Question

@satellite73, same question as earlier part 2

Answer 1

P of heads = 1/3, total of 5 tosses, Z = X - Y. Compute E(Z) without computing the PMF of Z. I think I got it, just need the answer checked.. one second..

Answer 2

\[E(Z) = E(2X-5) = 2E(X)-E(5)=2*(1*1/3)-5=2/3-5\]that is

Answer 3

i think there might be a mistake here

Answer 4

Hmm can you give me a hint, I want to try to understand what I'm doing wrong.

Answer 5

ok just a hint
how many heads do you expect if you flip 5 times ?

Answer 6

I'd expect 2.5, but we can't have that number, and the coin is biased so I'd lean towards 2 oh wait expected cant be negative lol

Answer 7

it can for sure, for Z but that is not the question i was asking

Answer 8

the fact that the coin is not fair should not throw you off

Answer 9

hang on i think i understand what you mean

Answer 10

E(X) = Sum over all possible x of x*(p(x)).. this will take me a while to do, I'm slow at prob

Answer 11

So that's the sum from 0 to 5

Answer 12

oh no takes no time at all!!

Answer 13

you are thinking way way too hard, these are "bernoulli" trials
you flip five times
you get heads one third of the time
how many heads do you expect??

Answer 14

TTHTT so just one H?

Answer 15

ok i think you are missing something here
expected value does not have to be a possible outcome, it is just an average

Answer 16

Oh is it just the probability itself? Sorry I googled lol

Answer 17

on average you get heads one third of the time
you flip 300 times, how many heads would you expect?

Answer 18

I'd expect 100 heads

Answer 19

right
now suppose you only flip 100 times?

Answer 20

33.33333

Answer 21

5/3?

Answer 22

for 5 tosses

Answer 23

right again

Answer 24

So \[E(Z)=E(2X-5)=2E(X)-E(5)=2*(5/3)-5=10/3-5=-5/3?\]

Answer 25

kind of went off the page, but looks good

Answer 26

cool, thanks sat, I'll just write this down, there's also a part 3 for this thing, where I need to compute the expected value by computing the pmf first. Not sure how to approach the pmf yet. I'll tag you if you still have time.

Answer 27

whatever \[\frac{10}{3}-5\] is

Answer 28

you just do it

Answer 29

the possible values of Z are what?

Answer 30

oh I see what you mean
-5 <= z < = 5 and then calculate corresponding probabilities?

Answer 31

right
this approach, which will kinda suck, is meant for you to see how much easier the other is

Answer 32

of course there are some values Z cannot take

Answer 33

oh maybe just zero which is unimportant any way

Answer 34

Yea it can only be -5, -3, -1, 1, 3, 5

Answer 35

ok looks like you are good to go

Answer 36

Ok, well I know how to finish this off, thanks sat. I have one last question a bit tougher, so I saved it for last. Can you take a look at that one please

Answer 37

sure if i can
what is it?