Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
The rule does not fail, you are right, the "solution" makes a mishmash of stochastic independence. –  Did Sep 20 '12 at 17:41
2  
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. –  celtschk Sep 20 '12 at 17:41
    
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. –  mjqxxxx Sep 20 '12 at 19:01

1 Answer 1

up vote 2 down vote accepted

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.

share|improve this answer
    
Thank you for the detailed answer! –  xorxorxor Sep 20 '12 at 18:12
    
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. –  Michael Chernick Sep 20 '12 at 20:16
    
@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. –  Carl Morris Sep 20 '12 at 21:38

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.