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?