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A scientist inoculates several mice, one at a time, with a disease until he finds 3 that have contracted the disease. If the probability of contracting the disease is 1/6, what is the probability that 8 or more mice are required?

At first, my initial instinct was that this is an example of a negative binomial, where X = the total number of trials to achieve the rth success (being 3 mice that have contracted the disease). I then realized if I had set up the problem as such, where $P(Y\geq 8)$, or $1-P(Y<8)$, I would have a negative and a zero value in my set up, which would not be possible.

This is the point where I find myself rather stuck at how to proceed.

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    $\begingroup$ Why do you think you'd have a negative number or a zero? $\endgroup$
    – jbowman
    Oct 9, 2019 at 19:47
  • $\begingroup$ Because (from my limited understanding), I assumed my set up would be: 1-[P(X=0)+P(X=1)+P(X=2)...] and so forth up until 7. In the negative binomial, the initial combination is (x-1) and (k-1) and if X equals 0 or 1, then it would be negative or zero. $\endgroup$
    – rrhodes
    Oct 9, 2019 at 19:51
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    $\begingroup$ You can model the number of failures using a negative binomial, then the total number of trials is just the number of failures + 3. The NB distribution can be used either way, but you have to define $Y$ differently, as you have discovered; the way you have written it in the response to my question is the "# of failures" way. $\endgroup$
    – jbowman
    Oct 9, 2019 at 20:13
  • $\begingroup$ Alternatively, can't I use the binomial distribution where I look for P(8 or more mice are required) = the probability that in 7 mice where less than or equal to 2 mice contracted the disease? $\endgroup$
    – rrhodes
    Oct 9, 2019 at 20:36
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    $\begingroup$ Look at the different formulations of the negative binomial here: en.wikipedia.org/wiki/Negative_binomial_distribution (compare the main "number of failures" one with the "number of trials" one in the table). $\endgroup$
    – Glen_b
    Oct 10, 2019 at 0:59

2 Answers 2

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you can do it like this:

$$P(Y \ge 8| succ == 3) = P(Y == 7| succ == 0) + P(Y == 7| succ == 1) + P(Y == 7| succ == 2) = \binom{7}{0}(\frac{5}{6})^7 + \binom{7}{1}\frac{1}{6}(\frac{5}{6})^6 + \binom{7}{2}(\frac{1}{6})^2(\frac{5}{6})^5$$

it's possible to model it with negative binomial distribution but you would end up with sumation over some condition anyway

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Considering the definition of each is the best way to tell. For binomial, you have independent, constant number of trials and the variable (what is random) is the number of successes. Just based on this definition, the situation you have already does not fit since in your situation, the number of trials is not constant and is in fact, the variable.

When we look at NB, one of its formulations (there are several), fits your situation, where we count the number of trials before some condition is met (3 successes). So NB does fit the distribution of your random variable.

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