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6.7: Geometric Distribution

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    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, ...)





    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.


    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:


    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.

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


    This page titled 6.7: Geometric Distribution is shared under a CC BY-SA 4.0 license and was authored, remixed, and/or curated by Maurice A. Geraghty via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.