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Probability Q

If you have 5 balls and 18 boxes, what are the chances (if you put the 5 balls into the 18 boxes randomly) that there will be no box with 2 or more balls inside it?

I've used an argument like: there should be 19 lines like this |
and there should be 5 balls like this O
arranged in a sequence like: |||O|||O|||||OO||O|||O|||

and did 1 - (# possibilities with only 1 or 0 balls in a box)/(#total possibilities)
= 1 - 18C5/22C5 = 0.67

but I have a feeling this number is incorrect. Please can someone give me a reason why?
Reply 1
Avatar for vc94
vc94
12
How did you get 22C5 ?
If each ball has a choice of 18 boxes then isn't the #total possibilities 18^5 ?
Reply 2
Avatar for jamesair99
jamesair99
OP
Original post by vc94
How did you get 22C5 ?
If each ball has a choice of 18 boxes then isn't the #total possibilities 18^5 ?


There are 22 spaces with the O and /'s in my diagram and there are 5 balls.

The reason don't think 18^5 works because it includes repeats. :frown: the balls are distinct
Reply 3
Avatar for JoMo1
JoMo1
12
Use simpler reasoning - the chance of getting 5 balls in individual boxes is the chance of getting each of them in an individual box sequentially, or the product of getting each ball in an individual box, given that you've already got the others right.

so effectively (17*16*15*14)/18^4 = 0.54
Reply 4
Avatar for vc94
vc94
12
Label each ball: A,B,C,D,E.
So ball A can be placed in any of the 18 boxes, so can ball B, C, D and E. Giving 18^5 possibilities?

I've probably misunderstood something...
Reply 5
Avatar for JoMo1
JoMo1
12
Original post by vc94
Label each ball: A,B,C,D,E.
So ball A can be placed in any of the 18 boxes, so can ball B, C, D and E. Giving 18^5 possibilities?

I've probably misunderstood something...


But you need to place them such that no balls are in the same box.
Reply 6
Avatar for vc94
vc94
12
1 - (# possibilities with only 1 or 0 balls in a box)/(#total possibilities)
= 1 - 18P5/(18^5) = 1-0.5441...= 0.456 ?

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