I'm getting a bit dumb on the probabilities. - page 8
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Look here, namesake, this is a simulation of a maths game (4 dice), a hundred million games:
Result:
The simulation of a uniform distribution from 1 to 6 is not very accurate, but the error is small, not more than 0.001.
The S.c.o. of the deviation of frequency from probability is MathSqrt( npq ) / n ~ 1/20000, so here too you have no chance of getting close to p=2/3.
The exact probability value (or... er... m.o. frequency) is 1 - (5/6)^4 ~ 0.517747.
Wow!
Need to read up on Bernoulli and solve some problems urgently. All forgotten...
PS: Your another namesake )
0.517747 is the probability of one in four throws, as far as my stupid brain goes. Or one throw with four cubes?
Six edges, 1 or 4 throws with 4 or 1 cubes.
0.517747 chick is like that.
How do you get the total balance from here?
I.e. one. 6 4 1 0.517747 times divide and add?
0.517747 is the probability of one in four throws, as far as my stupid brain goes. Or one throw with four cubes?
Six edges, 1 or 4 throws with 4 or 1 cubes.
0.517747 chick is like that.
How do you get the total balance from here?
I.e. one. 6 4 1 0.517747 times divide and add?
alexeymosc, you beat me to it, I'm erasing my answer.
It's all right, Alexei. The question was not for me personally, as I understood it.
2 Dersu: But what is the overall balance, I don't understand shit. What do you mean by that?
Sorry to interrupt the scientific debate, but going back to the original problem: there was no additional "if it rains on one day and it's dry on the rest of the days" in the problem. So no need to make them up. You are interested in the probability of rain on at least one day, and you are not interested in what happens on other days.
Well, you need to formulate the problem specifically, then there will be nothing to invent. And since your original wording is ambiguous, you may think or guess, but no one here has telepathic powers.
If the probability of rainfall is at least one day out of three, i.e. there cannot be a three-day drought, then: 1 - 0.9^3 = 0.271, i.e. subtract from the full probability the probability of three consecutive days without rainfall
4-online: 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?
This is your problem. As you can see, it wasn't what you just wrote, but rather like the "rain only on one day out of three" condition.
Now to the point: you did your calculations correctly in the first post.
If directly, the reasoning is as follows: count separately the probability of the events "rain on one day only", "rain on exactly two days", "rain three days out of three" and sum up.
C(3,1)*p^1*(1-p)^2 + C(3,2)*p^2*(1-p)^1 + C(3,3)*p^3*(1-p)^0 =
3*0.1*0.9^2 + 3*0.1^2*0.9^1 + 1*0.1^3*0.9^0 =
0.243 + 0.027 + 0.001 = 0.271.
But it is easier to do it the first way, because the sum of all probabilities is 1.
This is your problem. As you can see, it wasn't what you just wrote, it was more like the "rain on only one day out of three" condition.
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"Only" wasn't there. And there were no additional conditions. So it was more likely to be understood as "on any one day and the rest doesn't matter, and if it doesn't, then there's no need to write anything about it". But I agree that it is better to decipher such tasks in as much detail as possible.
Now to the point: you calculated everything correctly in the first post.
If directly, the reasoning is as follows: count separately the probability of the events "rain on one day only", "rain on exactly two days", "rain three days out of three" and sum up.
C(3,1)*p^1*(1-p)^2 + C(3,2)*p^2*(1-p)^1 + C(3,3)*p^3*(1-p)^0 =
3*0.1*0.9^2 + 3*0.1^2*0.9^1 + 1*0.1^3*0.9^0 =
0.243 + 0.027 + 0.001 = 0.271.
But it is easier to do it by the first method, because the sum of all probabilities equals 1.
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Got it. Thank you.