Problem (Combinatorics):
Let $T=$ 3. Suppose that $T$ fair coins are flipped. Compute the probability that at least one tails is flipped.

Solution:
The probability of flipping all heads is $\left(\frac{1}{2}\right)^{T}$, so the probability of flipping at least one tails is $1-\frac{1}{2^{T}}$. With $T=3$, the desired probability is $1-\frac{1}{8}=\frac{7}{8}$.

Final answer: $\frac{7}{8}$