Solve for p of Binomial distribution

Is there a way to solve for the probability p of a binomial distribution in a set of equations? With some arbitrary functions g[ _ ] and f[ _ ] such that:

eq1 = -g[y] + Sqrt[g[y]^2 + 4*5*5/200] (200^2/2*5) - x
eq2 := Sum[PDF[BinomialDistribution[15, y], i]*f [i], {i, 0, 15}] - 10*x
bif = NSolve[eq1 == 0, eq2 == 0, {x, y}]

I even simplified it to:

f[x_] := 0.34*Exp[-150*x]

eq1 := Sum[
Probability[BinomialDistribution[15, y], x]*f[x], {x, 0, 15}] - 10
bif = NSolve[eq1 == 0, {y}]

However the output I get is:

NSolve[-10 +
2.43953*10^-66 Probability[BinomialDistribution[15, y], 0] +
0.337692 Probability[BinomialDistribution[15, y], 1] +
0.34 Probability[BinomialDistribution[15, y], 2] +
0.34 Probability[BinomialDistribution[15, y], 3] +
0.34 Probability[BinomialDistribution[15, y], 4] +
0.34 Probability[BinomialDistribution[15, y], 5] +
0.34 Probability[BinomialDistribution[15, y], 6] +
0.34 Probability[BinomialDistribution[15, y], 7] +
0.34 Probability[BinomialDistribution[15, y], 8] +
0.34 Probability[BinomialDistribution[15, y], 9] +
0.34 Probability[BinomialDistribution[15, y], 10] +
0.34 Probability[BinomialDistribution[15, y], 11] +
0.34 Probability[BinomialDistribution[15, y], 12] +
0.34 Probability[BinomialDistribution[15, y], 13] +
0.34 Probability[BinomialDistribution[15, y], 14] +
0.34 Probability[BinomialDistribution[15, y], 15] == 0, {y}]

Effectively it is just solving this coupled system of equations. However, Mathematica does not provide a solution when I use the binomial function. The exact parameters don't matter, it's more a general question on how to solve these types of problems. I hope you can help :)

(Mathematica newbie here)

• Try NSolve~ or FindRoot`. Sep 29 '21 at 13:27
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• I think you need to change PDF[Binomial[15, y], i] to PDF[BinomialDistribution[15, y], i]. You can solve for x in both equations and then solve for y but solving for y probably requires knowing $g$ and/or $f$.
– JimB
Sep 29 '21 at 15:35
• You need to change Probability to PDF.
– JimB
Oct 1 '21 at 7:12
• I suspect you need the restriction $0<y\leq 1$ but if you plot eq1 (from your simplified example), there is no value of $y$ in that range where eq1==0.
– JimB
Oct 1 '21 at 16:16