# Thread: Probability - Finding P(X = 0)

1. ## Probability - Finding P(X = 0)

Suppose that P(X=0)=1-P(X=1). If E[X]=3Var(X), nd P(X=0).
Well the answer is 1/3 which i cannot seem to get.

I calculate P(X = 1) = p
thus P(X = 0) = 1 - p

Var(x) = (1 - p)/p^2
E(x) = 1 / p

Solving this u get

p = 3(1 - p) and thus p = 3/4...so P(x = 0) = 1/4 and not 1/3 which is the answer.

Any help would be appreciated.

Thank-you

2. Originally Posted by AshleyT
Well the answer is 1/3 which i cannot seem to get.
I calculate P(X = 1) = p
thus P(X = 0) = 1 - p
Var(x) = (1 - p)/p^2
E(x) = 1 / p
Those expressions in red are wrong.
$\displaystyle E(X)=1(p)+0(1-p)=p$
and $\displaystyle Var(X)=E(X^2)-E^2(X)$.
If you solve that you will get $\displaystyle p=\frac{2}{3}$

3. Originally Posted by Plato
Those expressions in red are wrong.
$\displaystyle E(X)=1(p)+0(1-p)=p$
and $\displaystyle Var(X)=E(X^2)-E^2(X)$.
If you solve that you will get $\displaystyle p=\frac{2}{3}$
How comes they are wrong?
I was assuming from the above it would be a Geometric distribution and thus the equations in the red.
If it's not a geometric distribution, what would it be?
Thankyou

4. Originally Posted by AshleyT
How comes they are wrong?
I was assuming from the above it would be a Geometric distribution and thus the equations in the red. If it's not a geometric distribution, what would it be?
You have no right to assume anything about the distribution.
It is just a distribution on two points $\displaystyle X=0~\&~X=1$.

This question is designed to test your knowledge of the definitions of $\displaystyle E(X)~\&~Var(X).$

5. Originally Posted by Plato
You have no right to assume anything about the distribution.
It is just a distribution on two points $\displaystyle X=0~\&~X=1$.

This question is designed to test your knowledge of the definitions of $\displaystyle E(X)~\&~Var(X).$
Sorry. previous questions was on geometric which is why i assumed so... Sorry.

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### suppose that p(x=0)=1-p(x=1). If E(x)-3var(x).find p(x=0)

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