Combinations and probability distributions

In summary, the probability of a success is 1/5 and the probability of a failure is 1-1/5 = 4/5. The probability of a success is when the first ball goes into a bowl and the probability of a failure is when the second and third balls go into the same bowl as the first one.
  • #1
SavvyAA3
23
0
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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  • #2
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:
 
  • #3
why reality check? The question did not state that this is not a possibility - please tell me why I can't assume this?

Thanks.
 
  • #4
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?
 
  • #5
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:
 
  • #6
Hi yourself TT!

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

But I read about this guy once...
 

1. What is the difference between combinations and permutations?

Combinations and permutations are both ways to determine the number of possible outcomes for a given set of items. The main difference is that combinations are used when the order of the items does not matter, while permutations are used when the order does matter.

2. How do you calculate the number of combinations?

The formula for calculating the number of combinations is nCr = n! / r!(n-r)!, where n represents the total number of items and r represents the number of items being chosen. For example, if you have 5 items and want to choose 3 of them, the number of combinations would be 5C3 = 5! / 3!(5-3)! = 10.

3. What is a probability distribution?

A probability distribution is a function that describes the likelihood of each possible outcome in a given set of data. It shows the probabilities of all possible outcomes occurring, and the sum of all the probabilities is equal to 1.

4. How do you calculate the mean of a probability distribution?

The mean of a probability distribution can be calculated by multiplying each outcome by its probability, and then adding all of these values together. This is also known as the weighted average. For example, if the probabilities of getting a 1, 2, 3, 4, or 5 on a dice roll are 1/6, the mean would be (1*1/6) + (2*1/6) + (3*1/6) + (4*1/6) + (5*1/6) = 3.5.

5. What is the difference between a discrete and continuous probability distribution?

A discrete probability distribution is one where the possible outcomes are finite and countable, such as the number of heads from flipping a coin. A continuous probability distribution is one where the possible outcomes are infinite and uncountable, such as the exact temperature on a given day. Discrete distributions are represented by probability mass functions, while continuous distributions are represented by probability density functions.

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