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: Conceptualization: The objective is to determine the minimum coin tosses, denoted by 'n', such that the probability of observing at least one head surpasses 90% (0.9). The complementary probability approach is efficient here: calculate the probability of the event's opposite (no heads, i.e., all tails) and subtract it from 1. Step 2: Foundational Formula: Let 'n' represent the number of coin tosses. For a fair coin, the probability of heads \(P(H)\) is 0.5, and the probability of tails \(P(T)\) is 0.5. The event "at least one head" is the complement of "no heads". Thus: \[ P(\text{at least one head}) = 1 - P(\text{no heads}) \] The given condition is: \[ P(\text{at least one head})>0.90 \] Step 3: Elaboration: The probability of obtaining "no heads" in 'n' tosses signifies getting tails on every toss. Due to the independence of tosses, this probability is: \[ P(\text{no heads}) = (P(T))^n = (0.5)^n = \left(\frac{1}{2}\right)^n \] Substituting this into the inequality: \[ 1 - \left(\frac{1}{2}\right)^n>0.90 \] Rearranging to isolate 'n': \[ 1 - 0.90>\left(\frac{1}{2}\right)^n \] \[ 0.10>\left(\frac{1}{2}\right)^n \] \[ \frac{1}{10}>\frac{1}{2^n} \] Taking the reciprocal of both sides reverses the inequality: \[ 10<2^n \] We now seek the smallest integer 'n' satisfying this. Testing values: If n = 1, \(2^1 = 2\), not>10. If n = 2, \(2^2 = 4\), not>10. If n = 3, \(2^3 = 8\), not>10. If n = 4, \(2^4 = 16\), which is>10. Step 4: Conclusion: The minimum number of coin tosses required is 4.