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

Introduction / Context:
This is identical in structure to the “odd heads” problem with a fair coin and an even number of tosses. By symmetry, the probability that the count of tails is odd equals 1/2.

Given Data / Assumptions:

• n = 100 tosses (even).
• P(T) = 1/2, P(H) = 1/2; independence across tosses.

Concept / Approach:
Parity splitting for binomial with p = 1/2 and even n yields P(odd) = 1/2 and P(even) = 1/2 for the count of the tracked outcome.

Step-by-Step Solution:
Let X be the number of tails.Using the identities (p + q)^n and (p − q)^n with p = q = 1/2 and n even implies P(X odd) = 1/2.

Verification / Alternative check:
Direct pairing of outcomes by flipping the first toss (head↔tail) creates a bijection between sequences with even and odd tails.

Why Other Options Are Wrong:
1/8 and 3/8 do not reflect parity symmetry; “None of these” is incorrect because 1/2 is valid; 1/4 is too small.

Common Pitfalls:
Forgetting the requirement that n is even and p = 1/2 for the parity split to hold exactly.

Final Answer:
1/2