Combinations and probability distributions

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The discussion revolves around calculating the probability distribution of empty bowls when three balls are thrown into five bowls randomly. Participants explore the idea of treating it as a success-failure problem, with the probability of a bowl receiving a ball being 1/5. The distribution is clarified, showing that the probabilities of having 2, 3, and 4 empty bowls are 0.48, 0.48, and 0.04, respectively. There is confusion regarding the possibility of a ball missing a bowl, but it is emphasized that the problem's parameters do not support this assumption. Ultimately, the focus remains on understanding the established probabilities based on the given scenario.
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...
 
The standard _A " operator" maps a Null Hypothesis Ho into a decision set { Do not reject:=1 and reject :=0}. In this sense ( HA)_A , makes no sense. Since H0, HA aren't exhaustive, can we find an alternative operator, _A' , so that ( H_A)_A' makes sense? Isn't Pearson Neyman related to this? Hope I'm making sense. Edit: I was motivated by a superficial similarity of the idea with double transposition of matrices M, with ## (M^{T})^{T}=M##, and just wanted to see if it made sense to talk...

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