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: Understanding the Concept:
The problem asks for the minimum number of coin tosses, 'n', required to make the probability of a certain event (getting at least one head) exceed a specific value (90% or 0.9). The key is to use the complement rule of probability, as calculating the probability of "at least one" is often easier by calculating 1 minus the probability of "none".
Step 2: Key Formula or Approach:
Let 'n' be the number of times the coin is tossed.
The probability of getting a head (H) in a single toss of a fair coin is \(P(H) = 0.5\).
The probability of getting a tail (T) is \(P(T) = 0.5\).
The event "at least one head" is the complement of the event "no heads" (i.e., all tails). \[ P(\text{at least one head}) = 1 - P(\text{no heads}) \] We are given the condition: \[ P(\text{at least one head})>0.90 \] Step 3: Detailed Explanation:
First, let's find the probability of getting "no heads" in 'n' tosses. This means getting tails on all 'n' tosses. Since the tosses are independent events, the probability of getting n tails in a row is: \[ P(\text{no heads}) = P(T \text{ and } T \text{ and } \dots \text{ n times}) = (P(T))^n = (0.5)^n = \left(\frac{1}{2}\right)^n \] Now, substitute this into our inequality: \[ 1 - \left(\frac{1}{2}\right)^n>0.90 \] Rearrange the inequality to solve for n: \[ 1 - 0.90>\left(\frac{1}{2}\right)^n \] \[ 0.10>\left(\frac{1}{2}\right)^n \] \[ \frac{1}{10}>\frac{1}{2^n} \] To get rid of the fractions, we can take the reciprocal of both sides. When we do this, we must reverse the inequality sign: \[ 10<2^n \] Now we need to find the smallest integer 'n' that satisfies this condition. We can test values of n: If n = 1, \(2^1 = 2\), which is not greater than 10.
If n = 2, \(2^2 = 4\), which is not greater than 10.
If n = 3, \(2^3 = 8\), which is not greater than 10.
If n = 4, \(2^4 = 16\), which is greater than 10.
Step 4: Final Answer:
The minimum number of times the coin must be tossed is 4.