a simple question but i can't solve

[sky31even]

known conditions:
$$P(b|a) = 0.6$$$$P(c|b) = 0.8$$$$P(c|\neg b) = 0.7$$
What I want to solve:
$$P(b\cup c|a )$$ (in the condition of "a" happen, at least one of b and c happen. )

I tried to break it up into:
$$P(b|a) + P(c|a) - P(bc|a)$$
but the final part $P(bc|a)$ stuck me so long ...

 

[AvgStudent]

known conditions:
$$P(b|a) = 0.6$$$$P(c|b) = 0.8$$$$P(c|\neg b) = 0.7$$
What I want to solve:
$$P(b\cup c|a )$$ (in the condition of "a" happen, at least one of b and c happen. )

I tried to break it up into:
$$P(b|a) + P(c|a) - P(bc|a)$$
but the final part $P(bc|a)$ stuck me so long ...

Hi @sky31even
Are the events independent?
I feel this is a smaller problem of a larger problem. The way you stated the problem seems incomplete.

 

[blamocur]

What about $P(c|a)$ -- have you figured that one out?

 

[pka]

known conditions:
$$P(b|a) = 0.6$$$$P(c|b) = 0.8$$$$P(c|\neg b) = 0.7$$
What I want to solve:
$$P(b\cup c|a )$$ (in the condition of "a" happen, at least one of b and c happen. )

I tried to break it up into:
$$P(b|a) + P(c|a) - P(bc|a)$$
but the final part $P(bc|a)$ stuck me so long ...

I think at you may have dropped us into the middle of a larger question without stating the whole.
Have you given the exact statement of the entire question?
Can you prove that $\mathcal{P}\left(A|B^c\right)=\dfrac{\mathcal{P}(A)-\mathcal{P}(A\cap B)}{1-\mathcal{P}(B)}~\bf?$

 

[sky31even]

I think at you may have dropped us into the middle of a larger question without stating the whole.
Have you given the exact statement of the entire question?
Can you prove that $\mathcal{P}\left(A|B^c\right)=\dfrac{\mathcal{P}(A)-\mathcal{P}(A\cap B)}{1-\mathcal{P}(B)}~\bf?$

I've confirmed twice with the guy give me this question and he said that's all the conditions we have.....

Not clarified those events are independent or not. All conditions listed.

 

[sky31even]

What about $P(c|a)$ -- have you figured that one out?

I broke it down to $P(c|b)*P(b|a)$ at the first time when I tried it , and that has be challenged somewhere else, now I dont know how to figure out that part either ?

 

[sky31even]

Hi @sky31even
Are the events independent?
I feel this is a smaller problem of a larger problem. The way you stated the problem seems incomplete.

there are no other condition = = I'm confused too.
The raw question is in natural language, and all I did is transforming them into the equation....
I've confirmed that indeed, they dont give any other condition.
BTW, not clarified those events are independent or not.