Question

@wio can you look over my probability problem here, thanks!

Answer 1

Find pmf of Z=X-Y where X is the number of heads obtained by tossing an unfair coin 5 times (where probability of getting a head is 1/3)

Answer 2

I know how to solve this, just want to make sure I don't make some silly mistakes along the way. So here we go...

Answer 3

Z can take on the values below:
-5 (T T T T T)
-3 (H T T T T)
-1 (H H T T T)
1 (H H H T T)
3 (H H H H T)
5 (H H H H H)

Answer 4

\[P(Z=-5)=(2/3)^5\]\[P(Z=5)=(1/3)^5\]

Answer 5

probability mass function?

Answer 6

\[P(Z=-3)=\binom{5}{1}(\frac{1}{3})^1*(1-\frac{1}{3})^{5-1}=\frac{5}{3}*(\frac{2}{3})^4\]
and yes prob mass function

Answer 7

Do they say what Y is?

Answer 8

oh yea sorry forgot to add X is heads, Y is tails

Answer 9

Long day lol

Answer 10

Oh, so Y and X are completely dependent on each other.

Answer 11

Yes, Y=5-X

Answer 12

Does what I'm doing look alright so far?

Answer 13

Hmmm, I guess so.

Answer 14

My thinking is a bit more different... \(Y=5-X\).

Answer 15

uh yea lol 5 is not random i meant y = 5 - x

Answer 16

Sorry, hahah

Answer 17

So we can say \(\Pr(Z=5x-x^2) = \Pr(X=x)\), though I guess this doesn't help much.

Answer 18

\[P(Z=3)=\binom{5}{4}*(\frac{1}{3})^{4}*(\frac{2}{3})^{5-4}\]

Answer 19

Yeah, it looks like you would use binomial distribution, but it is going to be discontinuous.

Answer 20

P(Z=-1) means we got 2 heads out of 5 (2-3=-1), so:\[P(Z=-1)=\binom{5}{2}(1/3)^2(2/3)^3\]

Answer 21

Yea and P(Z=z) is 0 for all z not in {-5, -3, -1, 1, 3, 5}

Answer 22

You have to write it as a piecewise function

Answer 23

like P(z) = { blah if z = -5, etc.?

Answer 24

Yeah

Answer 25

yea sure I'll do that in my notes lol, I'm just terrible at latex formatting for those things

Answer 26

Just want to make sure I'm not screwing up any of these values:
P(Z=1) means 3 heads out of 5:\[P(Z=1)=\binom{5}{3}(1/3)^3(2/3)^2\]

Answer 27

Do those look right?

Answer 28

Yeah, so far they're looking right.

Answer 29

And I think that's all we have right for possible Z values
0 heads => 0 - 5 = -5
1 head => 1 - 4 = -3
2 heads => 2 - 3 = -1
3 heads => 3 - 2 = 1
4 heads => 4 - 1 = 3
5 heads => 5 - 0 = 5

Answer 30

I think you've got this down.

Answer 31

Cool thanks for all the help wio!