How many fair-coin tosses are required so that the probability of at least one head is at least 99%?

Introduction / Context:
We seek the smallest n such that the probability of seeing at least one head in n fair-coin tosses is ≥ 0.99.

Given Data / Assumptions:

• P(head) = 1/2 per toss; independence between tosses.
• Target probability ≥ 0.99 for “≥ 1 head”.

Concept / Approach:
P(≥1 head) = 1 − P(no heads) = 1 − (1/2)^n. Set 1 − (1/2)^n ≥ 0.99 and solve for the minimal integer n.

Step-by-Step Solution:
1 − (1/2)^n ≥ 0.99 ⇒ (1/2)^n ≤ 0.01.Take logs: n ≥ log(0.01) / log(0.5) ≈ 6.6439.Minimal integer n = 7.

Verification / Alternative check:
For n = 6: 1 − (1/2)^6 = 1 − 1/64 ≈ 0.9844 < 0.99. For n = 7: 1 − 1/128 ≈ 0.9922 ≥ 0.99.

Why Other Options Are Wrong:
5 or 6 give probabilities below 0.99; 8 or 9 work but are not minimal.

Common Pitfalls:
Rounding the logarithmic threshold the wrong way (must take the ceiling to ensure ≥ 0.99).

Final Answer:
7