Minimum Coin Tosses to Achieve At Least Two Heads

Let $n$ be the number of times a fair coin is tossed.

Let $X$ be the random variable denoting the number of heads obtained in $n$ tosses. Since the coin is fair, the probability of getting a head in a single toss is $p = \frac{1}{2}$, and the probability of getting a tail is $q = 1 - p = \frac{1}{2}$.

Thus, $X$ follows a binomial distribution $X \sim \text{Binomial}\left(n, \frac{1}{2}\right)$, and the probability of obtaining exactly $k$ heads in $n$ tosses is given by: $P(X = k) = \binom{n}{k} \left(\frac{1}{2}\right)^k \left(\frac{1}{2}\right)^{n-k} = \binom{n}{k} \left(\frac{1}{2}\right)^n$

We are required to find the minimum value of $n$ such that the probability of obtaining at least two heads is at least $0.96$: $P(X \ge 2) \ge 0.96$

Using the complement rule, we can rewrite $P(X \ge 2)$ as: $P(X \ge 2) = 1 - P(X = 0) - P(X = 1)$

Substitute the probabilities for $X = 0$ and $X = 1$: $P(X = 0) = \binom{n}{0} \left(\frac{1}{2}\right)^n = \frac{1}{2^n}$ $P(X = 1) = \binom{n}{1} \left(\frac{1}{2}\right)^n = \frac{n}{2^n}$

Therefore, $P(X \ge 2) = 1 - \left(\frac{1}{2^n} + \frac{n}{2^n}\right) = 1 - \frac{n+1}{2^n}$

Now, set up the inequality: $1 - \frac{n+1}{2^n} \ge 0.96$ $\frac{n+1}{2^n} \le 1 - 0.96$ $\frac{n+1}{2^n} \le 0.04$ $\frac{n+1}{2^n} \le \frac{1}{25}$ $\frac{2^n}{n+1} \ge 25$

Now, we test positive integer values for $n$:

• For $n = 6$: $\frac{2^6}{6+1} = \frac{64}{7} \approx 9.14 < 25$

• For $n = 7$: $\frac{2^7}{7+1} = \frac{128}{8} = 16 < 25$

• For $n = 8$: $\frac{2^8}{8+1} = \frac{256}{9} \approx 28.44 \ge 25$

Since $n = 8$ is the smallest integer satisfying the inequality, the minimum number of tosses required is $8$.