2
$\begingroup$

"Events A and B are independent. Find the missing probability."

P(B)=9/20
P(A|B)=1/5
P(A)=?

Can anyone help me solve this? I've completed other problems similar to this already...

Here is my thinking so far: Since P(A|B) was given, I could use this formula to find A: 

P(B|A)=P(A,B)/P(A)

This has already worked for a previous problem, but to work for this problem I would need it to give A instead of B, or something along the lines of that. Also, is there a notable difference between P(A|B) and P(B|A), and what does that do exactly...?

Thanks so much!

  • 0
    If $A,B$ are independent then $P(A\,|\,B)=P(A)$. – 2017-01-14
  • 0
    @lulu thank you for the quick reply! – 2017-01-14

2 Answers 2

1

Using the fact that the events are independent we have that $P(A,B) = P(A)P(B)$. Then we can use the definition of conditional probability

$$P(A\mid B) = P(A,B)/P(B) = \frac{P(A)P(B)}{P(B)} = P(A).$$

This is a simple consequence of indepdence. Whether or not event $B$ occurs does not effect the probability of event $A$ happening, therefore the conditional probability $P(A\mid B) = P(A)$.

  • 1
    So the answer would just be 1/5 then, the same as P(A|B)? @David – 2017-01-14
  • 0
    @meowsome correct – 2017-01-14
  • 0
    @meowsome Also to give a little bit more intuition about $P(A|B)$ we have that it is the probability of event $A$ happening given event $B$ happens. For an example where these might not be independent probabilities, let us consider event $A$ to be a team winning a football game, and event $B$ being the team's quarterback gets injured. Then the probability of the team winning *given* the quarterback gets injured is written $P(A|B)$. In this case, it might be reasonable to assume that the probability $P(A|B) < P(A)$ because the team depends on the quarterback for success. – 2017-01-14
  • 0
    Thank you so much for the awesome explanations! I think I understand most of how it works now. :) Wish my teacher explained stuff that way! – 2017-01-14
1

Since $A$ and $B$ are independent, $\operatorname{P}(A)$ is the same as $\operatorname{P}(A\mid B).$