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

To determine the minimum number of times a fair coin needs to be tossed such that the probability of getting at least two heads is at least 0.96, we can use the concept of complementary probability. Start by calculating the probability of getting fewer than two heads and subtract that from 1.

Let n be the number of tosses. The probability of getting 0 or 1 head can be found by using the binomial probability formula:

P(X=k)=C(n,k)×p^k×(1-p)^(n-k), where p=0.5 for a fair coin.

1. Calculate the probability for 0 heads: P(X=0)=C(n,0)×0.5^0×0.5^n=0.5^n.

2. Calculate the probability for 1 head: P(X=1)=C(n,1)×0.5^1×0.5^(n-1)=n×0.5^n.

3. Sum these probabilities: P(fewer than 2 heads)=0.5^n+n×0.5^n.

4. We require that the probability for at least 2 heads satisfies: P(at least 2 heads)≥0.96.

Thus, 1-(0.5^n+n×0.5^n)≥0.96.

Solving: 0.5^n+n×0.5^n≤0.04 or (n+1)×0.5^n≤0.04.

Check values of n:

• n=5: (5+1)×0.5^5=6×0.03125=0.1875 (Not ≤0.04)
• n=6: (6+1)×0.5^6=7×0.015625=0.109375 (Not ≤0.04)
• n=7: (7+1)×0.5^7=8×0.0078125=0.0625 (Not ≤0.04)
• n=8: (8+1)×0.5^8=9×0.00390625=0.03515625 (≤0.04, satisfies the condition)

Hence, the minimum number of tosses required is 8.

Answer 2

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. The minimum number of tosses required is 8.