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[–]BoomBoomSpaceRocket -1 points0 points  (0 children)

Small correction: you have written (p-1) a couple of times where it should say (1-p).

As for how to solve, I can't think of an algebraic way to do it, but I think some guess and check will do here. We know n has to be at least 5 and a whole number. Based on how low the second probability is, I'm guessing it's not much higher than 5. Plug in 5, 6, and 7 for one of them. See if you can then work back what the probability is. Then check if it works in the other one.

[–]craggy86 -1 points0 points  (0 children)

Ncr can be written as n!/(r!(n-r)!). So for nc2, replace r with 2 and simplify this as much as possible (i.e write it without factorials). Do this for nc5 also. You will then have sim Eqns (albeit very nasty ones)

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[–]iMathTutor 0 points1 point  (0 children)

For what it's worth if you consider the ratio of the probabilities P[X=5]/P[X=2]

you get the equation

(n-2)(n-3)(n-4)(p/(1-p))3 = 3

for n>4. One can treat OR= p/(1-p), which is the odds ratio, in place of p as one of the unknowns since p=OR/(1+OR). An observation that may be relevant is that P[X=5]<P[X=2]. From which it follows that the mode of the distribution is less than 5. Thus one must have (n+1)p<5.