Probability of Independent Events

**yzhao5** · 08-02-2012 04:17 PM

Hi,

If hypothetical events A and B are independent, with P(B/A)=.4, and P(A)=.3

What is the P(B)? What formula would you use to figure this out?

Thanks.

**Dason** · 08-02-2012 04:19 PM

Hi!

We are glad that you posted here! This looks like a homework question though. Our homework help policy can be found here. We mainly just want to see what you have tried so far and that you have put some effort into the problem. I would also suggest checking out this thread for some guidelines on smart posting behavior that can help you get answers that are better much more quickly.

**yzhao5** · 08-02-2012 04:27 PM

P(B/A)=P(A intersection B)/P(A)?

am I on the right track?

**Dason** · 08-02-2012 04:32 PM

It's certainly a start. What do you know about $P(A \cap B)$ if A and B are independent?

**yzhao5** · 08-02-2012 04:34 PM

P(A intersection B) cannot exist if A and B are independent.

**Dason** · 08-02-2012 04:40 PM

Not quite true. If A and B are mutually exclusive then P(A intersect B) = 0. However, mutually exclusive and independent aren't the same thing. Maybe you want to go back and look at what you have for the definition of independent?

**yzhao5** · 08-02-2012 04:49 PM

Well if they are independent, than by definition P(B/A) has to equal P(B), thus P(B) also equals .4?

**Dason** · 08-02-2012 04:53 PM

That's certainly one way to do it.

**yzhao5** · 08-02-2012 05:03 PM

lol. so in conclusion if P(A)=.3, P(B)=.4, than P(B/A)=.4 ?

thanks.

**derksheng** · 08-05-2012 12:39 PM

To set formal definitions: $B \backslash A : = \{ x : x \in B \land x \notin A \}$ and $A : = \{ x : x \in A \}$ .

We then have $B = (B \cap A) \cup (B \cap A^c)$ , then $B = (B \cap A)\cup (B \backslash A)$ .

By the independence of A and B, $P(B \cap A) = P(B)*P(A) \Rightarrow P(B) = P(B)*P(A) + P(B \backslash A)$ .

The answer follows; $P(B) = \frac{P(B \backslash A)}{1-P(A)} = \frac{0.4}{1-0.3}$ .

Don't rely on this being correct.

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