# Math Help - Conditional Probabilities

1. ## Conditional Probabilities

I have a question on a review that I just cannot crack.
It goes as follows:
"Event A occurs with probability .4. The conditional probability that A occurs given B occurs is .5, while the probability that A occurs given that B does not occur is .2. What is the conditional Probability that B occurs given A occurs."

So:
P(A) = .4
P(A|B)=.5
P(A|Not B) = .2.

I've tried multiplying things out. I just can't get it .
Can anyone help?

2. $P(A)=P(B)\cdot P(A|B) + P(\bar{B})\cdot P(A|\bar{B})$
$0.4=P(B)\cdot 0.5 + P(\bar{B})\cdot 0.2$
$0.4=0.5\cdot P(B) + 0.2 - 0.2\cdot P(B)$
so $P(B)=\frac{2}{3}$

$P(B|A)=\frac{P(A\cap B)}{P(B)}$
so from $P(A|B)=\frac{P(A\cap B)}{P(A)}=0.5$We know that $P(A\cap B)=0.5\cdot 0.4 = 0.2$so $P(B|A)=\frac{0.2}{2/3}=0.3$

3. The answer that the packet says is correct is "5/8." I don't know why though. T_T

4. Originally Posted by tylerm874
The answer that the packet says is correct is "5/8."
It appears that the packet's supplied answer is incorrect.

5. The thing is, that answer isn't even an option on the packet ....

And I haven't learned to do the problem that way, so I'm still confused.
Thanks for helping though.

6. Perhaps I should have used Bayes Rule, sorry its been a little while since i did some stat, damn summer holidays.. Although this still doesn't help you?
$

P(B|A)=\frac{P(B)\cdot P(A|B)}{P(A)}=\frac{2/3\cdot 0.5}{0.4}=\frac{5}{6}
$

7. Not an answer option either.
Thanks a lot though :/ .
Unfortunately my teacher hasn't taught us Bayes Rule.
She taught us the basic addition and multiplication rules, and taught us conditional earlier this week.

8. Originally Posted by tylerm874
I have a question on a review that I just cannot crack.
It goes as follows:
"Event A occurs with probability .4. The conditional probability that A occurs given B occurs is .5, while the probability that A occurs given that B does not occur is .2. What is the conditional Probability that B occurs given A occurs."

So:
P(A) = .4
P(A|B)=.5
P(A|Not B) = .2.

I've tried multiplying things out. I just can't get it .
Can anyone help?
Construct a Karnaugh Table:

$\begin{tabular}{l | c | c | c} & A & A' & \\ \hline B & a & b & a + b\\ B' & c & d & c + d \\ \hline & 0.4 & 0.6 & 1 \\ \end{tabular}$

From the given data you have:

Pr(A | B) = 1/2 => a/(a + b) = 1/2 => a = b .... (1)

Pr(A | B') = 1/5 => c/(c + d) = 1/5 => d = 4c .... (2)

a + c = 2/5 ..... (3)

b + d = 3/5 .... (4)

Solve equations (1), (2), (3) and (4) simultaneously for a, b, c and d and complete the above table. The answer to your question is Pr(B | A) = a/0.4 = 5a/2.

I get Pr(B | A) = 5/6.