# 6.7: Geometric Distribution


Consider these two random variables, which both start with repeated Bernoulli trials:

1. Flip a fair coin 10 times. Let $$X$$ = the number of heads.
2. 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.

$$P(X \geq 3)=1-P(1)-P(2)=(0.7)(0.3)^{0}+(0.7)(0.3)^{1}=1-0.91=0.09$$

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