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One of the popular tourist attractions in Alaska is watching black bears catch salmon swimming upstream to spawn. Not all “black” bears are black, though— some are tan-colored. Suppose that six black bears and three tan-colored bears are working the rapids of a salmon stream. Over the course of an hour, six different bears are sighted. What is the probability that those six include at least twice as many black bears as tan-colored bears? I don't understand how to do this problem, i know i have to use hypo geometric distribution but i don't know how. There was another post like this question but the answer given unfortunately didn't help

kingston
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  • We want $6$-$0$ or $5$-$1$ or $4$-$2$. Find the probability for each, and add up. More easily, compute the probability of $3$ blacks and $3$ tans, and subtract from $1$. – André Nicolas Oct 04 '15 at 19:56
  • Would we not need to subtract 4 states from 1 if that should work: 3-3, 2-4,1-5 and 0-6? – mathreadler Oct 04 '15 at 20:03
  • @mathreadler - there are only three tan bears. – Marconius Oct 04 '15 at 21:45
  • Are the bears captured? It says in the question they are "sighted" which would be more like pulling teddy bears out of a basket on random but then putting them back again. – mathreadler Oct 05 '15 at 05:56

1 Answers1

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So a very partial solution and hint due to lack of time, not in unbearable danger though. Will expand later ( wished there was a way to save partial answers )

First part of the problem is to decide how much is required for "at least twice as many". We can write this as equation system

$$\begin{align}x+y =& 6\\y - 2x \geq& 0\end{align}$$

If we solve the system with equality, we get $x = 2, y = 4$. So $y\geq4$ which means $y\in\{4,5,6\}$

Then we check how the hypergeometric distribution works.

mathreadler
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