expectation

[hs]

Suppose that a person repeatedly tries to perform a certain
task until he is successful. Suppose also that the probability of success on each given
trial is p (0 <p <1) and that all trials are independent. If X denotes the number
of the trial on which the first success is obtained, then E(X) is ??

[Gayyam]

I think what you have mentioned is the geometric distribution.
So, let us assume that the person is successful on his r-th try. This implies he failed r-1 times before that. So, (with the understanding that the random variable X being defined as the number of trials for the first success) the probability of X=r equals the probability of r-1 failures and finally the 1st success in the r-th try. Thus,

P(X=r) = [(1-p)^(r-1)]*p = p*(1-p)^(r-1)

Thus, E(X) = the weighted average of all the possible outcomes of the distribution (all possible values of r, i.e. r can vary from 1 trials to ∞ trials), with the weight of the outcomes being of course defined by the probability of that outcome.
               = ∑r*P(X=r)
               = ∑r*p*(1-p)^(r-1)
               = p*∑r*(1-p)^(r-1)
This is an arithmetico-geometric progression. If you evaluate the sum, you'll get E(X)=1/p.

[Akshay Jain]

This is equal to the submation of i trial*prob of success on i trial nd failure on i-1 trials where i goes from 1 to infinity
E(X)=p*1 + q*p*2 + q^2*p*3 + q^3*p*4......and so on
now to solce this G.P.
q*E(X)=q*p*1 + q^2*p*2 + q^3*p*3......nd so on
now
E(X)-q*E(X)= p*1 + q*p + q^2*p + q^3*p.......
(1-q)*E(X)= p/(1-q)
E(X)= p/(1-q)^2=p/p^2=1/p

[hs]

thnx guys