1. ## Binomial distribution

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

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

use conditional distribution...

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

3. 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.

PS. Maiden rules!

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

5. 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?

6. 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.

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

8. 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?

9. 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?

10. Originally Posted by Plato
Look at the OP. Are those wrong?
Yet another typo from my side; I meant

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

11. You have yet another mistake.

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

12. 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?

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

This is the correct answer: $0.9078085$

Over and out.

14. Thanks!

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