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A hash table has space for 100 records .what is the probability of collision before it is 5% full??

a. 0.25

b. 0.10

c. 0.40

d. 0.20
closed as a duplicate of: Doubt
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Sir,  Then (1-(99/100)*(98/100)*(97/100)*(96/100))=.096 so option b
in that case chose the nearest one :) Which exam?
May be some test series Sir. actually someone posted the qs on FB group :)

required probability =   ((100*1 )/ 1002) + ((100*99*2 ) / 1003) +.....+((100*99*98*97*4 ) / 1005) = 0.096= 0.10( appox)

option B is correct

edited

Hash table will be 5% full.

Max collision possible,

In 1 insertion of record there is 0 collision.

" 2      "       "     "         "     "  1    "

"  3     "       "     "         "     "   2   "

"  4    "        "     "        "     "  3   "

Before 5% full there are maximum 6 collision

So, there are total collision possible 0+1+2+.....+6=21

here max 6 collision i got it but why u write 0+1+2+..+6 plz explain???
Depends on the collision resolution method to be used.If chaing it would be different i guess
@kishalas why not only 6 collision ..
Think of chaining Collision resolution technique...

Suppose while inserting all get same slot(lets say slot 1) and pushed in chain.Then in worst case all will be hashed to only one slot.And in that case there will be as many number of collision as the number of records to be hashed(almost) and only one slot get filled.
Actually before 5% full,
there is a case 0% full ,So, no collision
1%full=maximum 1 collision
2%full =maximum 2 collision
......
4% full=6 collision
right?

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