# Binomial distribution

• Jan 30th 2011, 10:42 AM
SvendMortensen
Binomial distribution
Hi all.

This is my first forum post, so please forgive me if I have done something wrong. (Headbang)

Now for my problem:
A stochastic variable $X$ is binomial distributed with number of trials $n = 18$ and success probability in each trial $p = 0.4$.
Calculate the probability that $X$ is less than $11$ when given the fact that $X$ is greather than or equal to $7$.

My suggestion:
$P(7 \leq X < 11) = P(X > 11) - P(X \geq 7) = P(X \geq 11) - P(X \geq 7) \simeq 0.4163$
• Jan 30th 2011, 10:58 AM
harish21
X is less than 11 when given the fact that X is greater than or equal to 7 .

use conditional distribution...

$P(X<11|X \geq 7)$
• Jan 30th 2011, 11:46 AM
SvendMortensen
Quote:

Originally Posted by harish21
X is less than when given the fact that X is greather than or equal to .

use conditional distribution...

$P(X<11|X \geq 7)$

Thank you for your time! I am really bad at probability and statistics, so I am fare from sure of what to do. I guess I have to use Bayes' formula:

$P(X < 11 \mid X \geq 7) = P(X \geq 7 \mid X < 11) \cdot \frac{P(X < 11)}{P(X \geq 7)} = P(X \geq 7 \mid X < 11) \cdot \frac{1 - P(X \geq 11)}{P(X \geq 7)}$

but how do I calculate $P(X \geq 7 \mid X < 11)$?

If it isn't correct, I would very much appreciate if you would please tell me the solution.

Thank you in advance!

PS. Maiden rules! (Rock)
• Jan 30th 2011, 11:53 AM
Plato
$P(X<11|X\ge 7)=\dfrac{P(7\le X <11)}{P(X\ge 7)}$
• Jan 30th 2011, 12:09 PM
SvendMortensen
Quote:

Originally Posted by Plato
$P(X<11|X\ge 7)=\dfrac{P(7\le X <11)}{P(X\ge 7)}$

So
$P(X < 11 \mid X \geq 7) = \dfrac{P(7 \leq X < 11)}{P(X \geq 7)} = \dfrac{P(X \geq 11) - P(X \geq 7)}{P(X \geq 7)} \simeq 0.7388$
or what?
• Jan 30th 2011, 12:32 PM
mr fantastic
Quote:

Originally Posted by SvendMortensen
So
$P(X < 11 \mid X \geq 7) = \dfrac{P(7 \leq X < 11)}{P(X \geq 7)} = \dfrac{P(X \geq 11) - P(X \geq 7)}{P(X \geq 7)} \simeq 0.7388$
or what?

No. You should know that $\Pr(7 \leq X < 11)$ is equal to $\Pr(X < 11) - \Pr(X < 7)$.

Edit: Corrected typo pointed out by Plato in order to avoid confusing the OP.
• Jan 30th 2011, 12:34 PM
Plato
Right idea. Wrong concept.
$P(7\le X < 11)=P(X<11)-P(X<7)$
• Jan 30th 2011, 12:55 PM
SvendMortensen
Quote:

Originally Posted by Plato
Right idea. Wrong concept.
$P(7\le X < 11)=P(X<11)-P(X<7)$

Of course--a typo from my side; is

$P(X < 11 \mid X \geq 7) = \dfrac{P(X \leq 11) - P(X \leq 7)}{P(X \leq 7)} \simeq \dfrac{0.9797-0.5634}{0.5634} \simeq 0.7388$

correct?
• Jan 30th 2011, 01:01 PM
Plato
Quote:

Originally Posted by SvendMortensen
Now for my problem:
A stochastic variable $X$ is binomial distributed with number of trials $n = 18$ and success probability in each trial $p = 0.4$.
Calculate the probability that $X$ is less than $$11$$ when given the fact that $X$ is greather than or equal to $7$.

Look at the OP. Are those wrong?
• Jan 30th 2011, 01:30 PM
SvendMortensen
Quote:

Originally Posted by Plato
Look at the OP. Are those wrong?

Yet another typo from my side; I meant

$P(X < 11 \mid X \geq 7) = \dfrac{P(7 \leq X < 11)}{P(X \geq 7)} = \dfrac{P(X \leq 11) - P(X \leq 7)}{1 - P(X \leq 7)} \simeq \dfrac{0.9797-0.5634}{1-0.5634} \simeq 0.9535$
• Jan 30th 2011, 01:35 PM
Plato
You have yet another mistake.

$P(X\ge 7)=1-P(X<7)$
• Jan 30th 2011, 01:51 PM
SvendMortensen
Now I am (even more) confused! If my suggestion in my latest post isn't correct, I don't know that to do! Is it then

$P(X < 11 \mid X \geq 7) = \dfrac{P(7 \leq X < 11)}{P(X \geq 7)} = \dfrac{P(X < 11) - P(X \leq 7)}{1 - P(X < 7)} = \dfrac{P(X \leq 10) - P(X \leq 7)}{1 - P(X \leq 6)} \simeq \dfrac{0.9424-0.5634}{1-0.3743} \simeq 0.6056$

or am I doing yet another thing wrong?
• Jan 30th 2011, 02:03 PM
Plato
$P(7\le X<11)=P(X<11)-P(X<7)$

This is the correct answer: $0.9078085$

Over and out.
• Jan 30th 2011, 02:19 PM
SvendMortensen
Thanks!

$P(X < 11 \mid X \geq 7) = \dfrac{P(X \leq 10) - P(X \leq 6)}{1 - P(X \leq 6)} \simeq 0.9078$