1. ## Probability help

Hi

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
$P[0] = P[!B!C] = (1-\theta)^2$
$P[1] = P[B!C \to B!C] + P[!BC \to !B C] = [\theta(1-\theta)](1-\theta) + [\theta(1-\theta)](1-\theta) = 2\theta (1-\theta)^2$
$P[2] = P[BC] + P[B!C \to BC] + P[!BC \to BC] = \theta^2 + [\theta(1-\theta)]\theta + [\theta(1-\theta)]\theta = \theta^2(1 + 2(1-\theta)) = \theta^2(3-2\theta)$
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
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.

9. ## Re: Probability help

Originally Posted by romsek
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 "