Combinations and probability distributions

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SUMMARY

The discussion centers on calculating the probability distribution of empty bowls when three balls are thrown into five bowls, with each ball having an equal chance of landing in any bowl. The established probabilities are P(R=2) = 0.48, P(R=3) = 0.48, and P(R=4) = 0.04, where R represents the number of empty bowls. Participants clarify that the problem does not consider the possibility of balls missing the bowls, as it is not stated in the question, and the focus is on the outcomes defined by the problem parameters.

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SavvyAA3
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Can someone please help with the method of how to solve this problem...

Question:
Three balls are thrown at random into 5 bowls so that each ball has the same chance of going into any bowl independently of wherever the other 2 balls fall. Determine the probability distribution of the numberof empty bowls.


The anwers are below, but i don't understand them. Can it be looked at as a success failure type poblem. with probability of success being that a bowl gets 1 ball: probability = 1/5, and the probability of a failure; 1-1/5 = 4/5. If so how do I use this to obtain the distribution.

Also can someone please tell me why in this scenario we don't look at the possibility of the ball missing the bowl - because i feel that this should be a possibility in this scenario, which then gives 6 equally likely outcomes, which would mean probabitlity of a success is then 1/6?

The anwer to this question (which I don't understand) is:
if we let R = no. of empty bowls

the distribution of R:
P(R=2) = 0.48, p(R=3) = 0.48, p(R=4) = 0.04

Please help!
 
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Hi SavvyAA3! :smile:
SavvyAA3 said:
Can it be looked at as a success failure type poblem. with probability of success being that a bowl gets 1 ball: probability = 1/5, and the probability of a failure; 1-1/5 = 4/5.

Yes, that's right! :smile:
If so how do I use this to obtain the distribution.

P(R=2) = 0.48, p(R=3) = 0.48, p(R=4) = 0.04

Well, try p(R=4) first … that's the probability that the second and third balls go into the same bowl as the first one … which is … ? :smile:
Also can someone please tell me why in this scenario we don't look at the possibility of the ball missing the bowl - because i feel that this should be a possibility in this scenario, which then gives 6 equally likely outcomes, which would mean probabitlity of a success is then 1/6?

erm … REALITY CHECK … ! :rolleyes:

When you're sitting in the examination room, THE ONLY REALITY IS WHAT'S ON THE QUESTION PAPER … anything else is just a mirage brought about by heat and lack of water and sleep. :wink:
 
why reality check? The question did not state that this is not a possibility - please tell me why I can't assume this?

Thanks.
 
The question also didn't state that aliens would not grab the balls with a tractor beam or that your dog would not eat them... do you see?
 
Hi montoyas7940! :smile:
montoyas7940 said:
The question also didn't state that aliens would not grab the balls with a tractor beam

hmm …
:smile: … is that what happened to you … ? :smile:
 
Hi yourself TT!

No, fortunately aliens didn't grab my balls. :approve:

But I read about this guy once...
 

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