Hashing

Question

Consider the following keys that are hashed into the hash table in the order given using the hash function

Where to handle the collision chaining is used, after inserting all the above keys in table if 2 new keys inserted into table the what is the probability new items hashed into empty slot ?

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I am getting 5/11 * 4/11.

Comments

Can someone please confirm, the correct answer?

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2*5/11 *4/11 is correct

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@VS,  I think we should have to multiply by 1/2 in both cases because first we have to pick a key at random from the 2 distinct keys and then insert. So, To pick one key initially from 2 distinct keys , probability is 1/2.

So, answer should be $(\frac{1}{2}* \frac{5}{11} * \frac{4}{11}) + (\frac{1}{2}* \frac{5}{11} * \frac{4}{11}) =\frac{5}{11} * \frac{4}{11}$

Answer 1

I think correct answer should be   $\frac{\binom{5}{1}}{\binom{11}{1}}*\frac{\binom{4}{1}}{\binom{11}{1}}$  = $\frac{5}{11}*\frac{4}{11}$ (same could be verified via case tree diagram)

In mentioned case it can not be $\frac{\binom{5}{2}}{\binom{11}{2}}$  =  $\frac{5}{11} * \frac{4}{10}$ (because insertion is sequential)