- #1

baseballfan_ny

- 92

- 23

- Homework Statement
- See below. I got parts (a) and (b) but still working on (c)

- Relevant Equations
- Total probability theorem, Baye's Rule for Mixed variables

I got (a) and (b) but I'm still working on (c). The solutions can be found here for your reference: https://ocw.mit.edu/courses/electrical-engineering-and-computer-science/6-041sc-probabilistic-systems-analysis-and-applied-probability-fall-2013/unit-ii/lecture-9/MIT6_041SCF13_assn05_sol.pdf. But I'm trying my hardest not to look at them :)

My first idea was that since the problem says successive tosses are independent, then the conditional probability of the 2nd toss being heads should just be the probability of the 1st toss being heads... I know that's wrong I'm not sure how to refute it though.

So then I did

Let Yi be the outcome of the ith toss.

$ p_{Y_i|P} = \begin{cases} p & \text{if } y_i = 1 \\ 1 - p & \text{if } y_i = 0 \end{cases} ${pif yi=11−pif

And then by Baye's rule,

$ P(Y_2 = 1 | Y_1 = 1) = \frac {P(Y_2 = 1) P(Y_1 = 1 | Y_2 = 1)} {P(Y_1 = 1)} $

But that doesn't do me too much good, since I believe P(Y2=1) should equal P(Y1=1), and both would simply be the answer to (b) and they would cancel each other, giving me

$ P(Y_2 = 1 | Y_1 = 1) = P(Y_1 = 1) $

Edit: LaTeX is giving me a lot of trouble (or I'm typing something wrong in the commands). Please let me know if you cannot see what I have written. Thanks.