Re: doubt

Posted by anonymouse on May 28, 2013; 11:25am
URL: http://discussion-forum.276.s1.nabble.com/doubt-tp7581374p7581379.html

Hi Maahi,

This is how I did it. Let p be the probability of success given by p = 1 - (5/6)^4.

Then, the probability that the nth trial has the first success is given by p(1-p)^n-1.

Then E(X) is given by the sum (n, 1, inf) np(1-p)^n-1.

I really didn't know how to solve this, so I used a widget online to get the answer, which is approximately 1.93.

Of course we can't do this in the exam, so I tried to look online for another way to solve the question, and found out that the question describes a geometric distribution (I don't think I've encountered it before), for which the expected value is simply 1/p.

So, the answer is 1296/691, which is approximately 1.93.


tl;dr: The question describes a geometric distribution of the number of trials needed to get a success (at least one 6). So, E(X) = 1/p = 1.93, where p = 1 - (5/6)^4.