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