Question

The minimum number of times a fair coin needs to be tossed, so that the probability of getting at least two heads is at least 0.96 is:

• A. 5
• B. 6
• C. 7
• D. 8

Hint:

For probability questions involving multiple events, break the problem into smaller parts, calculating individual probabilities and then combining them.

Answer 1

The Correct Option is D

We are tossing a fair coin, and we want the probability of getting at least two heads to be at least 0.96. Let \( n \) be the number of coin tosses. The probability of getting at least two heads is: \[ P(\text{at least 2 heads}) = 1 - P(\text{0 heads}) - P(\text{1 head}) \] Step 1: The probability of getting 0 heads (all tails) in \( n \) tosses is: \[ P(\text{0 heads}) = \left( \frac{1}{2} \right)^n \] Step 2: The probability of getting exactly 1 head is: \[ P(\text{1 head}) = \binom{n}{1} \left( \frac{1}{2} \right)^n = n \left( \frac{1}{2} \right)^n \] Step 3: We want: \[ 1 - \left( \frac{1}{2} \right)^n - n \left( \frac{1}{2} \right)^n \geq 0.96 \] Step 4: Solving this inequality will give us the minimum number of tosses required. After performing the calculations, we find that \( n = 8 \) tosses is the minimum number of tosses required for the probability to be at least 0.96. % Final Answer The minimum number of tosses required is 8.