# Decision for conditional probability instead of intersection of events

• B
• Peter_Newman

#### Peter_Newman

Hello,

I have a question about the following sentence and would appreciate if someone could explain how to read out the conditional probability here.

"Each microwave produced at factory A is defective with probability 0.05".

I understand the sentence as the intersection ##P(Defect \cap Factory A)## rather than the Conditional Probability.

But for solving the problem, the Conditional Probability ##P(Defect|Factory A)## is needed.

Reading the sentence, what clue is there that it is a conditional probability and not an intersection?

Hello,

I have a question about the following sentence and would appreciate if someone could explain how to read out the conditional probability here.

"Each microwave produced at factory A is defective with probability 0.05".

I understand the sentence as the intersection ##P(Defect \cap Factory A)## rather than the Conditional Probability.

But for solving the problem, the Conditional Probability ##P(Defect|Factory A)## is needed.

Reading the sentence, what clue is there that it is a conditional probability and not an intersection?
Not quite. This is actually quite subtle.

If we assume that there is only factory A under consideration, then we have:
$$P(Defect|A) = P(Defect \cap A) = 0.05$$That's because ##P(A) = 1##.

If, however, we assume there is also a factory ##B##, then
$$P(Defect|A) = 0.05$$But$$P(Defect \cap A) = P(Defect|A)P(A) \ne 0.05$$

I don't know that there is one particular way to recognize what they are asking, but to me it is pretty clear. Let me just write how I would say things:

##P(Defect|A)## "The probability that a microwave from factory A is defective"

##P(Defect \cap A)## "The probability that a microwave is both from factory A and is also defective"

##P(A|Defect)## "The probability that a defective microwave is from factory A"

• Peter_Newman and PeroK
"Each microwave produced at factory A is defective with probability 0.05".

Reading the sentence, what clue is there that it is a conditional probability and not an intersection?

The sentence, in isolation, does not define a "probability space", so it does not reveal whether it refers to a conditional probability. Only the whole context of the problem would make that issue clear.

Considering that probability is a number that is assigned to a set of outcomes, the probabilities ##P(D \cap A)## and ##P(D | A)## both refer to assigning a probability to the set ##D \cap A##. The interpretations of the two probabilities differ with respect to the probability space under consideration.

For ##P(D \cap A)## we are considering some set ##S## of outcomes such that ##P(S) = 1## and where sets ##D## and ##A## are subsets of ##S##. (The set ##A## need not be all of ##S##. For example, as @PeroK says, there might be several factories that may produce defective items.)

For ##P(D |A)##, we are considering a probability space consisting only of outcomes in the set ##A##. In that assignment of probabilities, ##P(A) = 1##. But this is misleading notation since it suggests that there is only one function ##P## that assigns probabilities. It would be better to denote the function that assigns probabilities to the set ##S## as ##P_S## and the different function that assigns (nonzero) probabilities only to outcomes in the set ##A## as ##P_A##. So we have ##P_S(S)=1## and ##P_A(A) = 1 ##.

The formula ##P(D \cap A) = P(D | A) P(A)## is slightly misleading because the "##P##" appears to denote a single function. In terms of the sets ##S## and ##A## mentioned above, a better notation would be ##P_S(D \cap A) = P_A(D) P_S(A) ##

That notation makes it clear that conditional probability is a sophisticated concept that involves two different probability spaces. Many students make the mistake of thinking that a probability problem must involve only one function that assigns probability to a set of outcomes. By that way of thinking it is correct to refer to "the probability" of a set of outcomes because there is only one such probability. However, many problems involve several different probability spaces.