# Binomial and geometric distributions

1. Dec 29, 2007

### sara_87

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.
Thank you very much

2. Dec 29, 2007

### HallsofIvy

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.

3. Dec 30, 2007

### sara_87

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(<or=to)2)=1-P(x=2)-P(x=1)-P(x=0)
=1-(1/3)(2/3)-(1/3)(2/3)^0-(1/3)(2/3)^-1=-1/18
thank u very much

4. Jan 2, 2008

### EnumaElish

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

5. Jan 9, 2008

### sara_87

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.

6. Jan 9, 2008

### EnumaElish

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

Last edited: Jan 9, 2008
7. Jan 9, 2008

### sara_87

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?

8. Jan 11, 2008

### EnumaElish

Shouldn't x=3 be part of your formula?

9. Jan 12, 2008

### sara_87

no it should only be for x=2 and x=1 as these are the ones strictly less than three

10. Jan 14, 2008

### EnumaElish

Your OP stated "P(x(>or = to)4)," doesn't that exclude x=3?