Probability Theory ; Binomial Distribution?

[aeubz]

Homework Statement

Now you and your fiend play a different game. You flip your coin until it comes up heads the first time. Let X denote the number of flips needed. Your friend rolls its die until it comes up "3" or "5". The first try let Y denote the number of rolls needed. Assume X and Y are independent; note that X >= 1 and Y >=1. a) Determine P(X=n), n>=1 b) Determine P(Y=n), n>=1 c) Determine E(x) d) Determine E(y) e) Determine P(x=y) f) Given that x = y, determine the expected value of this common value.

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Homework Equations

Binomial Distribution
nCk (P)^k (1-P)^n-k

The Attempt at a Solution

a) nC1 (1/2)(1/2)^n-1 (My friend did not include the nC1)
b) nC1 (1/3)(2/3)^n-1 (My friend did not include the nC1)
c) n*1/2
d) n*1/3
e) nC1 (1/2)(1/2)^n-1 * nC1 (1/3)(2/3)^n-1 (This is weird, my friend integrated from 1 to infinite) he got 1/4??
f) nsubx*psubx + nsuby*psuby (I definitely need help on this one.)

 

[Dick]

For a) and b), why do you think the nC1 should be in the expression? And for c) and d) the expectation value is a sum over all n. The final expression shouldn't have an 'n' in it. Maybe you'd better review the definition of 'expectation value'.

 

[aeubz]

For a) and b), why do you think the nC1 should be in the expression? And for c) and d) the expectation value is a sum over all n. The final expression shouldn't have an 'n' in it. Maybe you'd better review the definition of 'expectation value'.

Ok so the reason nC1 should not be there is because we only care about the n trial. Right? Still a bit confused about the Expected.. If X is geometric with probability p then E(X) = 1/p = 2 and E(Y) = 3? And e) is P(x=y)= sum(n=1 to +oo) 1/6(1/3)^(n-1) = 1/4

f) is still bothering me a little..

 

[Dick]

Right. The first n-1 trials have to be failures, only the last one needs to succeed. And, yes, now I think you have the expectation values right as well. I've got to confess, I'm a little vague on the last one. Probabilility isn't my field, but isn't it sum(n=1 to +oo) n*(1/6)*(1/3)^(n-1)?

 

[aeubz]

Right. The first n-1 trials have to be failures, only the last one needs to succeed. And, yes, now I think you have the expectation values right as well. I've got to confess, I'm a little vague on the last one. Probabilility isn't my field, but isn't it sum(n=1 to +oo) n*(1/6)*(1/3)^(n-1)?

Ok ok, so by definition, E[X] = sum(n=1 to +oo) n * f(n) where f(n) is the density function. Yes I agree. Thank you very very much for all your help!