6.7: Geometric Distribution
- Page ID
- 20893
Consider these two random variables, which both start with repeated Bernoulli trials:
- Flip a fair coin 10 times. Let \(X\) = the number of heads.
- Flip a fair coin repeatedly until you get a head. Let \(X\) = the number of total flips.
The first random variable is a binomial random variable where \(n =10\) and \(p=0.5\). The possible values of \(X\) are {0,1,2,3,4,5,6,7,8,9,10}
The second random variable is unusual in that there are an infinite number of possibilities for \(X\). The possible number of flips until you get a head are {1, 2, 3, ...}. This is called the geometric distribution and its features are shown in the box.
Geometric Probability Distribution (parameter= \(p\))
Two possible outcomes (Success/Failure) or (Yes/No)
\(p = P\)(yes/success) on one trial
\(q = 1‐p = P\)(no/failure) on one trial
\(X\) = Number of independent trials until the first success. (1, 2, 3, ...)
\(\mu=\dfrac{1}{p}\)
\(\sigma^{2}=\dfrac{1-p}{p^{2}}\)
\(\sigma=\sqrt{\dfrac{1-p}{p^{2}}}\)
\(P(x)=p(1-p)^{x-1}\)
Example: Free throw shooting
Let’s again return to the example of Draymond Green, a 70% free throw shooter. Now let \(X\) = the number of free throws Draymond takes until he makes a shot. \(X\) follows a geometric distribution.
Solution
The expected number of shots: \(\mu=\dfrac{1}{p}=1.43\) shots
The variance: \(\sigma^{2}=\dfrac{1-0.7}{0.7^{2}}=0.612\)
The probability that Draymond Green takes exactly 3 shots to make a free throw:
\(P(X=3)=0.7(0.3)^{2}=0.063\)
The probability that Draymond Green takes 3 or more shots to make a free throw:
Since \(P(X \geq 3)=P(3)+P(4)+\ldots\) is an infinite sum, is better to use Rule of Complement.
\(P(X \geq 3)=1-P(1)-P(2)=(0.7)(0.3)^{0}+(0.7)(0.3)^{1}=1-0.91=0.09\)