A biologist captures $21$ grizzly bears during the spring, and fits each with a radio collar. At the end of summer, the biologist is to observe $15$ grizzly bears from a helicopter, and count the number that are radio collared. This count is represented by the random variable X.

Suppose there are $114$ grizzly bears in the population.

The biologist gets back from the helicopter observation expedition, and was asked the question: How many radio collared grizzly bears did you see? The biologist cannot remember exactly, so responds " somewhere between $4$ and $9$ (inclusive) ".

Given this information, what is the probability that the biologist saw $7$ radio-collared grizzly bears?

Would this just be conditional probability? I.E

$A$ = Event that biologist saw between $4$ and $9$ Radio collared bears

$B$ = Event that the biologist saw $7$ radio collared bears

And the probability is just $P( A|B )$?


What you're given is $A$. So you want

$$P(B|A)=\frac{P(B\cap A)}{P(A)}=\frac{P(B)}{P(A)}$$

since if $B$ happens then $A$ is guaranteed to happen.

The hypergeometric pmf is given by

$$P(X=k)=\frac{{K \choose k}{N-K \choose n-k}}{N \choose n}$$


$$N \text{ is the number of grizzly bears in the population}$$

$$K \text{ is the number of collared grizzly bears}$$

$$n \text{ is the number of observed grizzly bears}$$

$$k \text{ is the observed number of collared grizzly bears}$$

You'll want to sum the probabilities where $$k\in\{4,5,6,7,8,9\}$$

  • $\begingroup$ so $21 \choose 4$ * $93 \choose 11$ / $114 \choose 15$ and so on for k = 4, 5, 6, 7, 8 ,9? $\endgroup$ – MattyS11 Mar 5 '18 at 6:30
  • $\begingroup$ Yes, that's correct. As a check of your work, R statistical software gives $P(A)=0.2857015$ $\endgroup$ – Remy Mar 5 '18 at 6:31
  • 1
    $\begingroup$ you're a great dude, thanks so much! $\endgroup$ – MattyS11 Mar 5 '18 at 6:40

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