Finding the probability

Stats · #1 $Old$ May 17th, 2009, 03:44 AM

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?

mr fantastic · #2 $Old$ May 17th, 2009, 08:16 AM

Quote:

Originally Posted by Stats $View Post$

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.

Stats · #3 $Old$ July 6th, 2009, 03:33 AM

Quote:

Originally Posted by mr fantastic $View Post$

where X ~ Binomial(n, p).

And it's trial, not trail.

Please let me know where if i am right or no?

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

Probability of atleast 1 success is

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

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

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

matheagle · #4 $Old$ July 6th, 2009, 09:59 AM

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

Quote:

Originally Posted by Stats $View Post$

Please let me know where if i am right or no?

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

Probability of atleast 1 success is

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

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

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

malaygoel · #5 $Old$ July 6th, 2009, 10:21 AM

Quote:

Originally Posted by Stats $View Post$

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

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