# Segment 1 Sanmit Narvekar

## Segment 1

#### To Calculate

1. Prove that **Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle P(ABC) = P(B)P(C|B)P(A|BC)}**
.

**Proof**

**Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle [P(B)P(C|B)]P(A|BC) = P(BC) P(A|BC) = P(ABC)}**

2. What is the probability that the sum of two dice is odd with neither being a 4?

**Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle P(}**
sum is odd AND neither die is 4**Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle ) = 1/3 }**

This problem can be solved similar to the problem in the slides and as done in class: create a table of valid configurations and use it to compute the desired probabilities. Of the 36 possible rolls of the die, only 18 of them lead to an odd sum. Of those 18, six use the number 4. Thus, there are 12 valid configurations of the dice, and the probability is **Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 12/36 = 1/3}**
.

Alternatively, notice that in order to receive an odd sum, one die (Die A) must be odd and the other (Die B) must be even. Consider the even die. Since we can't use 4, there are only 2 possibilities: 2 and 6. For each of these, there are 3 possibilities for the odd die: 1, 3, and 5. Thus there are 6 possibilities for each even/odd dice pair. Since we can swap Die A and Die B (i.e. make Die A the even and Die B the odd), that leads to 12 valid configurations out of a possible 36, again leading to 1/3 probability.

#### To Think About

3. For the trout/minnow problem, what if you want to know the probability that the Nth fish caught is a trout, for N=1,2,3,... What is an efficient way to set up this calculation? (Hint: If you ever learned the word "Markov", this might be a good time to remember it!)

Not really sure what this one is asking for...

P(nth fish caught is trout) = **Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sum_{fish_1 \ldots fish_{n-1}} P(trout | fish_1 \ldots fish_{n-1})\prod_{i=1}^{n-1} P(fish_i | fish_1 \ldots fish_{i-1}) }**

This can be simplified by observing that **for conditioning**, we only need to know how many trouts have been drawn up to the previous state (hence, Markov property).

**Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle =\sum_{fish_1 \ldots fish_{n-1}} P(trout | \# \text{trout drawn in n-1 rounds} )\prod_{i=1}^{n-1} P(fish_i | \# \text{ trout drawn in i-1 rounds}) }**