# /Segment1

# Segment 1 (16 January 2013)

**Jan. 16th Wed, Segment 1**

*Problems:*

**1. show <math>P(ABC)=P(B)P(C | B) P(A | BC)</math>**

right side:<math>P(B)P(C | B) P(A | BC) = P(BC)P(A|BC)</math> <math>=P(ABC)</math>

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

P(neither being 4 & sum of two dice is odd) =P(neither being 4 | sum of two dice is odd)* P(sum of two dice is odd) =12/18 * 18/36 =1/3

*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!)**

if N=1, P=2/5;

if N=2, P=P(1st is trout & 2nd is trout) + P(1st is minnow & 2nd is trout)=2/5*1/4 +3/5*2/5=0.34;

if N>2, P=P(Nth is trout & didn't have a trout before) + P(Nth is trout & had a trout before)

P(Nth is trout & didn't have a trout before) = (3/5)^(N-1) * (2/5)

P(Nth is trout & had a trout before) =P (Nth is a trout when there's one left) * [ P(1st trial got trout) +P( 2nd trial got trout) + ... +P (N-1 th trial got trout) ]

<math>\begin{align}

= (1/4) * [ (2/5) * (3/4)^{N-2} + (3/5) * (2/5) * (3/4)^{N-3} + (3/5)^2 * (2/5) * (3/4)^{N-4} + ... + (3/5)^{N-2}* (2/5) ] \\

=(1/4) * \sum_{i=0}^{N-2} (3/5)^i * (2/5) * (3/4)^{N-2-i} \\

\end{align}</math>

So, <math> P = (3/5)^{N-1} * (2/5) + (1/4) * \sum_{i=0}^{N-2} (3/5)^i * (2/5) * (3/4)^{N-2-i} </math>

## in class problem

**1. you are an oracle that, when asked, says "yes" with probability P and "no" with probability 1 − P. How do you do this using only a fair, two‐sided coin?**

**2. In the programming language of your choice, how would you simulate flips of a fair, two‐ sided coin? Do it. Simulate 106 flips. How many heads do you get?**

**In the programming language of your choice, how would you simulate draws from two weird 7‐sided dice whose faces (showing 1 through 7 spots) have probabilities proportional to: **
<math> 1:e: \pi :4:5:6:e^ \pi </math> respectively ?
**– Do it. Simulate 106 throws of the dice. How many times is the sum of the two dice equal to 8? – What should it be (in expectation) analytically?**

To calculated the expected probability, I summed across the probability of all possible combinations:

I got 62606 times with the sum being 8, over the 10**6 tossing.

So, the observed probability is 0.62606.