# repeated application of binomial distribution on a set of random variables

I have four solutions which are termed: A1, A2, A3, A4. These are actually the results of a searching algorithm. I know that A1 is the best solution, A2 is next to A1, A3 is next to A2 and A4 is the worst solution. Thus they are arranged in descending order on the basis of their being best or worst. Let the probability of selection of A1 is 0.94. Using binomial distribution can I do the following:

     Probability


A1 ...... 0.94

A2 ...... ?

A3 ...... ?

A4 ...... ?

But since its binomial and A2, A3, A4 can be regarded false compared to A1. Thus P(A2 and A3 and A4) = 1- 0.94 = 0.06. The problem is to find P(A2), P(A3) and P(A4). Since A2 is best among A3 and A4 so the 0.94x0.06 should be its chances, Thus P(A2)=0.0564 Now P(A3 and A4) would be 0.06-0.0564 = 0.0036.

Again A3 is best compared to A4 so 0.94x0.0036=0.003384 should be the chances of A3. Thus P(A4) = 0.0036-0.003384=0.000216. To summarize this way I am able to calculate:

P(A1) = 0.94

P(A2) = 0.0564

P(A3) = 0.003384

P(A4) = 0.000216.

I just want to ask is this method correct?

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It doesn't look correct without some huge assumptions. If you are still confused, work on clarifying your question, and see the FAQ for other sites where clarified versions of this question might fit better. –  Douglas Zare May 21 '13 at 8:28
Actually I am breaking the problem into sub-problems. The first time having 0.94 probability for the first r.v. Then breaking the problem so this time having only thress variables but the probability of first is again 94% of the probabilities of the other two. The reason of this assumption is that the first variable in each sub-problem is the best one. –  Syed May 21 '13 at 8:32
Your reasoning seems logical to me, but even so it does not dispense the care suggested in Douglas Zare comment. Strictly from a practical point of view however maybe you can do an experiment: run your algorithm enough times to make it statistically significant and test whether the solutions probability profile agrees or not with your model. –  Fernando Pimentel May 21 '13 at 16:12