1) We have n white balls and 2n black balls. In how many ways we could put it into n boxes if in every box it have to be at least one black ball have to be in every box.

My answer is: First of all let put one black ball into each box. So now we have a problem how to put n white and n black balls into n boxes which we can be done in $\binom{N+N-1}{N}\binom{N+N-1}{N}$ ways.

2) In the lake we have got k - A-fishes, 2k - B-fishes, 4-k - C-fishes Jack caught 7 fishes. Calculate the probability that caught at least one A-fish.

I'm not sure which Discrete probability distribution I could use...


Your analysis of the first problem is fine, and the answer $\binom{n+n-1}{n}\binom{n+n-1}{n}$ is correct.

For the second, we will assume that the number of Type A fish caught has Hypergeometric Distribution. It is easier to find the probability of catching no Type A fish.

There are $\binom{7k}{7}$ ways to choose $7$ fish. There are $\binom{6k}{7}$ ways to choose $7$ fish none of Type A. So the probability of catching no Type A is $\frac{\binom{6k}{7}}{\binom{7k}{7}}$. The probability of at least one Type A is $1-\frac{\binom{6k}{7}}{\binom{7k}{7}}$.

Remarks: In order to use the hypergeometric distribution, we needed to assume first that we do not replace the fish we catch. We also needed to assume that all $\binom{7k}{7}$ possible outcomes are equally likely. This is unreasonable, some types of fish are more easily caught. Also, we do not have independence, fish often gather in schools.

If $k$ is at all large, the no replacement and replacement models give very similar answers, and the probability of at least one Type A fish is approximately $1-\left(\frac{6}{7}\right)^7$.

  • $\begingroup$ since you are not replacing the fish back into the pond after catching, you are suggesting a hypergeometric distribution. Since the size of available fishes are large compared to the ones caught, could we not use multinomial distribution with $P(A) = \frac{1}{7}$,$P(B) = \frac{2}{7}$, and $P(C) = \frac{4}{7}$ and that he has to find 1-[$P(A=0,B=0,C=7)+P(A=0,B=7,C=0)+ P(A=0,B=1,C=6)+P(A=0,B=6,C=1)+P(A=0,B=2,C=5)+P(A=0,B=5,C=2)+P(A=0,B=3,C=4)+P(A=0,B=4,C=3)$] Correct me if I am wrong. $\endgroup$ – Satish Ramanathan Jun 13 '14 at 20:48
  • $\begingroup$ I assumed we are not replacing the fish after catching them. In this problem, we are only interested in Type A versus others. If $k$ is at all large, we will get about the same answer if we use sampling with replacement. Thus for large $k$ the number of Type A fish caught has approximately Binomial distribution, and the probability of no Type A is approximately $(6/7)^7$. $\endgroup$ – André Nicolas Jun 13 '14 at 20:54
  • $\begingroup$ just curious, I was thinking that k means 1000, how wierd can my brain think, Thanks for your explanation. $\endgroup$ – Satish Ramanathan Jun 13 '14 at 20:58
  • $\begingroup$ You are welcome. It is possible that $k$ is intended to be $1000$. Then certainly the Binomial model would give a very accurate approximation. $\endgroup$ – André Nicolas Jun 13 '14 at 21:04

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