100 identical biased coins: Each coin shows heads with probability p (0 < p < 1). If P(50 heads) equals P(51 heads), find p.

Introduction / Context:
For a Binomial(n, p) with n = 100, we are told two adjacent probabilities match: P(X = 50) = P(X = 51). This determines p via the ratio of consecutive binomial terms.

Given Data / Assumptions:
X ~ Binomial(100, p); 0 < p < 1; equality of probabilities at 50 and 51.

Concept / Approach:
Use the identity P(k+1)/P(k) = [(n − k)/(k + 1)] * [p/(1 − p)] and set it equal to 1 at k = 50.

Step-by-Step Solution:

Verification / Alternative check:
Numerically, p ≈ 0.50495 is slightly > 1/2, which makes P(51) slightly larger than P(50) when p exceeds this threshold, matching equality at p = 51/101.

Why Other Options Are Wrong:
1/2 is close but not exact; 49/101 and 50/101 arise from flipping or off-by-one ratios.

Common Pitfalls:
Setting p = 1/2 by symmetry without checking the discrete correction embodied in the ratio formula.

Final Answer:
51/101