rabindranath72 said:Yes, that was the cumulative. What we want is the survival function 1-P(X). See my previous post.
I am sorry, but your result is still wrong. I know this sounds cocky, but I know my method (which takes longer to do, but uses math that I can get most high school students to do) works, and produces the correct answer.
To try and explain in other words than the OP, hopefully making it easier to follow:
We want to know the probability of getting 4 successes before we get 2 failures. This can only happen within 5 attempts, as we will always have reached 2 failures or 4 successes by then.
To keep the math easy to understand I will calculate the probability of the two possible successful cases, and then add them together.
Case 1:
We get 4 successes in a row, wohoo!
0.55^4=0.0915
Case 2:
We get a failure in 1 of our 4 first rolls, but luckily succeed using our 5th roll.
0.55^4*0.45*4
I guess I should stop and explain this one... Out of 5 rolls 4 are successful. Those four rolls are described just as in Case 1. Then we take into account that one roll is a failure, which means we have to multiply the result by 0.45. Finally all we know as that 1 of the first 4 rolls failed. To take into account that there are 4 such cases, we multiply the result by 4.
Total result is 0.55^4+0.55^4*0.45*4=0.256125
Edit: I see that Eldorian, who is clearly my superior when it comes to math, reached exactly the same answer as I did.
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