# Conditional probability.

• Apr 1st 2011, 12:59 AM
staralfur
Conditional probability.
Hey guys, I'm having trouble with figuring this out:

What is P(V) given the following information?

P(F|V) = 0.02, P(F^c|V) = 0.98 P(F|V^c) = 0.4, P(F) = 0.2, P(F^c) = 0.8

Where I've written F^c and V^c I mean 'not F' and 'not V'.

I tried figuring it out using the probability rules, i.e.
P(V) = P(F and V) + P(not F and V)
but it seems like I don't have enough information to figure it out. Am I missing something?

Thanks.
• Apr 1st 2011, 03:37 AM
SpringFan25
This is probably not the simplest way, but consider focusing on P(F):
P(F) = p(F and Not V) + P(F and V)

$P(F)=P(V^c)P(F|V^c) + P(V)P(F|V)$

$0.2=P(V^c)0.4 + P(V)0.02$

$0.2=(1-P(V))0.4 + P(V)0.02$
...
$P(V)=\frac{10}{19}$
• Apr 1st 2011, 03:56 AM
mr fantastic
Quote:

Originally Posted by staralfur
Hey guys, I'm having trouble with figuring this out:

What is P(V) given the following information?

P(F|V) = 0.02, P(F^c|V) = 0.98 P(F|V^c) = 0.4, P(F) = 0.2, P(F^c) = 0.8

Where I've written F^c and V^c I mean 'not F' and 'not V'.

I tried figuring it out using the probability rules, i.e.
P(V) = P(F and V) + P(not F and V)
but it seems like I don't have enough information to figure it out. Am I missing something?

Thanks.

Drawing a tree diagram (the first two brances are V and V') makes it triival:

Let Pr(V) = x so that Pr(V') = 1 - x.

Then (x)(0.02) + (1 - x)(0.4) = 0.2 => x = 10/19.
• Apr 1st 2011, 05:58 AM
staralfur
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• Apr 1st 2011, 06:03 AM
staralfur