Binomial distribution

**SvendMortensen** · January 30th 2011, 09:42 AM

Hi all.

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

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:
$P(7 \leq X < 11) = P(X > 11) - P(X \geq 7) = P(X \geq 11) - P(X \geq 7) \simeq 0.4163$

**harish21** · January 30th 2011, 09:58 AM

X is less than 11 when given the fact that X is greater than or equal to 7 .

use conditional distribution...

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

**SvendMortensen** · January 30th 2011, 10:46 AM

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!

**Plato** · January 30th 2011, 10:53 AM

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

**SvendMortensen** · January 30th 2011, 11:09 AM

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?

**mr fantastic** · January 30th 2011, 11:32 AM

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.

**Plato** · January 30th 2011, 11:34 AM

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

**SvendMortensen** · January 30th 2011, 11:55 AM

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?

**Plato** · January 30th 2011, 12:01 PM

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 [tex]11[/MATh] when given the fact that $X$ is greather than or equal to $7$ .

Look at the OP. Are those wrong?

**SvendMortensen** · January 30th 2011, 12:30 PM

Originally Posted by Plato

Look at the OP. Are those wrong?

Yet another typo from my side; I meant

$P(X < 11 \mid X \geq 7) = \dfrac{P(7 \leq X < 11)}{P(X \geq 7)} = \dfrac{P(X \leq 11) - P(X \leq 7)}{1 - P(X \leq 7)} \simeq \dfrac{0.9797-0.5634}{1-0.5634} \simeq 0.9535$

**Plato** · January 30th 2011, 12:35 PM

You have yet another mistake.

$P(X\ge 7)=1-P(X<7)$

**SvendMortensen** · January 30th 2011, 12:51 PM

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?

**Plato** · January 30th 2011, 01:03 PM

$P(7\le X<11)=P(X<11)-P(X<7)$

This is the correct answer: $0.9078085$

Over and out.

**SvendMortensen** · January 30th 2011, 01:19 PM

Thanks!

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

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