A fair coin is tossed n-times such that the probability of getting at least one head is at least 0.9

Q: Wait, why do we reject the direct approach of calculating the probability of 1 head, 2 heads, etc.?

You absolutely could calculate the probability of getting exactly 1 head, 2 heads, and so on, up to $n$ heads, and sum them up. However, that is mathematically exhausting! The beauty of the complement rule is that it collapses all those complex scenarios into one simple case: the probability of getting zero heads. It is the most elegant and efficient path to the solution.

Q: When I rearrange the inequality $0.1 \ge (\frac{1}{2})^n$, how do I handle the flip?

Great question! When you convert the decimal $0.1$ to the fraction $\frac{1}{10}$, you get $\frac{1}{10} \ge \frac{1}{2^n}$. When you take the reciprocal of both sides of an inequality involving positive numbers, the inequality sign must flip. Think of it this way: $\frac{1}{2} 1$. The same logic applies here, turning $\frac{1}{10} \ge \frac{1}{2^n}$ into $10 \le 2^n$.

Q: Is there a formula for this, or do I just guess values for $n$?

For small values of $n$, testing integers is perfectly valid and often faster than using logarithms. However, if the question asked for a much higher probability threshold (like 0.999), you would use logarithms: $n \ge \log_2(10)$. For this problem, testing $n=1, 2, 3, 4$ is the most intuitive way to see the threshold being crossed.

Question

A fair coin is tossed $n$-times such that the probability of getting at least one head is at least 0.9. Then the minimum value of $n$ is

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing before a table, a fair coin in your hand. You are about to toss it $n$ times. The question asks for the minimum number of tosses required to ensure the probability of getting at least one head is at least $0.9$.

At first glance, this might seem like a daunting task. You might be tempted to calculate the probability of getting exactly one head, then two, then three, all the way up to $n$, and sum them up. But stop! That is the long, winding road.

In probability, whenever you encounter the phrase "at least one," a bell should ring in your mind. It is a signal to use the **complement rule**. The total probability of all possible outcomes is always $1$. If we want the probability of getting at least one head, we can simply subtract the probability of the only outcome we do not want: getting zero heads.

### The Power of Independence

What does it mean to get zero heads in $n$ tosses? It means every single toss resulted in a tail. Since each coin toss is an independent event, the outcome of one toss does not influence the next.

The probability of getting a tail on any single toss is $P(T) = \frac{1}{2}$. Therefore, the probability of getting tails $n$ times in a row is simply the product of the individual probabilities:

$$P(	ext{All T}) = P(T_1) 	imes P(T_2) 	imes \dots 	imes P(T_n) = \left( \frac{1}{2} ight)^n$$

This is the core geometric reality of the problem. We have reduced a complex summation into a single, clean exponential term.

### Solving the Inequality

Now, let us bring in the condition given in the problem: the probability of getting at least one head must be at least $0.9$. Using our complement rule, we write:

$$1 - P(	ext{All T}) \ge 0.9$$

Substituting our expression for $P(	ext{All T})$, we get:

$$1 - \left( \frac{1}{2} ight)^n \ge 0.9$$

Now, let us perform the algebraic dance to isolate $n$. Subtracting $0.9$ from both sides and moving the exponential term to the right, we find:

$$0.1 \ge \left( \frac{1}{2} ight)^n$$

To make this easier to handle, let us convert the decimal $0.1$ into the fraction $\frac{1}{10}$. So:

$$\frac{1}{10} \ge \frac{1}{2^n}$$

Here is where many students stumble. When you take the reciprocal of both sides of an inequality involving positive numbers, the inequality sign must flip. It is a fundamental rule of algebra. Thus, we arrive at:

$$10 \le 2^n \quad 	ext{or} \quad 2^n \ge 10$$

### The Final Leap

We are now at the finish line. We need the smallest integer $n$ such that $2^n$ is at least $10$. Let us test the values:

For $n=1$, $2^1 = 2$, which is less than $10$. For $n=2$, $2^2 = 4$, still less than $10$. For $n=3$, $2^3 = 8$, getting closer, but still less than $10$.

Finally, for $n=4$, $2^4 = 16$. Sixteen is clearly greater than or equal to $10$. The condition is satisfied!

The **minimum number of tosses required is 4**. This journey shows that even in the face of complex-sounding probability problems, the right strategy—the complement rule—can turn a mountain of work into a simple, elegant calculation.

Animated Solution for Mathematics - Probability: A fair coin is tossed $n$ -times such that the probability of getting at least one head is at least 0.9. Then the minimum value of $n$ is

Enter Numerical Value:

Visualized Solution

Define the Experiment

The 'At Least One' Strategy

Calculate Probability of All Tails

Formulate the Inequality

Rearrange the Inequality

Invert the Fractions

Test Values for $n$

Find the Minimum $n$

The Sigma Insight: Classical Definition of Probability

Analyzing the Setup

The Power of Independence

Solving the Inequality

The Final Leap

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