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Slap2018-02-20 03:11:26
Mathematics
Slap, 2018-02-20 03:11:26

Probability Theory. What's wrong with her?

Suppose there are two independent events A and B. Also, there is an event C, which means the occurrence of at least one of these two events. The probability of this event is C \u003d A + B.
Let us denote the negation of an event by the sign "!" for lack of opportunity to underline the letter from above. So, event C will occur if and only if at least one event from A and B occurs. Therefore, the event C will not occur when both events A and B do not occur. That is, the notation C = A + B is equivalent to the notation !C = !A * !B. And also, 1 - C = !C, since these are opposite events.
Let's try after all the conclusions made to work with numbers.
Take, for example, the numbers 0.4 and 0.3. If you use stupid addition of probabilities, you get 0.7. If you do seemingly absolutely equivalent actions through the multiplication of opposite events, then you get the following ... We take opposite probabilities, 0.7 and 0.6. 0.7 * 0.6 = 0.42, so the answer is 0.58. The answers are different with different equivalent solutions, although this should not be. What's the matter?

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2 answer(s)
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Vladimir Olohtonov, 2018-02-20
@dr-Bright

"Therefore, event C will not occur when both events A and B do not occur." is the wrong transition.
Event C will occur when either event A or event B occurs, P(C) = P(A) + P(B) - P(A&B), since we have double-counted their intersection, event A&B.

0
0xD34F, 2018-02-20
@0xD34F

Suppose there are two independent events A and B. Also, there is an event C, which means the occurrence of at least one of these two events. The probability of this event is C = A + B.

Let's try.
Let A = 0.8, B = 0.7. Accordingly, in your opinion C = 0.8 + 0.7 = 1.5. Strange, I heard that the probability takes values ​​from 0 to 1. Where does 1.5 come from then?
You try to open, I don't know, some textbook on terver.

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