The minimum number of times a fair coin needs to be tossed, so that the probability of getting at le

Q: Wait, why do we use the complement $1 - P(X < 2)$? Can't we just calculate $P(X \ge 2)$ directly?

You *could*, but it would be incredibly inefficient. To calculate $P(X \ge 2)$ directly, you would need to sum $P(X=2) + P(X=3) + \dots + P(X=n)$. As $n$ increases, this sum becomes longer and more complex. By using the complement, you only need to calculate two simple terms: $P(X=0)$ and $P(X=1)$. It is the classic 'work smarter, not harder' approach in probability.

Q: Why does the inequality sign flip when we take the reciprocal in the final steps?

That is a fundamental rule of inequalities! If you have an inequality like $a \ge b$ where both $a$ and $b$ are positive, then $\frac{1}{a} \le \frac{1}{b}$. Think of it this way: $4 > 2$, but $\frac{1}{4} < \frac{1}{2}$. Since both sides of our inequality are positive, flipping the fraction forces the inequality sign to reverse.

Q: Is there a direct algebraic way to solve for $n$ in $2^n \ge 25(n+1)$ without testing values?

Not with standard elementary algebra! This is a transcendental inequality because $n$ appears both in the exponent ($2^n$) and as a linear term ($n+1$). In competitive exams like JEE, the most efficient and expected method is to test integer values. Since $n$ must be a small integer, testing values like $n=7$ and $n=8$ is much faster than trying to solve it analytically.

Question

The minimum number of times a fair coin needs to be tossed, so that the probability of getting at least two heads is at least 0.96, is .........

Sigma Percentile · Accepted Answer

### The Art of the Coin Toss: Mastering Probability

Imagine you are standing in a quiet room, holding a fair coin. You are tasked with a challenge: toss this coin $n$ times, and ensure that the probability of getting at least two heads is at least $0.96$.

It sounds simple, almost like a game, but beneath this simplicity lies a beautiful mathematical structure. Today, we are going to peel back the layers of this problem and see how we can turn a potentially tedious calculation into an elegant, swift solution.

### The Complementary Strategy: The Shortcut to Success

When we see the phrase 'at least two heads', our first instinct might be to calculate the probability of getting exactly two heads, then three, then four, and so on, all the way up to $n$. But stop! That is the trap.

If $n$ is large, that path is a dead end. Instead, we use the power of the complementary event. The total probability of all possible outcomes is always $1$.

Therefore, the probability of getting at least two heads is simply $1$ minus the probability of getting *fewer* than two heads. Mathematically, we write this as:

$$P(X \ge 2) = 1 - P(X < 2)$$

This is our North Star. It transforms a massive summation into a tiny, manageable problem.

### The Binomial Engine

Now, what does 'fewer than two heads' actually mean? It means we either get zero heads or exactly one head.

So, our condition $P(X \ge 2) \ge 0.96$ becomes:

$$1 - [P(X=0) + P(X=1)] \ge 0.96$$

To find these probabilities, we invoke the Binomial Distribution. For a fair coin, the probability of success (heads) is $p = \frac{1}{2}$, and the probability of failure (tails) is $q = \frac{1}{2}$. The formula for exactly $k$ successes in $n$ trials is:

$$P(X=k) = \binom{n}{k} p^k q^{n-k}$$

For $k=0$, we have:

$$P(X=0) = \binom{n}{0} \left(\frac{1}{2}\right)^0 \left(\frac{1}{2}\right)^n = \frac{1}{2^n}$$

For $k=1$, we have:

$$P(X=1) = \binom{n}{1} \left(\frac{1}{2}\right)^1 \left(\frac{1}{2}\right)^{n-1} = \frac{n}{2^n}$$

Adding these together, we get the probability of getting fewer than two heads: $\frac{n+1}{2^n}$.

### The Algebraic Dance

Now, let us substitute this back into our inequality:

$$1 - \frac{n+1}{2^n} \ge 0.96$$

Subtracting $1$ from both sides gives $-\frac{n+1}{2^n} \ge -0.04$. Multiplying by $-1$ flips the inequality sign:

$$\frac{n+1}{2^n} \le 0.04$$

Converting $0.04$ to a fraction, we get $\frac{n+1}{2^n} \le \frac{4}{100}$, which simplifies to:

$$\frac{n+1}{2^n} \le \frac{1}{25}$$

To make this easier to handle, we take the reciprocal of both sides. Remember, when you invert positive fractions, the inequality sign flips again! We arrive at:

$$\frac{2^n}{n+1} \ge 25$$

This is the heart of the problem. We need the smallest integer $n$ that satisfies this condition.

### The Final Verdict

Let us test our values. If $n=7$, we get:

$$\frac{2^7}{7+1} = \frac{128}{8} = 16$$

Since $16 < 25$, $n=7$ is not enough. But if we try $n=8$, we get:

$$\frac{2^8}{8+1} = \frac{256}{9} \approx 28.4$$

Since $28.4 \ge 25$, the condition is satisfied! The **minimum number of tosses required is 8**.

You see? By breaking the problem down and using the right tools, we turned a daunting probability question into a simple, logical journey. Keep practicing, and soon, these patterns will become second nature to you.

Animated Solution for Mathematics - Probability: The minimum number of times a fair coin needs to be tossed, so that the probability of getting at least two heads is at least 0.96, is .........

Enter Numerical Value:

Visualized Solution

Problem Statement

Mathematical Translation

The Complementary Approach

Breaking Down $P (X < 2)$

Binomial Distribution Setup

Substituting Values

Forming the Equation

Rearranging the Inequality

Simplifying the Constant

Inverting the Inequality

Testing Values: $n = 7$

Testing Values: $n = 8$

The Sigma Insight: Random Variables and Probability Distributions

The Art of the Coin Toss

The Complementary Strategy

The Binomial Engine

The Algebraic Dance

The Final Verdict

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