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Lets say, few independent events are happening. $E_1, E_2,...,E_n$. The probability of each of these events happening is given as $P_1,P_2,...,P_n$. Each of these events carry weightage, say $W_1,W_2,...,W_n$ respectively. I tried building a tree like the one shown here But its highly inefficient, for even small value of n, eg: $200$, the number of leaf nodes will be $2^{200}$.

Now, how can I find the probability of getting at least weight $W$?


Probabilities = [0.2, 0.8]
Weights       = [3, 5]

It means that, for the first event, there are 20% chances for that to happen. If it happens, the weightage will become 3, 0 otherwise. In the second case, there are 80% chances for that to happen. If it happens, the weightage will become 5, 0 otherwise.

Now if I want to find the probability of getting atleast Weight 4 would be like this

0.2 * 0.8 = 0.16 -> Total weight 8 (1)
0.8 * 0.8 = 0.64 -> Total weight 5 (2)
0.2 * 0.2 = 0.04 -> Total weight 3 (3)
0.8 * 0.2 = 0.16 -> Total weight 0 (4)

So, the total probability of getting atleast 4 weight is 0.16 (From (1)) + 0.64 (From (2)) = 0.80


The sum of all the probabilities $P_i$ need not be 1 always

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For me your question is very unclear. What do you exactly mean by "the probability of getting at least weight $W$"? And whether there is dependency between the singular events/probabilities? – al-Hwarizmi Aug 5 '13 at 10:10
@al-Hwarizmi Please check now. I have included an example. – thefourtheye Aug 5 '13 at 10:41
yes thanks, would it be possible that you make also clear what is functional relation between weights and events/probabilities? Or are they arbitrarily assigned numbers? – al-Hwarizmi Aug 5 '13 at 11:05
Yes. They are arbitrarily assigned numbers, there is no relationship between the probabilities and the weights and W would be half of sum of all the weights. – thefourtheye Aug 5 '13 at 11:14
In that case $W_i$ are identical with events and you do not need $E_i$, so you can directly associate to every probability directly the event weight. If the sum of all probabilities commonly $1$ then your $W$ is the average/mean. – al-Hwarizmi Aug 5 '13 at 11:23

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