# Thread: Probability help

1. ## Probability help

Hi

Please find the question attached

I did attempt the question, thinking it was binomially distributed with a probability of theta.

For probability of 0, i thought it would be $\displaystyle {{3}\choose{0}}\theta^{0}1-\theta^{3}$

But this doesn't match the correct show that answer given in the question.

2. ## Re: Probability help

$P[0] = P[!B!C] = (1-\theta)^2$

$P[1] = P[B!C \to B!C] + P[!BC \to !B C] = [\theta(1-\theta)](1-\theta) + [\theta(1-\theta)](1-\theta) = 2\theta (1-\theta)^2$

$P[2] = P[BC] + P[B!C \to BC] + P[!BC \to BC] = \theta^2 + [\theta(1-\theta)]\theta + [\theta(1-\theta)]\theta = \theta^2(1 + 2(1-\theta)) = \theta^2(3-2\theta)$

3. ## Re: Probability help

Thank you so much for your help, much appreciated

Can you please explain, why the P(A) isn't included in any of your calculations above?

4. ## Re: Probability help

Originally Posted by cooltowns
Thank you so much for your help, much appreciated

Can you please explain, why the P(A) isn't included in any of your calculations above?
the whole problem is predicated upon $A$ getting sick. $P[A]=1$

when they refer to $P[0]$ they mean that neither $B$ nor $C$ got sick.

this is why they don't ask for $P[3]$, they are just interested in $B$ and $C$

5. ## Re: Probability help

Originally Posted by romsek
$P[0] = P[!B!C] = (1-\theta)^2$
$P[1] = P[B!C \to B!C] + P[!BC \to !B C] = [\theta(1-\theta)](1-\theta) + [\theta(1-\theta)](1-\theta) = 2\theta (1-\theta)^2$
$P[2] = P[BC] + P[B!C \to BC] + P[!BC \to BC] = \theta^2 + [\theta(1-\theta)]\theta + [\theta(1-\theta)]\theta = \theta^2(1 + 2(1-\theta)) = \theta^2(3-2\theta)$
Please explain the unusual notation.
I guess that !B means not B. But what does $\to$ mean? In mathematics it usually means to as in mapping.
Because probability is taught using set notation, shouldn't we try to use notation student understand?
Even though I have written text material on and taught probability thirty years, I have no idea what "$P[2] = P[BC] + P[B!C \to BC] + P[!BC \to BC] = \theta^2 + [\theta(1-\theta)]\theta + [\theta(1-\theta)]\theta = \theta^2(1 + 2(1-\theta)) = \theta^2(3-2\theta)$" means.

6. ## Re: Probability help

Originally Posted by Plato
Please explain the unusual notation.
I guess that !B means not B. But what does $\to$ mean? In mathematics it usually means to as in mapping.
Because probability is taught using set notation, shouldn't we try to use notation student understand?
Even though I have written text material on and taught probability thirty years, I have no idea what "$P[2] = P[BC] + P[B!C \to BC] + P[!BC \to BC] = \theta^2 + [\theta(1-\theta)]\theta + [\theta(1-\theta)]\theta = \theta^2(1 + 2(1-\theta)) = \theta^2(3-2\theta)$" means.
if you read the problem it says that there is a 2nd stage of the experiment.

$!BC \to !BC$ and similar means that the first stage resulted in B healthy and C sick, and remained that way in the 2nd stage.

It's not customary notation but I feel it fit this problem nicely and the OP understood exactly what I meant.

And at the end of the day helping the OP is what this place is about.

7. ## Re: Probability help

Originally Posted by romsek
It's not customary notation but I feel it fit this problem nicely and the OP understood exactly what I meant. And at the end of the day helping the OP is what this place is about.
Well, I absolutely agree with that. But if the notation is not customary and then you do not explain its meaning, how does that help the OP.

If you do understand standard notation, then post the solution in readable form please. Maybe then it will help the OP to see what you have done.

8. ## Re: Probability help

Originally Posted by Plato
Well, I absolutely agree with that. But if the notation is not customary and then you do not explain its meaning, how does that help the OP.

If you do understand standard notation, then post the solution in readable form please. Maybe then it will help the OP to see what you have done.
Post #3. OP Had no issues with my answer.

9. ## Re: Probability help

Originally Posted by romsek
Post #3. OP Had no issues with my answer.
Originally Posted by cooltowns
Thank you so much for your help, much appreciated
Can you please explain, why the P(A) isn't included in any of your calculations above?
@cooltowns,
Did you really understand the notations in post #2?

10. ## Re: Probability help

Originally Posted by Plato
@cooltowns,
Did you really understand the notations in post #2?
is it really so difficult to interpret

"Thank you so much for your help, much appreciated "

cooltowns is long gone having had his or her question answered.

11. ## Re: Probability help

romsek, thank you for your explanation about probability of A. Thank you for your help

@Plato, normally when it comes to probability I don't really see the ! notation. I've only see ! in factorials. However, I did understand what Romsek was talking about and !B meant that B was healthy (not sick).

Thanks guys