I'm having some trouble really understanding this formula when applied to a problem:
$ P(A \cap B) = P(A) + P(B) - P(A \cup B) $
The problem I'm given is this:
Suppose we draw one card at random from a standard deck of 52 cards. Let $A$ denote the event that the card is an ace. Let $B$ denote that the card is not a club. Find the probabilities of $P(A)$, $P(B)$, and $P(A \cap B^c)$
This is pretty easy and I can get the answer from just basic logic. The problem is I have to actually go through the formula he gives for $P(A \cap B)$ above and show what I'm doing.
Before going into it, I've come up with these answers:
$P(A) = 4/52$
$P(B) = 13/52 = 1/4$
$P(A \cap B^c) = 1/52$
I'm having a problem understanding how to do $P(A \cap B^c)$. I'm doing this:
$P(A \cap B^c) = P(A) + P(B^c) - P(A \cup B^c)$
$ = P(A) + (1 - P(B)) - P(A \cup (1 - P(B))$
$ = P(A) + (1 - P(B)) - (P(A) + P(B) - P(A \cap (1 - P(B))$ [since $P(A \cup B) = P(A) + P(B) - P(A \cap B)$]
and it gets more and more complicated as I go along trying to substitute things in. So what am I doing wrong with this?
$\endgroup$ 13 Answers
$\begingroup$In basic probability you need 3 things:
1)Drawing graphs (at start) to understand the sets better.
2)The formula $P(A\cap B)=P(A)+P(B)-P(A\cup B)$ which you can simply see from the graph.
3)The formula $P(A\cap B)=P(A)P(B)$ for disjoint sets (which you should apply in your case).
$\endgroup$ 2 $\begingroup$Instead of trying to reduce the computation to something else, notice that $P(A \cap B^c)$ is the probability that the drawn card is an ace AND is a club. Thus, $P(A\cap B^c) = P(\{\text{ace of clubs}\}) = 1/52.$
$\endgroup$ $\begingroup$One of the mistakes that you are making is trying to substitute the probability $1-P(B)$ for the set $B^c$. These things are simply not equivalent. It is easiest just to recognize the meaning of these sets and their probabilities. If you must use the equation, then it should be done like this:
Let $A$ denote the event that the card is an ace.
Let $B$ denote that the card is not a club.
Find the probabilities of $P(A)$, $P(B)$, and $P(A∩B^c)$
First, since there are four aces and $52$ possible cards, $P(A) = 4/52$
Next, since there are $13$ clubs, there are $52-13=39$ non-clubs. Thus, $P(B) = 39/52 = 3/4$
If $B$ is the event that the card is not a club, then $B^c$ is the event that the card is a club. Thus $P(B^c)=13/52$.
Now, we must calculate $P(A \cap B^c)= P(A) + P(B^c) - P(A \cup B^c)$. Then we need to know $P(A \cup B^c)$. This is the probability that the card is either an ace or a club. In general, we would find this by using the principle of inclusion-exclusion, which is just the above equation in the form $P(A \cup B^c) = P(A) + P(B^c) - P(A \cap B^c)$. We do this because it is easier to compute $P(A \cap B^c)$. However, one may also compute $P(A \cup B^c)$ by recognizing that if $A \cup B^c=C$ then $C^c$ is the set of events where we do not choose an ace or a club. Thus, $C^c$ is the set $A^c \cap B$ which is the number of outcomes which are both not aces and not clubs. There are $39$ cards which are not clubs, and $3$ of those which are aces. Thus $P(C^c) = 36/52$. Then $P(C) = 1 - P(C^c) = P(A \cup B^c)$ = 52/52 - 36/52 = 16/52.
Finally, $P(A \cap B^c)= P(A) + P(B^c) - P(A \cup B^c) = 4/52 + 13/52 - 16/52 = 1/52$. I hope this has been somewhat helpful.
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