Question

How many minimum number of times must a man toss a fair coin so that the probability of having at least one head is more than 90%?

• A. 3
• B. 4
• C. 5
• D. 10

Hint:

Problems involving "at least one" are strong candidates for using the complement rule: P(A) = 1 - P(not A). Calculating the probability of the event *not* happening is often much simpler. In this case, 'not getting at least one head' means 'getting zero heads', which is a single, easy-to-calculate outcome.

Answer 1

The Correct Option is B

Step 1: Problem Definition: The objective is to determine the minimum coin toss count, 'n', required for the probability of obtaining at least one head to exceed 0.90. The complementary probability approach, P(event) = 1 - P(complement of event), is the most efficient for "at least one" scenarios.
Step 2: Core Mathematical Framework: Let 'n' represent the number of coin tosses. For a fair coin, P(Head) = P(Tail) = 0.5. The event "at least one head" is the complement of "no heads" (all tails). The governing equation is: \[ P(\text{at least one head}) = 1 - P(\text{no heads}) \]. The target condition is: \[ P(\text{at least one head}) > 0.90 \].
Step 3: Calculation Process: The probability of achieving "no heads" in 'n' tosses (i.e., all tails) is calculated as the product of individual probabilities: \[ P(\text{no heads}) = (P(T))^n = (0.5)^n = \left(\frac{1}{2}\right)^n \]. Substituting this into the inequality: \[ 1 - \left(\frac{1}{2}\right)^n > 0.90 \]. Rearranging to isolate the term with 'n': \[ 0.10 > \left(\frac{1}{2}\right)^n \implies \frac{1}{10} > \frac{1}{2^n} \]. Inverting both sides and reversing the inequality: \[ 10 < 2^n \]. We now seek the smallest integer 'n' satisfying this. Testing values: n=1 yields \( 2^1 = 2 \) (not >10); n=2 yields \( 2^2 = 4 \) (not >10); n=3 yields \( 2^3 = 8 \) (not >10); n=4 yields \( 2^4 = 16 \) ( >10).
Step 4: Conclusion: The coin must be tossed a minimum of 4 times.