Is it possible to get scipy.stats.binom.pmf(x, n, p) to return the number of trials (n) with the probabaility, number of successes (x), and probability of successes (p) known?
Example Problem:
How many throws does Alex need to do in order to be 90% sure of hitting the target at least 10 times?
where:
You can do something like this. What you want to compute is what scipy.stats calls the survival function (sf) which is 1-cdf. Since you are interested in greater than or equal to 10 success--that is the sum of the probabilities of 10 or more successes, that is precisely 1-cdf. The sf function can take a numpy array for the arguments, so we pass in an array for n (varying the number of trials). We then look for the number of trials that gave us a value greater than the defined confidence.
import numpy as np
import scipy.stats
import matplotlib.pyplot as plt
# Defining model parameters
p = 0.5
k = 10
confidence = 0.9
n = np.arange(k, 5*k)
# Generating survival function distribution and computing the number of required trials
binomSurvivalDist = scipy.stats.binom.sf(k, n, p)
nrequired = n[binomSurvivalDist >= confidence][0]
# Plotting the results to Verify that this works
fig, ax = plt.subplots(1, 1, figsize=(12, 10))
x = np.arange(0, nrequired+1, 1)
ax.plot(x, scipy.stats.binom.sf(x, nrequired, p), lw=2)
ax.set_xlabel("Number of Successes", fontsize=16)
ax.set_ylabel("Probability of Getting At Least this Many Success", fontsize=16)
ax.set_title("Distribution for %i Trials" % nrequired, fontsize=18)
print(nrequired)
plt.show()
The plot is not necessary for the computation, but allows us to verify that the procedure does give us the correct answer.