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. $P(make) =0.25\newline P(miss, make) =(0.75)(0.25) = 0.1875 \newline P(miss, miss, make) = (0.75)(0.75)(0.25) = 0.140625\newline P(M<4) = 0.25 + 0.1875 + 0.140625 = 0.578125$

3. We could also find the probability that $M<4$ by taking the complement of the probability that he missed the first $3$.

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

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.

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