Geometric Distribution

I. Introduction

The geometric distribution is similar to the binomial distribution, but unlike the binomial distribution, which calculates the probability of observing a fixed number of success in \(n\) observations, the geometric distribution allows us the probability of observing our first success on a given observation. In other words, there is no fixed \(n\)

For the Geometric Distribution, \(X\) is the number of trials required to obtain the first success.

Problems involving the geometric distribution will ask you to flip a coin UNTIL you get the FIRST tail, or ask you for the probability of getting your FIRST tail ON the 5th flip, etc. Look for key words such as until, first, on, and after. 


II. Geometric Settings

1) There are two outcomes called success or failure.
2) Observations are independent.

3) The probability of success, \(p\), is constant.


III. Calculating Geometric Probabilities (Geometric Formula)

When interested in finding the probability that your first success occurs on the \(k^{th}\) trial, one needs to use the following formula:

Geometric Formula (ON):    ​\(P(X = k) = q^{k-1}p \)  where \(q = 1 - p\)


For calculating more than problems – The probability that it takes more than \(n\) trials to see the first success is:

Geometric Formula (MORE THAN):   \(P(X > n) = q^{n}\)


IV. Calculating the Mean and Variance of a Geometric Distribution

Use the following formulas to calculate the mean, variance, or standard deviation of a geometric distribution:

Mean/Expected Value Variance Standard Deviation
\(E(X) = \mu_{x} = \frac{1}{p}\) \(VAR(X) = \sigma^{2} = \frac{q}{p^{2}}\) \(SD(X) = \sigma = \sqrt{\frac{q}{p^{2}}}\)




V. Calculator!

For P(X = k) 

Hit 2nd Vars →  Scroll to E:geometpdf → Fill in (n, p, k)

For P(X ≤ k)

Hit 2nd Vars   Scroll to F:geometcdf → Fill in (n, p, k)


Everything must be entered in the form of "less than or equal to" (≤). If the problem is asking you for "after" or "more than", draw a number line and shade in what is included. Then you can set up a "less than or equal to" (≤) problem using what is not included, as long as you remember to subtract the calculator's answer from 1. 


VI. Example

An individual decides to roll a fair 6-sided die until he observes a 4. Calculate the following:

a. Find the probability that the first time he observes a 4 is on his 3rd role of the die.

Before beginning with the full solution, we must first label our outcomes. Based on the problem, rolling a 4 can be labeled as a success, and rolling any number other than a 4 can be labeled as a failure. The probability of a success in this instance is 1/6 or 0.167.

\(P(X=4) = (1-0.167)^3(0.167)\)

\(P(X=4) = 0.097\)

b. What is the probability that individual must roll more than 5 times before he observes his first 4?

Without a calculator, you can use the formula to solve: \(P(X > 5) = \left(\frac{5}{6}\right)^5 = 0.401\) 

 With a calculator, it will help to start by drawing a number line: ← 1 2 3 4 5 6 7 8 9 10 → 

As this number line shows, "more than 5" is equal to 1 - "less than or equal to 5". Using your calculator, you can solve for P(x ≤ 5), and subtract this from 1.

1 - P(x ≤ 5) = 1 - 0.599 = 0.401

c. What is the expected number of rolls before observing the first 4?

Remember that "expected" is another term for "mean." Using the formula above, you know that the mean is equal to \(1\over p\) = \( \frac{1}{1/6} = 6\) 

d. Find the standard deviation pertaining to the the number of rolls needed before observing the first 4.

Using the formula above, you know that the standard deviation is equal to \( \sqrt{\frac{q}{p^{2}}}\) = \( \sqrt{\frac{5/6}{(1/6)^2}} = 5.477\)