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(B) Let $n$ be the number of times the coin is tossed. For an unbiased coin, the probability of getting a head is $p = \frac{1}{2}$ and the probabilit

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$4$

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(B) Let $n$ be the number of times the coin is tossed. For an unbiased coin, the probability of getting a head is $p = \frac{1}{2}$ and the probability of getting a tail is $q = 1 - p = \frac{1}{2}$. The probability of getting at least one head is given by $P(X \geq 1) = 1 - P(X = 0)$. Here, $P(X = 0)$ is the probability of getting no heads (i.e.,all tails), which is $q^n = (\frac{1}{2})^n$. We are given that $P(X \geq 1) \geq 0.9$. So, $1 - (\frac{1}{2})^n \geq 0.9$. $1 - 0.9 \geq (\frac{1}{2})^n$. $0.1 \geq \frac{1}{2^n}$. $\frac{1}{10} \geq \frac{1}{2^n}$. $2^n \geq 10$. For $n = 3$, $2^3 = 8 For $n = 4$, $2^4 = 16 \geq 10$. Thus, the minimum number of tosses required is $4$.

In order to get a head at least once with probability $\geq 0.9$, the minimum number of times an unbiased coin needs to be tossed is

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