# Doubt in Expected number of probes in successful seach in open address hashing

My attempt:

I need to find exected number of probes in case of successful search. I am assuming, n elements and m slots in hash table

E(# of probes) = average of {1st probe success , 2nd probe success, .... nth probe success} over n

$$i^{th}$$ probe success = $$(i-1)$$ probes unsuccessful and ith probe successful = probability that (i-1) probes unsuccesful and last probe successful = $$(\alpha^{(i-1)})* \alpha$$ --> Is this correct???

I am taking, probability that i-1 slots are filled up and those elements inside them are not equal to key k(unsuccesful) = $$(\alpha^{(i-1)})$$

taking forward,

E(# of probes) = $$\frac{1}{n} \sum_{i=1}^n i*\alpha^i \neq \frac{1}{\alpha} \ln \frac{1}{1-\alpha}$$