1. ## Finding the probability

An Infinit sequence of independent trails is to be performed. Each trail results in a success with probability 'p' and failure '1-p' . What is the probability that

1) at least 1 success occurs in the first 'n' trails
2) exactly 'k' successes occurs in the first 'n' trails.
3) all trails result in successes?

2. Originally Posted by Stats
An Infinit sequence of independent trails is to be performed. Each trail results in a success with probability 'p' and failure '1-p' . What is the probability that

1) at least 1 success occurs in the first 'n' trails

Mr F says: 1 - Pr(X = 0).

2) exactly 'k' successes occurs in the first 'n' trails.

Mr F says: Pr(X = k).

3) all trails result in successes?

Mr F says: 0.
where X ~ Binomial(n, p).

And it's trial, not trail.

3. ## Is this correct?

Originally Posted by mr fantastic
where X ~ Binomial(n, p).

And it's trial, not trail.
Please let me know where if i am right or no?

Binomial P(x;n,$\displaystyle \theta$) = $\displaystyle \binom{n}{x} \theta ^x (1- \theta)^{n-x})$

Probability of atleast 1 success is

$\displaystyle 1-P(X=0)$ = $\displaystyle 1 -\binom{n}{0} \theta ^0 (1- \theta)^{n-0}$ = $\displaystyle 1 - (1 - \theta)^n$

Probability of k success occuring in first n trials is
$\displaystyle P(X=k) = \binom{n}{k} \theta^k (1- \theta)^{n-k}$

Probability of all trials being success is
$\displaystyle P(X=n) = \binom{n}{n} \theta^n (1- \theta)^{n-n} = 0$

4. The last one is off....
That's for n trials, but letting n go to infinity you will get 0 as long as $\displaystyle \theta\ne 1$
the rest looks ok

Originally Posted by Stats
Please let me know where if i am right or no?

Binomial P(x;n,$\displaystyle \theta$) = $\displaystyle \binom{n}{x} \theta ^x (1- \theta)^{n-x})$

Probability of atleast 1 success is

$\displaystyle 1-P(X=0)$ = $\displaystyle 1 -\binom{n}{0} \theta ^0 (1- \theta)^{n-0}$ = $\displaystyle 1 - (1 - \theta)^n$

Probability of k success occuring in first n trials is
$\displaystyle P(X=k) = \binom{n}{k} \theta^k (1- \theta)^{n-k}$

Probability of all trials being success is
$\displaystyle P(X=n) = \binom{n}{n} \theta^n (1- \theta)^{n-n} =\theta^n$

5. Originally Posted by Stats
Probability of all trials being success is
$\displaystyle P(X=n) = \binom{n}{n} \theta^n (1- \theta)^{n-n} = 0$
$\displaystyle P(X=n) = \binom{n}{n} \theta^n (1- \theta)^{n-n} =\theta^n$