# Kai's Segment 1

## Segment 1. Let's Talk about Probability 1/16

### To calculate

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

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

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

By drawing two dice, we can get $6*6=36$ possible situations.

If the 1st is odd, the 2nd must be even to make the sum of them odd. But the 2nd can not be 4. Thus, when the 1st is odd, $3*2=6$ situations are possible.

If the 1st is even, the number of the possible situations is the same as the previous one by exchanging the number of the 1st dice and the number of the 2nd dice as their numbers are different(one is even, the other is odd).

Thus, there are 12 possible situations satisfying the problem. And the probability is 1/3.

1. use first-order logic as a calculus of inference

I don't know what 'a calculus of inference' means. Hope someone could help...

2. The probability that the Nth fish caught is a trout, for N=1,2,3,...

Providing the Nth fish caught is a trout, there are two situation to consider.

In the first situation, this trout from the Nth caught is the first trout caught until now, i.e. the fish caught from 1th caught to the N-1th caught are all minnow. The probability of this situation is $({\frac{3}{5}})^{N-1}\frac{2}{5}$.

In the second situation, this trout from the Nth caught is the second trout caught, i.e. the first trout is caught in one of the caughts from 1th caught to N-1th caught. Suppose the first trout is caught at the ith caught($1\leq i \leq {N-1}$), then its probability is $({\frac{3}{5}})^{i-1}\frac{2}{5} ({\frac{3}{4}})^{N-1-i} \frac{1}{4}$. The total probability in the second situation is

$\sum_{i=1}^{N-1}({\frac{3}{5}})^{i-1}\frac{2}{5} ({\frac{3}{4}})^{N-1-i} \frac{1}{4} = \frac{2}{3} \left[(\frac{3}{4})^{N-1}-(\frac{3}{5})^{N-1}\right]$.

Then the total probability is

$\frac{2}{3} (\frac{3}{4})^{N-1} - \frac{4}{15}(\frac{3}{5})^{N-1}$.