A fair coin is tossed 100 times. What is the probability that the number of heads obtained is an odd number?

Introduction / Context:
When a fair coin is tossed repeatedly, the binomial distribution with parameters n and p applies. Here, n = 100 independent tosses and p = 1/2 for heads. We are asked for the probability that the count of heads is odd.

Given Data / Assumptions:

• Number of trials n = 100 (even).
• Per-trial head probability p = 1/2; tail probability q = 1/2.
• Outcomes across trials are independent.

Concept / Approach:
For a symmetric Bernoulli trial with p = q = 1/2 and an even number of trials n, the binomial mass is perfectly split between “even” and “odd” counts. This follows from the binomial theorem applied to (p + q)^n and (p − q)^n.

Step-by-Step Solution:
Let S_even = P(even heads) and S_odd = P(odd heads).(p + q)^n = 1 = S_even + S_odd.(p − q)^n = 0 (since p = q and n is even) = S_even − S_odd.Solving gives S_even = S_odd = 1/2.

Verification / Alternative check:
This parity-splitting result holds for any even n with p = 1/2 and is a standard identity in probability.

Why Other Options Are Wrong:
1/4 and 3/4 would imply asymmetry; 2/3 does not arise from binomial parity symmetry; 0 is impossible.

Common Pitfalls:
Forgetting that the parity-splitting property requires p = 1/2 and n even; it does not generally hold for p ≠ 1/2 or odd n.

Final Answer:
1/2