I also am having an extremely rough time with this question:

There are 20 elk in a forest. You catch and tag 5 of them and release them. A little while later you catch 4. What is the probability that 2 of the 4 newly caught elk are tagged from before?


If I have 4 elk the probability that 2 of the 4 are tagged should just be (5/20)(4/19)(15/18)(14/17)

The first two numbers represent the probability of catching the elk and the last two are the probability that the last elk aren't tagged. This makes sense to me but it isn't the answer in the book which is 70/323.


I figured out just by punching numbers into my calc. that the answer I have multiplied by (20P5)/(20P4)*3/8=70/323 but I have no sense as to why. I am thinking that it is either something to do with the sample space, OR it is just some freak event that it works out, OR since (20P5)/(20P4)*3/8= 6 that 6 is reallly the key number I am missing. I really have no clue and anyone who could offer me a hint or a tutorial would really be helping a struggling boy.

Thanks,
Josh

Well you have a 1:4 to get at least one out of 20. Im going to guess it would be like a 12.5% chance.

Quote:

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Originally Posted by Joshua0317

There are 20 elk in a forest. You catch and tag 5 of them and release them. A little while later you catch 4. What is the probability that 2 of the 4 newly caught elk are tagged from before?

If I have 4 elk the probability that 2 of the 4 are tagged should just be (5/20)(4/19)(15/18)(14/17)
Josh

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You almost have it. You need to multiply what you have here by the number of different ways that the 4 elk of 2 different varieties (tagged or not) can be arranged. That would be 4C2. Hence you get:

5/20 * 4/19 * 15/18 * 14/17 * (4*3/2) = .2167 = 70/323