Here is the original question.

Of three possible events, event A is independent of the other two, and events B and C are mutually exclusive. The probabilities that the individual events A, B, and C will occur are 0.5, 0.3, and 0.2, respectively. What is the probability that both event A and event C will occur?

The answer to this question is:

Start with the “mutually exclusive” events, as this is the most restrictive statement. If event B happens, event C cannot happen. Likewise, if event C happens, event B cannot happen. It is possible that neither event B or C will happen, but they can’t both happen.

Consider the possibilities, starting with whether event B happens.

If event B occurs, event C cannot occur, so there is no way for both event A and event C to happen. (i.e. Probability of both A and C is zero if B occurs.) If event B does not occur, event C might happen, as might event A.

Thus, the probability that both event A and event C will occur is the probability that B will NOT happen, A will happen, and C will happen. {NOTE: "and" means multiply probabilities, "or" would mean add probabilities.}

P(A and C) = P(not B) × P(A) × P(C) P(A and C) = [1 – 0.3] × 0.5 × 0.2 = 0.7 × 0.5 × 0.2 = 0.07 = 7%

The correct answer is 7%

Source: http://www.manhattanprep.com/gre/ChallengeProblems/LastWeek/

My question is:

1) P(A,C) = 0.07 in this question. However, this is not P(A)*P(C)=0.10, despite A and C are independent. Why does the rule P(A,C) = P(A)*P(C) fail even though A and C are independent? Is there a certain restraint that applies to this rule?

2) Event B and C are not independent. However, the problem states that P(A,C,~B) = P(A)*P(C)*P(~B). I thought this was possible only if C and ~B are independent. Can you please explain if this is valid?

Your help is greatly appreciated. Have a wonderful day.

  • $\begingroup$ The rule does not fail, you are right, the "solution" makes a mishmash of stochastic independence. $\endgroup$
    – Did
    Sep 20, 2012 at 17:41
  • 2
    $\begingroup$ They really multiply P(not B) with P(C)? Since C implies not B (they are mutually exclusive!), P(C)=P(C and not B). "and" only means "multiply probabilities" if the events are independent. In short, their solution is wrong. $\endgroup$
    – celtschk
    Sep 20, 2012 at 17:41
  • $\begingroup$ I love solutions that make things more complicated than they really are. If $A$ and $C$ are independent, then $P(A \cap C)=P(A)P(C)$. That's not a "rule"; that's the definition of independence. The only way this solution could be defended is if "event A is independent of the other two" means something different than "event A is independent of event B, and event A is independent of event C". For instance, maybe event A is supposed to be independent of the event "B or C". If so, the problem doesn't make that clear. $\endgroup$
    – mjqxxxx
    Sep 20, 2012 at 19:01

1 Answer 1


The stated answer of 0.07 is wrong. I will endeavor to show the mistake in the answer's logic.

They are attempting to use the rule of total probability to calculate $P(A\cap C)$. (The expression $A\cap C$ is like saying $A$ and $C$ happen simultaneously.) This procedure is as follows

$$P(A\cap C) = P(A\cap C\cap B^c) + P(A\cap C\cap B) = P(A\cap C\cap B^c).$$

(Noted that $B^c$ is the same as $\tilde{}B$.) We can eliminate $P(A\cap C\cap B)$ because $C$ and $B$ are mutually exclusive. As the OP noted, $C$ and $B^c$ are not independent so we cannot use the product rule. This is where the quoted solution goes wrong.

The correct answer can be gotten directly via the product $P(A)P(C)$ by using independence, but I will also show how the calculation of $P(A\cap C\cap B^c)$ can be done using conditional probability. Using the definition of conditional probability, we have

$$P(A\cap C\cap B^c) = P(A | C\cap B^c)P(C\cap B^c) = P(A)P(C\cap B^c).$$

We were able to simplify $P(A | C\cap B^c)$ because $A$ is independent of $B$ and $C$. However, now we must calculate $P(C\cap B^c)$. We can rewrite this expression using the rule of total probability as

$$P(C\cap B^c) = P(C) - P(C\cap B) = P(C)$$

Because $B$ and $C$ are mutually exclusive, $P(C\cap B)=0$. And so we conclude that $P(A\cap C) = P(A)P(C)$ as given by the definition of independence.

For independent events, the calculation $P(A\cap C)=P(A)P(C)$ is always true. This result is given as a definition in most cases.

  • $\begingroup$ Thank you for the detailed answer! $\endgroup$
    – xorxorxor
    Sep 20, 2012 at 18:12
  • $\begingroup$ Welcome to the site Carl Morris. You may be interested in the CV site which is primarily on statistics and is where I spend most of my SE time. When I saw your name I had to check out your personal page. Did you know that there is a famous statistician at Harvard named Carl Morris? Probably you have heard that more often than you care to. Anyway I know Carl pretty well and when i saw your nice answer I thought that maybe he had joined SE. $\endgroup$ Sep 20, 2012 at 20:16
  • $\begingroup$ @MichaelChernick, you are actually one of the few that have mentioned my like named counterpart at Harvard, and you are certainly the first to actually know the other Carl :). I will take a look at the CV site. I am not very experienced in statistics, but its an area of growing interest to me. I look forward to learning more. SE is invaluable to me. $\endgroup$ Sep 20, 2012 at 21:38

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