Cosmin has three ping pong balls, and ten cups in front of him. If he throws the ping pong balls into the cups, what is the probability that they will land in different cups? (A ping pong ball has an equal probability of landing in any cup.)
Assuming you mean they go into three different cups (not two in one and the third in a second):
The first will cerainly go into a cup.
The second will have probabilty of 9/10 of going into a different cup than the first.
The third will have probability of 8/10 of going into a different cup than either of the firsr two.
Probability all in different cups is 9/10 x 8/10 = 72/100 = 18/25.
+2440 This is the the wrong blòódy solution for this question! (MAYBE?)
You know, Mr. BB, I didn’t recognize you in this post until I realized you plagiarized the answer verbatim from Quora. Like usual, you demonstrated your impeccable knack for plagiarizing WRONG answers.
Despite the credentials of the answerer, Carl Bryan, B.S. Mathematics, Carnegie Mellon University (1972), his answer is wrong. He answered the question nine days ago, and it seems his skills have atrophied in the past 46 years.
The reason it’s wrong: This assumption:
Assuming you mean they go into three different cups (not two in one and the third in a second)
changes the parameters of the probability. This is the same as saying “The cup is covered after a ball occupies it, so no more balls can enter. This changes the natural probability of the question. It’s not the question asked!
The question clearly indicates that 0, 1, 2, or 3 balls can occupy one cup. This is a necessary consideration when determining the probability of that NOT happening.
I can post the correct solution to this Mr. BB, but perhaps you can plagiarize a correct solution, or you could actually post your own solution. (I’m sure that will happen!)
Edit: Strike out snarky troll comment and wrong reason. (see below)
GA
+2440 You’re right; I did misunderstand it. Moreover, I jumped to criticism to soon. The reason I gave for his incorrect solution is wrong. His solution does imply a probability for a ball going into an already occupied cup.
There is still a problem though.
In my initial analysis, I calculated the probabilities based on indistinguishable balls.
Carl Bryan’s solution method assumes distinguishable balls. These are two different probabilities.
(Note that the cups are always distinguishable, else there is only one-way to distribute the balls for a success.)
Carl Bryan’s solution (where both the balls and cups are distinguishable) in binomial form is
\(\dfrac {C!}{(C-B)!} \div (C^B)\\ \dfrac {10!}{(7!)} \div (10^3) = \frac{18}{25} = 0.72 \)
My method is
The number of ways to select three (3) cups from a set of 10 (one for each ball) is 10C3 = 120
The number of ways to select two (2) cups from a set of 10 (one for two balls and one for a single ball) is 10C2 =45
The number of ways to select two (2) cups from a set of 10 (one for single a ball and one for two balls) is 10C2 =45
(Note the selection of two cups is counted twice because the cup with the single ball can be switched with the pair in the other cup)
The number of ways to select one (1) cup from a set of 10 (one for all three balls) is 10C1 = 10
Total number of distributions = 220
Number of successes = 120
120/220 = (6/11) ≈ 0.5455
In pure binomial form it is:
\(\dfrac {C!}{(C-B)!(B!)} \div \dfrac {(C+B-1)!}{( C-1)!(B!)}\\ \dfrac {10!}{(7!)(3!)} \div \dfrac {12!}{(9!)(3!)} = \dfrac{6}{11}\)
Now the question: what is the correct way to interpret the balls? Should they be indistinguishable or distinguishable? Is it mathematician’s choice?
I choose indistinguishable because Ping-pong balls are indistinguishable and there is nothing in the question to indicate they are distinguishable.
Carl Bryan’s solution method is for distinguishable balls, he may not have considered the indistinguishable.
What is the default? Or is there one?
Comments ................Alan, CPhill, Melody and/or anyone who is versed on these questions and their solutions.
GA
