Learning Outcomes
- State geometric probabilities using mathematical notation
- Calculate the mean and standard deviation of a geometric random variable
- Calculate a geometric probability using technology
Notation for the Geometric: [latex]G=[/latex] Geometric Probability Distribution Function
[latex]X{\sim}G(p)[/latex]
Read this as “[latex]X[/latex] is a random variable with a geometric distribution.” The parameter is [latex]p[/latex]; [latex]p=[/latex] the probability of a success for each trial.
Example
Assume that the probability of a defective computer component is 0.02. Components are randomly selected. Find the probability that the first defect is caused by the seventh component tested. How many components do you expect to test until one is found to be defective?
Show Answer
Let [latex]X=[/latex] the number of computer components tested until the first defect is found.
[latex]X[/latex] takes on the values 1, 2, 3, … where [latex]p = 0.02[/latex].
[latex]X{\sim}G(0.02)[/latex]
Find [latex]P(x=7)[/latex]. [latex]P(x=7)=0.0177[/latex].
USING THE TI-83, 83+, 84, 84+ CALCULATOR
To find the probability that [latex]x=7[/latex],
- Enter 2nd, DISTR
- Scroll down and select geometpdf(
- Press ENTER
- Enter 0.02, 7); press ENTER to see the result: [latex]P(x=7)=0.0177[/latex]
To find the probability that [latex]x\leq7[/latex], follow the same instructions EXCEPT select E:geometcdf(as the distribution function.
The probability that the seventh component is the first defect is 0.0177.
The graph of [latex]X{\sim}G(0.02)[/latex] is:
The [latex]y[/latex]-axis contains the probability of [latex]x[/latex], where [latex]X=[/latex] the number of computer components tested.
The number of components that you would expect to test until you find the first defective one is the mean, [latex]\displaystyle{\mu}={50}[/latex].
The formula for the mean is [latex]\displaystyle{\mu}=\frac{{1}}{{p}}=\frac{{1}}{{0.02}}={50}[/latex]
The formula for the variance is [latex]\displaystyle{\sigma}^{{2}}={(\frac{{1}}{{p}})}{(\frac{{1}}{{p}}-{1})}={(\frac{{1}}{{0.02}})}{(\frac{{1}}{{0.02}}-{1})}={2},{450}[/latex]
The standard deviation is [latex]\displaystyle{\sigma}=\sqrt{{{(\frac{{1}}{{p}})}{(\frac{{1}}{{p}}-{1})}}}=\sqrt{{{(\frac{{1}}{{0.02}})}{(\frac{{1}}{{0.02}}-{1})}}}={49.5}[/latex]
Try It
The probability of a defective steel rod is 0.01. Steel rods are selected at random. Find the probability that the first defect occurs on the ninth steel rod. Use the TI-83+ or TI-84 calculator to find the answer.
Show Solution
[latex]P(x=9)=0.0092[/latex]
Example
The lifetime risk of developing pancreatic cancer is about one in 78 (1.28%). Let [latex]X=[/latex]the number of people you ask until one says he or she has pancreatic cancer. Then [latex]X[/latex] is a discrete random variable with a geometric distribution: [latex]\displaystyle{X}~{G}{(\frac{{1}}{{78}})}{\quad\text{or}\quad}{X}~{G}{({0.0128})}[/latex]
- What is the probability of that you ask ten people before one says he or she has pancreatic cancer?
- What is the probability that you must ask 20 people?
- Find the (i) mean and (ii) standard deviation of [latex]X[/latex].
Show Solution
- [latex]P(x=10)=\text{geometpdf}(0.0128,10)=0.0114[/latex]
- [latex]P(x=20)=\text{geometpdf}(0.0128,20)=0.01[/latex]
- [latex]\text{Mean}={\mu}=\frac{{1}}{{p}}=\frac{{1}}{{0.0128}}={78}[/latex]
- [latex]\text{Standard Deviation}={\sigma}=\sqrt{{\frac{{{1}-{p}}}{{{p}^{{2}}}}}}=\sqrt{{\frac{{{1}-{0.0128}}}{{0.0128}^{{2}}}}}\approx{77.6234}[/latex]
Try It
The literacy rate for a nation measures the proportion of people age 15 and over who can read and write. The literacy rate for women in Afghanistan is 12%. Let [latex]X=[/latex] the number of Afghani women you ask until one says that she is literate.
- What is the probability distribution of [latex]X[/latex]?
- What is the probability that you ask five women before one says she is literate?
- What is the probability that you must ask ten women?
- Find the (i) mean and (ii) standard deviation of [latex]X[/latex].
Show Solution
- [latex]X{\sim}G(0.12)[/latex]
- [latex]P(x=5)=\text{geometpdf}(0.12,5)=0.0720[/latex]
- [latex]P(x=10)=\text{geometpdf}(0.12,10)=0.0380[/latex]
- [latex]\text{Mean}={\mu}=\frac{{1}}{{p}}=\frac{{1}}{{0.12}}\approx{3333}[/latex]
- [latex]\text{Standard Deviation}={\sigma}=\sqrt{{\frac{{{1}-{p}}}{{{p}^{{2}}}}}}=\sqrt{{\frac{{{1}-{0.12}}}{{{0.12}^{{2}}}}}}\approx{7.8174}[/latex]
