> For the complete documentation index, see [llms.txt](https://stephanosterburg.gitbook.io/scrapbook/llms.txt). Markdown versions of documentation pages are available by appending `.md` to page URLs; this page is available as [Markdown](https://stephanosterburg.gitbook.io/scrapbook/math/statistics-and-probability/binomial-variables/geometric-random-variable.md).

# Geometric random variable

#### Binomial Random Variable

*X = # of 6's after 12 rolls of fair die*

* trial outcome success or failure
* trial results independent
* fixed # of trials
* same probability on each trial

#### Geometric Random Variable

*Y = # of* *rolls until get 6 on fair die*

* ***NO*** fixed # of trials -> How many trials until success?

#### Example:

Caterina scans animals brought to the shelter to check for microchips that will help locate their owners. There is a $$0.05$$ probability that a stray dog brought to the shelter will have a microchip. Let $$D$$ be the number of stray dogs Caterina scans until she finds one with a microchip. Assume the probability of each dog having a microchip is independent.

**Find the probability that the** $$4^{\text{th}}$$ **dog Caterina scans will be the first to have a microchip.**

$$P(D=4) = (1-0.05)^3 \cdot 0.05 = 0.04$$

### Cumulative geometric probability (greater than a value)

$$P(V>4)=P(V=5)+P(V=6)+P(V=7)+....$$

same as

$$P(V \text{not} \leq 4) = P(\text{first 4 cars not SUVs}) = (0.88)^4 = 0.5997$$

#### **Quiz**

Jeremiah makes $$25%$$ of the three-point shots he attempts. For a warm up, Jeremiah likes to shoot three-point shots until he makes one. Let $$M$$ be the number of shots it takes Jeremiah to make his first three-point shot. Assume that the results of each shot are independent.

**Find the probability that it takes Jeremiah fewer than** $$4$$ **attempts to make his first shot.**

1. On each shot:

   $$P(\blueD{\text{make}})=0.25P$$

   $$P(\purpleD{\text{miss}})=0.75P$$(

   If it takes Jeremiah fewer than 444 attempts to make his first shot, here are the possible sequences of shots:

   * make
   * miss, make
   * miss, miss, make

   We can find the probability of each sequence and add those probabilities together.
2. $$P(make) =0.25\newline P(miss, make) =(0.75)(0.25) = 0.1875 \newline P(miss, miss, make) = (0.75)(0.75)(0.25) = 0.140625\newline P(M<4) = 0.25 + 0.1875 + 0.140625 = 0.578125$$<br>
3. We could also find the probability that $$M<4$$ by taking the complement of the probability that he missed the first $$3$$.

   $$\begin{aligned} P(M<4) &= 1-P(\text{missed }3)\\\ &=1-(\purpleD{0.75})^3 \\\ &=1-0.421875\\\ &=0.578125 \end{aligned}$$<br>

#### Quiz

Anand knows from experience that if he does not review a new vocabulary word that he has learned, that he has a $$70%$$ chance of forgetting it each day. Let $$D$$ be the number of days Anand goes without reviewing a word until he forgets it.

**Find the probability that it takes Anand** $$4$$ **or more days to forget the word.**

On each day:

$$P(\blueD{\text{forget}})=0.7$$\
$$P(\purpleD{\text{remember}})=0.3$$

If it takes Anand $$4$$ or more days to forget the word, then he must remember for each of the first $$3$$ days.

$$\begin{aligned} P(D\geq 4)&=P(\text{remember first 3}) \\\ &=(\purpleD{0.3})^3 \\\ &= 0.027 \end{aligned}$$<br>