# Probability Theory; Binonial Distribution?

• Nov 29th 2009, 07:32 PM
aeubz
Probability Theory; Binonial Distribution?
The problem 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.

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.)

Update:
Can someone confirm these?
a) (1/2)(1/2)^n-1
b) (1/3)(2/3)^n-1
c) E(X)=1/(1/2)=2
d) E(Y)=1/(2/6)=3
e) P(x=y)= sum(n=1 to +oo) 1/6(1/3)^(n-1) = 1/4
f) sum(n=1 to +oo) n*(1/6)*(1/3)^(n-1) = 3/8
• Nov 30th 2009, 12:46 AM
Moo
Hello,

Quote:

Update:
Can someone confirm these?
a) (1/2)(1/2)^n-1
b) (1/3)(2/3)^n-1
c) E(X)=1/(1/2)=2
d) E(Y)=1/(2/6)=3
e) P(x=y)= sum(n=1 to +oo) 1/6(1/3)^(n-1) = 1/4
f) sum(n=1 to +oo) n*(1/6)*(1/3)^(n-1) = 3/8
(Yes) (I hope you understood where these formulas came from in e and f)

The binomial distribution counts the number of successes. The geometric distribution denotes the first time a success happens (Wink)
• Nov 30th 2009, 05:12 AM
aeubz
Quote:

Originally Posted by Moo
Hello,

(Yes) (I hope you understood where these formulas came from in e and f)

The binomial distribution counts the number of successes. The geometric distribution denotes the first time a success happens (Wink)

Wow. Thank you very much! That really cleared things up. I was wondering why we used the nCk in the nCk (P)^k (1-P)^n-k. And now it makes sense, that is the case for the number of successes and without nCk means the first time a success happens. Thank you! (Rofl)