What is the cumulative probability? - page 2
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Why isn't it certain?
Bull says: -Event X will happen with a 35% probability.
The bear says: -No. Event X will happen with a 51% probability.
Of course I'm going to believe the Bull. But how much should I believe him? After all, witchdoctors don't have definitively vague predictions. (Foggy is 50/50).
The arithmetic average has to be calculated here.
There is not enough data for a solution.
For example, the conditions are:
-if a man has a ring on the ring finger of his right hand, he is married p=0.5 (women are married)
-any man is married with p=0.5 (there are singles, children, widowers)
but if both conditions are met - a man has a ring on his right ring finger, he is married. The probability of such an event is close to 1. That is, the probability p(X/A) and p(X/B) cannot be calculated from the probabilities p(X/AB)
The formula p(x) = 1 - (1-p(A))*(1-p(B)) for two consecutive independent events, and the result is the probability that at least one of events A or B will occur. For example, the probability of hitting an enemy missile with the first line of defense =0.7, with the second line of defense 0.5. What is the probability of hitting one of the lines? p=1-(1-0.7)*(1-0.5)=0.85
In the case of dependent events, we need conditional probabilities in the formula, but that's still not it. It's all about calculating the probability of at least one event occurring in successive outcomes.
Also, in the case of the market there is such a thing as robustness, which results in the problem having a different solution.
For example, from New Market Magicians (Erckhardt):
"...Are there other practical implications of robust methods that would differ from the results of studies assuming a normal probability distribution?
- An important application concerns the situation where you have multiple indicators for a particular market. The question arises: how to combine several indicators in the most efficient way? Based on certain precise statistical measurements, it is possible to assign weights to different indicators. However, the choice of weights assigned to each indicator is often subjective.
You will find in the robust statistics literature that in most cases the best strategy is not to weight, but to assign a value of 1 or 0 to each indicator. In other words, to accept or reject an indicator. If an indicator is good enough to be used in principle, it is also good enough to be assigned a weight equal to the others. And if it does not meet this standard, it is not worth bothering with.
The same principle applies to the selection of trades. How do you best allocate your assets to different trades? Again, I will argue that the allocation should be even. Either the trade idea is good enough to execute - in which case it should be executed in full - or it is not worthy of attention at all."
In your first example the number of events is discrete. More precisely: there are only three of them (unmarried without a ring, unmarried with a ring, married with a ring). That's why you get the corresponding results. I was referring to the analogue series.
For the second example I might add that indeed the problem can be understood in different ways. I meant: one missile flies over the southern boundary, another missile flies over the northern boundary. What is the probability of these missiles hitting both milestones. (Each line has a missile, and you need a total probability).
As far as weight is concerned, the weight of A is equal to the weight of B.
Here the arithmetic mean should be counted.
Probabilities of 100% and 0% do not do it.
Why not....There's another example!!!
Given: - a car with a maximum speed of 40 km/hour
- asphalt
-ground
When the car travels on asphalt, its speed is P(A)=0.4 or 40
When the car is driving on the ground, its speed is P(B)=0.2 or 20
Conclusion:
If the car were to drive on a dirt road, its speed would be 30 km. or P(A && B) =0.3
In your first example, there are a discrete number of events. To be more precise: there are only three (unmarried without a ring, unmarried with a ring, married with a ring). That's why you get the corresponding results. I was referring to the analogue series.
For the second example I might add that indeed the problem can be understood in different ways. I meant: one missile flies over the southern boundary, another missile flies over the northern boundary. What is the probability of these missiles hitting both milestones. (Each line has a missile, and you need a total probability).
As for weight, the weight of A is equal to the weight of B.
No. I wrote missiles wrong. That, of course, is also an option, but the wrong one. I can't think of anything about missiles.
Why not....There's another example!!!
Given: - a car with a maximum speed of 40 km/hour
- asphalt
-ground
When the car travels on asphalt, its speed is P(A)=0.4 or 40
When the car is driving on the ground, its speed is P(B)=0.2 or 20
Conclusion:
If the car travels on a dirt road, its speed will be 30 km. or P(A && B) =0.3
I'm not in the mood for jokes. Can you tell speed from probability?
Probabilities of 100% and 0% do not make it so.
Why? Petya says YES! and stomps his feet insisting he is right. Vasya also stomps his feet and claims NO!!! What will the observer think? He'll think it's 50-50.
Maybe we need to use some clever function of each opinion's participation in the overall vote.
Why? Petya says YES! and stomps his feet insisting he is right. Vasya also stomps his feet and says NO!!! What will the observer think? He'll think it's 50-50.
Maybe we need to use some clever function of each opinion's participation in the overall vote.
I find myself in an awkward position because I can't clearly express the goal in words. The situation you cite cannot happen in this case, because it is logically contradictory, or at most it can only happen once. For, after the key event X has passed, someone (either Petya or Vasya) will no longer be able to stomp their feet 100%. And I think you've already got the gist of it. And I am still in contemplation, how to express this problem more clearly through rockets or something else. Perhaps you will be able to formulate the problem's condition better.
I have a question for the mathematicians. Although it looks like an off-topic, it is applicable to MTS.
Problem:
Let there be an event X whose probability of occurrence is equally dependent separately on two events A and B independent of each other.
If the probability of A-dependent event X is P(A)=0.4,
and the probability of an event X depending on B is P(B)=0.2,
then question:
What is the resulting probability of occurrence of event X: P(A && B) ???
P(not A) = 1 - A // Negation of event A
P(A | B) = P(A) + P(B) - P(A) * P(B) // If event A or event B or both occur simultaneously
P(A & B) = P(A) * P(B) // if event A and event B occur at the same time
P(A xor B) = P(A) + P(B) - 2 * P(A) * P(B) // If only one of the events A or B occurs
Assuming independence between P(A) and P(B)
P(not A) = 1 - A // Negation of event A
P(A | B) = P(A) + P(B) - P(A) * P(B) // If event A or event B or both occur simultaneously
P(A & B) = P(A) * P(B) // if event A and event B occur at the same time
P(A xor B) = P(A) + P(B) - 2 * P(A) * P(B) // When either A or B occur.
Assuming independence between P(A) and P(B)
Thanks for the formulas. Only in the output I don't get the correct answer by any of the formulas.
Below p1 and p2 are probability values in the range (0;1) not included:
1.1 If P(A)=1 and P(B)=p1, then P(A && B)=1.
1.2. If P(A)=p1 and P(B)=1, then P(A && B)=1.
2.1 If P(A)=0 and P(B)=p1, then P(A && B)=0.
2.2 If P(A)=p1 and P(B)=0, then P(A && B)=0.
3.1. If P(A)=p1 and P(B)=p1, then P(A && B)=p1.
3.2 If P(A)=0.5-p1/2 and P(B)=0.5+p1/2, then P(A && B)=0.5.
4.1 The option P(A)=0 and P(B)=1 is not possible.
4.2 The option P(A)=1 and P(B)=0 is impossible.
5. If P(A)=p1 and P(B)=p2, then P(A && B)=???