Is this a trick question in probability?

Question

Calculate the probability of $P(A|B')$ given that $A$ and $B$ are independent and $P(A)=\frac{1}{3}$

My intuition tells me since $P(A)= \frac{1}{3}$ and $A$ does not depend on $B$ the answer should just be $\frac{1}{3}$ and even more since I looked at this answer.

Is this really just a trick question or is there more to it?

Is $B'$ a typo or does it mean something? Possibly the non-occurence of $B$? In that case you answer is still right of course. — Kasper, Jan 24 '19 at 22:30
@kmm $B', B^c, \neg B$ and others are common ways to write the complementary event to $B$, i.e. $\Omega \setminus B$ where $\Omega$ is the sample space. — JMoravitz, Jan 24 '19 at 22:31
No need to tell me. Curious, I've seen $B^c$, $\lnot B$, $\overline{B}$ but never before $B'$ — Kasper, Jan 27 '19 at 17:10

score 2 · Accepted Answer · edited Jan 24 '19 at 22:27

2

You need to use Bayes theorem for these kind of exercises :

$$\begin{align}P(A \mid B) &= \frac{P(A \cap B)}{P(B)} \\&= P(A)& \text{since A et B are independent}\end{align}$$

And you are right here intuition is quite clear, yet be carefull with intuition when working with conditional porbabilities since some situations can be somehow tricky. But yes that's it.

edited Jan 24 '19 at 22:27

Graham Kemp

133,231

answered Jan 24 '19 at 22:22

Thinking

1,310

I was trying to use total probability $P(A)=P(A|B)P(B)+P(A|B')P(B')$ but I couldn't figure out how to solve for P(B) which would give me the answer. I quickly realized however, that this was likely a trick. – Colin Hicks Jan 24 '19 at 22:32

Is this a trick question in probability?

1 Answers1