The minimum number of times one has to toss a fair coin so that the probability of observing at least one head is at least 90% is:

We want the probability of getting at least one head to be at least 90 %. A fair coin has probability $$\tfrac12$$ of showing a head and the same probability $$\tfrac12$$ of showing a tail on every toss.

It is often easier to work with the complementary event. The complementary event to “at least one head in $$n$$ tosses” is “no head at all in $$n$$ tosses”, i.e. all the $$n$$ tosses show tails.

For a single toss, the probability of getting a tail is $$\tfrac12$$. Because successive tosses are independent, the probability that all $$n$$ tosses are tails is

$$\left(\tfrac12\right)^n.$$

The required probability of getting at least one head is therefore

$$1-\left(\tfrac12\right)^n.$$

We are told this probability must be at least 90 %, that is $$0.90$$. Translating the verbal condition into an inequality, we write

$$1-\left(\tfrac12\right)^n \;\ge\; 0.9.$$

Rearranging, we subtract 1 from both sides and then multiply by −1; remembering that multiplying an inequality by a negative reverses the inequality sign, we get

$$\left(\tfrac12\right)^n \;\le\; 0.1.$$

To solve for $$n$$ we take natural logarithms (any logarithm base could be used, but using one base consistently is essential). First we state the logarithm rule we are about to use:

If $$a>0$$ and $$0<b<1$$, then $$b^a=c$$ implies $$a=\dfrac{\ln c}{\ln b}.$$

Applying this rule to our inequality

$$\left(\tfrac12\right)^n \le 0.1,$$

we obtain

$$n \,\ln\!\left(\tfrac12\right) \;\le\; \ln(0.1).$$

Because $$\ln\!\left(\tfrac12\right)$$ is negative, dividing by it reverses the inequality once again:

$$n \;\ge\; \dfrac{\ln(0.1)}{\ln\!\left(\tfrac12\right)}.$$

Now we evaluate the logarithms numerically:

$$\ln(0.1) = -2.302585\ldots,$$

$$\ln\!\left(\tfrac12\right) = -0.693147\ldots.$$

Substituting these values, $$ n \ge \dfrac{-2.302585\ldots}{-0.693147\ldots} = 3.321928\ldots $$

The number of tosses $$n$$ must be an integer, and it must be at least the value just obtained, so we round up to the next whole number:

$$n = 4.$$

We can verify quickly: with 4 tosses the probability of at least one head is

$$1-\left(\tfrac12\right)^4 =1-\dfrac1{16} =\dfrac{15}{16} =0.9375,$$ which indeed exceeds 0.90. For 3 tosses the probability is only $$1-\tfrac18=0.875,$$ which is insufficient.

Hence, the correct answer is Option B.