Binomial Probability

Compute the probability of exactly k successes in n independent trials, each succeeding with probability p.

P(k) = C(n, k) * p^k * (1 - p)^(n - k)

where C(n, k) = n! / (k! * (n - k)!) counts the orderings.

Return 0.0 when k is negative or larger than n.

Use math.comb for the binomial coefficient. Computing factorials separately overflows fast, and for large n the intermediate values are enormous even when the answer is tiny.

Function signature

def binomial_probability(n, k, p):

Example

Input

binomial_probability(n=5, k=2, p=0.5)

Output

0.3125

Reasoning

C(5, 2) = 10 orderings, each with probability 0.5^2 * 0.5^3 = 1/32. So the answer is 10 / 32 = 0.3125.

Constraints and notes

• Return 0.0 when k < 0 or k > n.
• Use math.comb for the coefficient.