If the case of 1 occurs and has no effect on the second case, then the two cases are independent of each other.

Mutually exclusive events has several related formulas, such as P(A+B)=PA+PB.

Or more intuitively, P(AB)=0, which means that two things cannot happen at the same time, and there is no intersection between them.

But independence is different. The most typical independence formula is P(AB)=PAPB.

In other words, two events have an intersection, and the size of this intersection is equal to the product of these two probabilities.

From the above points, it is easy to distinguish between mutual exclusion and independence. First, look at whether there is intersection, and then calculate whether PA*PB is equal to PAB.

In the poker example you mentioned, PAB is not equal to 0. The probability of drawing 1 A card is 1/52, and the probability of drawing a card B is also 1/52. There is no interaction between drawing A and drawing B, but the probability PAB of requesting event AB should be expressed as drawing back two cards and AB respectively. It's not what you think. PAB=0

Your view of P(A+B) in natural language should be interpreted as the probability that only 1 card is drawn and the card obtained is one of A and B, P(A+B)=PA+PB, thus it can be proved that PAB=0 has no intersection, (P(A+B)=PA+PB-PAB, and there is only PA+Pb here. )

Generally speaking, independence and mutual exclusion are two unrelated concepts. Under the guidance of an inertial thinking, many students often associate two events and compare them unconsciously, which leads to mistakes. Moreover, many questioners like to drag these two unrelated concepts together.

The key to dealing with this kind of problem is to look at PAB. If PAB=PAPB, then it must be independent.

As for mutual exclusion, we can see whether the events intersect.

Give you a few conclusions:

(1) If A and B are independent, they must not be mutually exclusive.

(2) If A and B are mutually exclusive, they must not be independent of each other.

(3) If A and B are not independent of each other, they may or may not be mutually exclusive.

(4) If A and B are not mutually exclusive, they may or may not be independent.

Just write (1) during the demonstration! Other examples can be understood.

Syndrome (1)

It is proved that if A and B are mutually exclusive, then PA=0 or PB=0 when PAB=0.

It contradicts what is known, so AB must not be mutually exclusive.

I revised it five times and finally answered your question. I hope it helps you. The key is to watch PAB.

Finally, I saw that you added the sentence 1 Judging from my 1 argument, the sentence you added is reasonable, but it is not standardized. Rather, it must not be mutually exclusive. If the implied conditions PA and Pb are not 0, you are right.