Animated Solution for Mathematics - Probability: 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 :

My dear student, welcome to the beautiful world of probability. Today, we are going to dissect a problem that seems simple on the surface but carries a profound lesson about how we approach mathematical challenges. We are asked to find the minimum number of tosses, $n$, required to ensure that the probability of observing at least one head is at least 90%.

Many students, when they see 'at least one', immediately try to calculate the probability of getting exactly one head, plus the probability of getting exactly two heads, and so on. Stop! If you do that, you are walking into a classic JEE trap. You are making the problem harder than it needs to be. In the realm of competitive exams, efficiency is just as important as accuracy. Let us learn the elegant way.

Whenever you encounter the phrase 'at least one', your first instinct should be to look at the complement. What is the opposite of 'at least one head'? It is 'no heads at all'.

Think about it: if you toss a coin $n$ times, there are only two possibilities for the outcome regarding heads: either you get at least one head, or you get absolutely zero heads. Since these two events cover all possibilities, their probabilities must sum to 1. Mathematically, we write this as:

This is the key that unlocks the door. Instead of summing up probabilities for 1, 2, 3, ..., $n$ heads, we only need to calculate the probability of the single event where every single toss results in a tail. This is the power of the complement rule.

Now, let us define the probability of getting 'no heads'. If we toss a fair coin, the probability of getting a tail in a single toss is $\frac{1}{2}$. If we toss it $n$ times, and we want a tail every single time, we are looking at independent events. Therefore, we multiply the probabilities:

Now, we substitute this back into our inequality. We want the probability of at least one head to be at least 90%, or 0.9. So, our inequality becomes:

This is where the algebra begins. We need to isolate $n$. Let us move the terms around carefully. Subtract 1 from both sides, or better yet, move the 0.9 to the left and the ${\left(\frac{1}{2}\right)}^n$ to the right:

We are almost there, but we must be careful with the next step. We have $0.1\ge {\left(\frac{1}{2}\right)}^n$. Let us write 0.1 as a fraction, $\frac{1}{10}$. So we have:

Now, we want to solve for $n$. To do this, we take the reciprocal of both sides. But remember the golden rule of inequalities: when you take the reciprocal of both sides of an inequality involving positive numbers, the inequality sign flips! It is a fundamental property of the function $f(x)=\frac{1}{x}$. Thus, our inequality becomes:

or, written more conventionally:

Now, we simply test integer values for $n$. This is the moment of truth. We are looking for the smallest integer $n$ such that $2^n$ is at least 10.

If $n=1$, $2^1=2$. This is less than 10. We are nowhere near our target.

If $n=2$, $2^2=4$. Still less than 10.

If $n=3$, $2^3=8$. We are getting closer! The probability of at least one head here is $1-\frac{1}{8}=1-0.125=0.875$, which is 87.5%. It is very close to 90%, but it is not quite there.

If $n=4$, $2^4=16$. Finally! 16 is greater than 10. The probability of at least one head here is $1-\frac{1}{16}=1-0.0625=0.9375$, which is 93.75%. This crosses our 90% threshold.

So, my friend, we have our answer. We need a minimum of 4 tosses.

This problem teaches us more than just probability; it teaches us strategy. It teaches us to look for the complement, to handle inequalities with care, and to verify our results. When you face the JEE, remember this: don't just calculate—think. Visualize the problem, choose the most efficient path, and execute with precision. You have the tools; now go out there and master the physics and math of the universe!