I'm getting a bit dumb on the probabilities. - page 2
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It will be a pain to sort things out, because the topicwriter, and not the first, cannot clearly set out the conditions of the problem. It's useless and time-consuming to get anything out of them. That's why in my reply I give one of the possible variants of the problem's conditions.
And in general there is nothing to appeal to, because everything is calculated by the banal Bernoulli formula: the probability of one success in three trials.
All the more it's impossible to solve the wrong problem - the probability can't be 10% as there are only seven days... :)
Topikstarter - set the problem correctly and it will solve itself... :)
I once had to study Boolean algebra. This knowledge is still useful to this day. Remember, the logical union "or" in Boolean algebra is marked by the addition symbol (the "+" sign); the logical symbol "and" is marked by the multiplication sign "*". Actually, in symbolic logic these two operations are called "logical addition" and "logical multiplication". Well, these principles are also true for probability theory. Wherever event probabilities are joined by the conjunction "or", their probabilities should be added. And those that are joined by the conjunction "and" must be multiplied. Therefore:
В понедельник вероятность дождя равна 10%. Во вторник вероятность дождя равна 10%. В среду вероятность дождя равна 10%. Какова вероятность того, что дождь пойдет в один из этих трех дней?
10%+10%+10% = 30%
Note, however, that this is not a strict "or". If you want to calculate the probability of a strict "or" - the probability of rain on one and only one of these days - then the reasoning is different. Since the condition does not specify that the probability of rain must be calculated strictly, this or-conjunction on autopilot is calculated as not strict.
The problem is solved. The answer is 30 percent.
Equivalent problem. There is a die with numbers from 1 to 6. What is the probability that in 1 roll a 2 or a 3 will fall?
The probability of 2 = 1/6. Probability of getting a 3 = 1/6. Probability of 2 or 3 = 1/6+1/6 = 2/6 = 1/3.
The problem I just gave you does not allow you to see a situation where two twos, or a deuce and a three are rolled at the same time. To do this, take two dice and roll them at the same time. Then it works out:
probability of a deuce = 2/12.
Probability of falling three = 2/12
Probability of falling either or three = 2/12 + 2/12 = 4/12 - 1/3.
Why, you may ask, is this the case? It's simple. A cycle of three days is as simultaneous as rolling three dice. The cubes are just ten-sided and only one side of each cube is shaded. That's how the probability of the painted side = 10 percent.
Then the obvious question is: What is the probability of rain on one of the eleven days?
To answer this question, you need to know the probability of rain on each of the eleven days.
The other conditions do not change
Alexei, got it. :) It turns out that it will rain with a probability of 110 per cent. But we know that the full field of events always = 1 (100 per cent). So taking a sample of 11 days puts us beyond the limit of one. Something is not right here.
Oh, shit, but I'm right - when you connect the probabilities of events with a non-strict "or", their probabilities add up. You can't get away from that either. I'm missing something here.
I seem to have forgotten that the events must be able to fall at the same time.
My answer is that rainfall depends solely on the location of my umbrella (hood, car...).
If the wording of the problem is "on exactly one of the three days", then the answer is obvious. It is a Bernoulli scheme with probability p = 0.1 and probability q = 1 - p = 0.9.
And you have to calculate the probability of one success. Bernoulli's formula:
p = C(3,1) * p^1 * q^2 = 3!/(1!*2!) * 0.1 * 0.9^2 = 0.243.
Yura is right.
For the problem about 11 days under the same condition ("exactly one in 11 days") it is similar:
p = C(11,1) * p^1 * q^10 = 11!/(1!*10!) * 0.1 * 0.9^10 ~ 0.3836.
P.S. You correctly calculated the probability under the condition "at least once every three days".
The probability addition theorem for incompatible events:
P(A + B) = P(A) + P(B) - the probability of at least one of two incompatible events occurring as a result of an experiment is equal to the sum of the probabilities of these events.
In the case of calculating the probability of event C, which occurs either when event A occurs or when event B occurs, if A and B are not incompatible, the following theorem can be used:
2. the general theorem for the addition of probabilities:
P(C)=P(A)+P(B)-P(AB), where P(AB) is the probability of both event A and event B occurring simultaneously.
http://www.mathelp.spb.ru/book2/tv6.htm
I think it's only Thursday...