Hashing Doubt

Question

In this question :

 

https://gateoverflow.in/2272/gate1997-12

Had it been "Open addressing" only then for part A would it be :

$\frac{n-1}{n}*\frac{n-2}{n-1}*.....*\frac{n-k-1}{n-k}$

?

(as every time we occupy a bucket , the probability will be changed as all slots are equiprobable.)

?

Comments

I think its correct in case of linear probing

but not sure about quadratic probing(it should to be true in this case as well because the probe sequence should cover all the buckets for n probes)

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@Hemanth_13

Yes coz its indpendeind of hashing function.