a simple question but i can't solve

[sky31even]

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

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

 

[AvgStudent]

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

I tried to break it up into:
[math]P(b|a) + P(c|a) - P(bc|a)[/math]
but the final part [imath]P(bc|a)[/imath] 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 [imath]P(c|a)[/imath] -- have you figured that one out?

 

[pka]

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

I tried to break it up into:
[math]P(b|a) + P(c|a) - P(bc|a)[/math]
but the final part [imath]P(bc|a)[/imath] 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 [imath]\mathcal{P}\left(A|B^c\right)=\dfrac{\mathcal{P}(A)-\mathcal{P}(A\cap B)}{1-\mathcal{P}(B)}~\bf?[/imath]

 

[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 [imath]\mathcal{P}\left(A|B^c\right)=\dfrac{\mathcal{P}(A)-\mathcal{P}(A\cap B)}{1-\mathcal{P}(B)}~\bf?[/imath]

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 [imath]P(c|a)[/imath] -- have you figured that one out?

I broke it down to [imath]P(c|b)*P(b|a)[/imath] 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.