# Conditional Probability

• Apr 5th 2014, 09:07 PM
ebehop
Conditional Probability
Hi there, I'm currently trying to figure out a probability question..

A patient with a failing heart has been told that the probability of having a heart transplant is 0.38, the probability of surviving more than 12 months given that a transplant is performed is 0.85, and the probability of surviving more than 12 months given that a transplant is not performed is 0.30

determine the probability that a transplant was performed given that the patient survived more than 12 months.

I figured it's a conditional probability thing, so the thought the equation would be:

Pr (A|B) = Pr(A intersect B)/Pr (B)

with, A being the patient has a transplant and B being the patient surviving >12 months

One of the prior questions is find the probability the patient survives more than 12 months, which I calculated as 0.509, which I think plays a part in it but I can’t seem to calculate a right answer….A friend gave me his and he got .635, but I can't figure out how as it doesn't fit into the equation above...Just wondering if anyone could help at all?

Thanks http://www.freemathhelp.com/forum/im...icon_smile.gif
• Apr 5th 2014, 09:21 PM
romsek
Re: Conditional Probability
Let \$T\$ stand for a transplant, and \$S\$ stand for surviving 12 months or more.

\$Pr[T|S]=\dfrac{Pr[S|T]Pr[T]}{Pr[S]}\$

\$Pr[S]=Pr[S|T]Pr[T]+Pr[S|!T]Pr[!T]\$

\$Pr[T]=0.38\$

\$Pr[!T]=1-Pr[T]=1-0.38=0.62\$

\$Pr[S]=(0.85)(0.38)+(0.30)(0.62)=0.509\$

\$Pr[T|S] = \dfrac{(0.85)(0.38)}{0.509}=0.634578\$
• Apr 5th 2014, 09:32 PM
ebehop
Re: Conditional Probability
Ahhhh thank you so much romsek, I can now see why I was so off with my original calculations. I Really really appreciate it.