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Suppose there are $11$ slots in a Hash Table. At an instant there are $5$ empty slots and a new element $x$ is inserted in Hash Table. What is the the probablity that new element fits without collision?

I think it should be $\frac{5}{11}$ but answer given says :::

EDIT :: Actual question is

edited | 213 views

Given number of free slots available = 5

Total number of slots  =  11

So probability that collision will not occur on insertion  = n(Favourable outcome) / n(Total outcome)

= n(free slots) / n(total slots)

= 5 / 11

Hence 5 / 11 is correct..

habib you are wrong ...5/11 is the probabilty of choosing a free slot among the 11 slots

but when you want to insert a key without collision you must insert it in one of the five free slots which gives probability of 1/5
But bro actually the thing is the key may be anything..It is  not  fixed..So it may go to any of the entries of the table..

So how r u guaranteeing that only 5 slots will be total outcome??
So what is the correct answer guys ? 5/11 or 1/5???
5/11 is correct