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In an oyster experiment, an oyster predator is placed in the center of a tray with two oysters and two of other prey nearby. The experimenter notes the order of the preying on the oysters - for example, if the predator preys upon the two oysters first, the order would be $1, 2$; if on the oysters last, it would be $3, 4$.

I wrote out the sample space for the question which gave me, $$\{(1,2),(1,3),(1,4),(2,3),(2,4),(3,4)\}$$

Let $X$ denote the sum of the orders of preying on the two oysters. Find the probability distribution of $X$ under the assumption that the predator chooses randomly among its available prey at each state.(Note that after preying upon an animal, that animal is killed, so it is not involved at the next choice stage.)

what can I do for this?

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1 Answer 1

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Guide:

You have listed out all the possibilities. Each of the possibilities happen with probability $\frac16$.

For each of the possibility, compute their sum, for example the sum for $(1,2)$ would be $3$. Do it for all the $6$ possibilities.

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  • $\begingroup$ Note that after preying upon an animal, that animal is killed, so it is not involved at the next choice stage. This is what the lat line of the question says. How does this fall into the answer? $\endgroup$
    – Mal
    Commented Sep 28, 2017 at 3:53
  • $\begingroup$ That is why the tuple-length is always $2$ as there are exactly two oysters. $\endgroup$ Commented Sep 28, 2017 at 3:55
  • $\begingroup$ so X can take the values 3,4,5,5,6.7. Now for the probability distribution for say X=3 , do I just take 1/6? and for 5 take it 1/6*1/6? $\endgroup$
    – Mal
    Commented Sep 28, 2017 at 3:59
  • $\begingroup$ The probability of $5$ would be $2 \times \frac16$. $\endgroup$ Commented Sep 28, 2017 at 3:59
  • $\begingroup$ You won't have cases like (oyster, oyster, oyster again, other prey). That is only possible if the oyster is still alive and it could be attacked again and you will end up with $(1,2,3)$. $\endgroup$ Commented Sep 28, 2017 at 4:00

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