# Binomial and geometric distributions

by sara_87
Tags: binomial, distributions, geometric
 P: 773 i was doing some exercises nut i'm not sure if my answers are correct 1) X~B(5,0.25) i have to find: a) E(x^2) and my answer was 2.5, is this correct? b) P(x(>or=to)4) and my answer was 0.0889, is this correct? 2) X~Geom(1/3) i have to find: a) E(x) my answer is 1/3 b) E(x^2) c) var(x) d) P(X=4) my answer is 0.988 e) P(X>2) my answer is..... -(1/18) i dont know how to compute b) and c) and i think d) is wrong and i know e) is wrong. can someone help please. Thank you very much
 Math Emeritus Sci Advisor Thanks PF Gold P: 39,559 Would "yes" or "no" answers help you? Show what you did to get those answers and we can point out exactly where, if at all, you went wrong.
 P: 773 my working out is: 1) a) E(x^2)= var(x)+E(x)^2 = np(1-p+np)= 5(1/4)(1-1/4-5/4)=2.5 b) P(x(>or = to)4)= 1-P(x=3)-P(x=2)-P(x=1)-P(x=0) using the formula P(x=m)=(nCm)P^m(1-p)^(n-m) =1-(15/1024)-(135/512)-(405/1024)-(243/1024)=91/1024=0.0889 2) a)E(x)=1/p=1/(1/3)=3 b) and c) i have no idea d) using the formula P(x=m)=P(1-P)^(m-1) P(x=4)=1/3(1-1/3)^3=8/81=0.988 e) P(x>2)=1-P(x(
 Sci Advisor HW Helper P: 2,481 Binomial and geometric distributions 1a is correct except you meant +5/4) not -5/4). For 1b can you write out the nCm terms? 2a is correct. For 2b use the same formula as in 1a. 2c: Var(Geometric(p)) = (1-p)/p^2; see http://en.wikipedia.org/wiki/Geometric_distribution 2d: 1/3(1-1/3)^3 = 8/81 < 0.1 so it cannot be > 0.9 (i.e., you have made a fraction computation error). 2e: if the exponent is k-1 then k = 1, 2, .... OTOH, if the exponent is k, then k = 0, 1, 2, ... In either case, exponent term > 0 so it cannot be -1. See http://en.wikipedia.org/wiki/Geometric_distribution
 P: 773 for 1b: my working including the ncm terms: P(x(>or = to)4)= 1-P(x=3)-P(x=2)-P(x=1)-P(x=0) using the formula P(x=m)=(nCm)P^m(1-p)^(n-m) =1-[(5C3)(1/4)^3(1-1/4)^2]-[(5C2)(1/4)^2(1-1/4)^3]-[(5C1)(1/4)(1-1/4)^4]-[(5C0)(1/4)^0(1/1/4)^5] =1-(15/1024)-(135/512)-(405/1024)-(243/1024) =91/1024=0.0889 for 2e: i looked at the link but i still havent figured out what i had done wrong with my working out. thank u very much for everything else.
HW Helper
P: 2,481
 Quote by sara_87 for 2e: i looked at the link but i still havent figured out what i had done wrong with my working out.
You have the exponential term -1, which shouldn't be there under either definition of the geometric distribution (referenced on the page I linked to).

My guess is you should have 1 - (1/3)(2/3)^2 - (1/3)(2/3)^1 - (1/3)(2/3)^0, but you should verify that.

For 1b, you have:
1 - Binomial[5, 3] (1/4)^3 (1 - 1/4)^2 - Binomial[5, 2] (1/4)^2 (1 - 1/4)^3 - Binomial[5, 1] (1/4) (1 - 1/4)^4 - Binomial[5, 0] (1/4)^0 (1/1/4)^5, where Binomial[n,m] stands for nCm.

Binomial[5, 3] = Binomial[5, 2] = 10
Binomial[5, 1] = 5
Binomial[5, 0] = 1

so the expression is:
1 - 10 (1/4)^3 (1 - 1/4)^2 - 10 (1/4)^2 (1 - 1/4)^3 - 5 (1/4) (1 - 1/4)^4 - 1 (1/4)^0 (1 - 1/4)^5
=1 - 10 (1/4)^3 (3/4)^2 - 10 (1/4)^2 (3/4)^3 - 5 (1/4) (3/4)^4 - 1 (1/4)^0 (3/4)^5
= 1 - 10 (9/1024) - 10 (27/1024) - 5 (81/1024) - 1 (243/1024)
= 1- (90 + 270 + 405 + 243)/1024
= 1 - 1008/1024
= 1 - 504/512
= 8/512
= 1/64
= 0.015625
 P: 773 oh okay i see so the formula i used will only work if k=1,2,3... (as u said before) so if i used the formula P(x=m)=P(1-P)^(m-1) and did 1-P(x=2)-P(x=1) =1-(1/3)(2/3)-(1/3)(2/3)^0 =4/9 would that work?
 Sci Advisor HW Helper P: 2,481 Shouldn't x=3 be part of your formula?
 P: 773 no it should only be for x=2 and x=1 as these are the ones strictly less than three
 Sci Advisor HW Helper P: 2,481 Your OP stated "P(x(>or = to)4)," doesn't that exclude x=3?

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