B Decision for conditional probability instead of intersection of events

[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?

 

[PeroK]

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$$P(Defect \cap A) = P(Defect|A)P(A) \ne 0.05$$

 

[Dale]

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"

 

[Stephen Tashi]

"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.