Segment 1 Sanmit Narvekar

Segment 1

To Calculate

1. Prove that $\displaystyle P(ABC) = P(B)P(C|B)P(A|BC)$ .

Proof

$\displaystyle [P(B)P(C|B)]P(A|BC) = P(BC) P(A|BC) = P(ABC)$

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

$\displaystyle P($ sum is odd AND neither die is 4$\displaystyle ) = 1/3$

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 $\displaystyle 12/36 = 1/3$ .

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.

P(nth fish caught is trout) = $\displaystyle \sum_{fish_1 \ldots fish_{n-1}} P(trout | fish_1 \ldots fish_{n-1})\prod_{i=1}^{n-1} P(fish_i | fish_1 \ldots fish_{i-1})$
$\displaystyle =\sum_{fish_1 \ldots fish_{n-1}} P(trout | \# \text{trout drawn in n-1 rounds} )\prod_{i=1}^{n-1} P(fish_i | \# \text{ trout drawn in i-1 rounds})$