end{align*} represent?
5. What does the expected value of a geometric random variable represent?
6. You play a game that you can either win or lose. Your probability of winning is .38. What is the probability that it takes 6 games until you win?
7. Suppose you are looking for a friend to go to the movies with you. The probability that a friend will agree to go with you is .27. What is the probability that the fifth friend you ask will be the first one to agree to go with you?
8. Given a geometric probability distribution with probability of success .3. Compute each of the following:
1. P(X=6)\begin{align*}P(X=6)\end{align*}
2. P(X>4)\begin{align*}P(X>4)\end{align*}
3. P(X7)\begin{align*}P(X \le 7)\end{align*}
4. P(X>9)\begin{align*}P(X>9)\end{align*}
5. P(X8)\begin{align*}P(X \ge 8)\end{align*}
6. P(3X10)\begin{align*}P(3\le X \le 10)\end{align*}
7. P(3<X<10)\begin{align*}P(3 < X < 10)\end{align*}

To view the Review answers, open this PDF file and look for section 4.7.

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### Vocabulary Language: English

TermDefinition
geometric distribution The geometric distribution focuses on the number of trials before the first success occurs.
mean The mean, often called the average, of a numerical set of data is simply the sum of the data values divided by the number of values.
probability distribution In a probability distribution, you may have a table, a graph, or a chart that shows you all the possible values of X (your variable), and the probability associated with each of these values P(X).
variance A measure of the spread of the data set equal to the mean of the squared variations of each data value from the mean of the data set.

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