I'm terribly out of practice in even the most basic rules of probability, any help with the following question is appreciated. Essentially, I am having an issue with the application of the conditional probability calculaiton $P(A|B) = \frac{P(A \cap B)}{P(B)}$ towards finding the probability of $P(A|B^c)$. When I implement the formula, the $P(B^c)$'s are cancelling out in the numerator/denominaor, resulting in a probability value for $P(A|B^c)$ which betrays my intuition.

As an example, 3 events involving cupcakes:

$P(A) = 0.7$ - A cupcake passes frosting inspection.

$P(B) = 0.2$ - The cupcake is sent for re-frosting. For example, the cake part was alright, but the frosting failed to pass inspection, so the cupcake gets returned to the decorating table.

$P(C) = 0.1$ - The cupcake is thrown away entirely. It may have been burned or crushed or dropped on the floor, etc.

If we ask the question, Given that a cupcake is not sent for re-frosting, what is the probability it will be thrown away entirely?, and I fill in the terms of the formula by reflex, using the value of $P(B^C) = 0.8$, I get:

$$P(C|B^C) = \frac{P(C \cap B^C)}{P(B^C)} = \frac{(0.1 )(0.8)}{0.8} = 0.1$$

The solution above is of course incorrect, it is underestimating the true share of cupcakes thrown away in the subset of cupcakes not sent for re-frosting. In a situation where $10$ cupcakes are undergoing frosting inspection, the subset of $0.8$ which are not sent for re-frosting would experience throw away at the same rate as the rest of the population, at a probability of $0.1$.

Intuitively, and possibly with the help of a Venn Diagram, we know that the actual probability value we are looking for surrounding these 10 cupcakes is:

$$P(C|B^C) = \frac{1}{8} = 0.125$$

Where, how, and how badly have I butchered the application of this otherwise simple approach? Any help assisting me in applying formal probability rules towards this issue is greatly appreciated. The general hope for this question is that some formalized routine for this exists, and I am simply missing it. Thanks.

  • $\begingroup$ How did you come up with the equation $P(A\cap B^C)=(0.1)(0.8)$? What axiom of probability allows such an equation and how do you know it applies? (And how are those numbers related to the chances of $A$ and $B^C$?) $\endgroup$
    – whuber
    May 27 '16 at 20:30
  • $\begingroup$ Your statement and formula are mismatched. "Given that a cupcake is not sent for refrosting" is event $B^c$. "thrown away entirely" is event $C$. So you're interested in $P(C|B^c)$ $\endgroup$
    – Alex R.
    May 27 '16 at 20:32
  • $\begingroup$ @AlexR., sorry this is a typo - I will fix it now. $\endgroup$
    – Jim
    May 27 '16 at 20:33
  • $\begingroup$ @whuber Sorry for this, this was a typo. We were looking for $P(C|B^C)$ per Alex R's comment. $\endgroup$
    – Jim
    May 27 '16 at 20:35
  • $\begingroup$ Fine: but it's the rest of my question that is important. Please tell us what rules of probability you used at this juncture and why they apply. It would also be useful to display the Venn Diagram to demonstrate that it actually matches the information you have. $\endgroup$
    – whuber
    May 27 '16 at 20:38

Your "of course incorrect" comments are...incorrect. With 10 cupcakes, you subset to the 8 which are not sent for refrosting. By definition of independence, the chance these are thrown out is $0.1$, so that $0.8$ cupcakes are expected to be thrown out. However this $0.8$ is the NUMBER of cupcakes thrown out (out of 8), whereas you're interested in the probability of a given non-refrosted cupcake being thrown out, which is $0.8/8=0.1$.

Also you are declaring that 1 cupcake out of 8 was thrown out. This is wrong because $C$ is independent of $B$, which means that there's a $0.1$ chance a cupcake gets thrown out, regardless of whether or not it is re-frosted. In other words, out of the 10 cupcakes, you can't just say that 1 cupcake gets thrown out AND that cupcake was not refrosted.

  • $\begingroup$ Ok - there are obvious logical dead-ends in the question, however your note of the difference between the NUMBER of cupcakes thrown away and the PROBABILITY that they would be thrown away clarified things. I am seeing in your answer that the probability of one independent event happening does not change given a subset of the other events. The probability of trashing a cupcake does not change it is applied to a full set of cupcakes going for inspection or just a subset cupcakes coming in for inspection. I had asked a critical piece of this question incorrectly. Thanks. $\endgroup$
    – Jim
    May 27 '16 at 21:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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