Probability of collision in a hash table

Question

Hello, I have these 3 short questions that I am not sure how to approach. These are in my Computer Science class, specifically regarding hash table and probability of collision/ non-collision in a hash table. I would really appreciate your help if you can explain step by step how to compute the probability in each case. Thank you so much in advance!

Question 1: Imagine we have a hash table with 10 slots. However, 4 of them are filled already (without any collision). What is the probability that we will have at least 1 collision in the next 2 insertions?

Question 2: Imagine we have a hash table with 10 slots. However, 4 of them are filled already (without any collision). Assuming we use linear probing to resolve collisions, what is the probability that we will have exactly 1 collision in the next 2 insertions?

Question 3: Imagine we have a hash table with 10 slots. However, 4 of them are filled already (without any collision). Assuming we use separate chaining to resolve collisions, what is the probability that we will have exactly 1 collision in the next 2 insertions?

Answer 1

Exactly 1 collision in two insertions:

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1st insertion, the probability is 4/10.

If the collision does not occur on this first insertion,

then we must not have a collision on the 2nd insertion

, which has probability 5/10,

so 4/10*5/10 = 20/100 = 20%

Exactly two insertions:

4/10*5/10 = 20/100 = 20%

so 40% of getting at least one collision over the next 2 insertions