Zack Chase

2022-09-17

Let $X\sim Geom\left(0.75\right)$. Find the probability that X is divisible by 3

unfideneigreewl

Step 1
There are two versions of the geometric distribution. In one of them, the random variable X measures the number of trials until the first success, including the trial that gave the success. In the other version, the random variable Y counts the number of failures until the first success.
In elementary courses, the first version is more common than the second. So we use that one. Thus the possible values of X are 1,2,3,4,….
So X is divisible by 3 if the number of trials is 3,6,9,12,….
Step 2
We have $Pr\left(X=3\right)=\left(0.25{\right)}^{2}\left(0.75\right)$ (2 failures and then success). Similarly, $Pr\left(X=6\right)=\left(0.25{\right)}^{5}\left(0.75\right)$, and $Pr\left(X=9\right)=\left(0.25{\right)}^{8}\left(0.75\right)$, and so on. Thus the probability that X is divisible by 3 is $\left(0.25{\right)}^{2}\left(0.75\right)+\left(0.25{\right)}^{5}\left(0.75\right)+\left(0.25{\right)}^{8}\left(0.75\right)+\left(0.25{\right)}^{11}\left(0.75\right)+\cdots .$
This is an infinite geometric series, with first term $a=\left(0.25{\right)}^{2}\left(0.75\right)$ and common ratio $r=\left(0.25{\right)}^{3}$. By a standard formula, the sum of the geometric series is equal to $\frac{a}{1-r}.$

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