A student appeared in an examination consisting of 8 true-false type questions. The student guesses

Q: Wait, why is the distribution symmetric? Does this only happen when $p=1/2$?

Yes! This is the 'magic' of the fair coin toss. When $p=1/2$, the binomial coefficient $\binom{n}{k}$ is equal to $\binom{n}{n-k}$, and the probability terms $(p)^k(1-p)^{n-k}$ become $(1/2)^n$ for all $k$. This creates a perfect mirror image around the mean $\mu = np = 4$.

Q: Why can't we just calculate the values for $n=5, 6, 7, 8$ and add them up?

You absolutely could! That is the brute-force method. But in JEE Advanced, time is your currency. Understanding the symmetry allows you to solve this in seconds without calculating massive binomial coefficients like $\binom{8}{5}$ or $\binom{8}{6}$. Always look for the structural shortcut first.

Q: Why is $P(X \ge 4)$ greater than $1/2$?

Think of it this way: $P(X \ge 4)$ includes the 'middle' value $P(X=4)$ plus the right tail $P(X \ge 5)$. Since the total probability is $1$, and the distribution is symmetric, the middle value $P(X=4)$ pushes the sum of the right side just over the $1/2$ threshold. It is the inclusion of the mean that tips the scale.

Question

A student appeared in an examination consisting of 8 true-false type questions. The student guesses the answers with equal probability. The smallest value of $n$, so that the probability of guessing at least '$n$' correct answers is less than $1/2$, is

Options:

(A) 5

(B) 6

(C) 3

(D) 4

Sigma Percentile · Accepted Answer

### The Art of the Guess: Unlocking the Binomial Symmetry

Imagine you are sitting in the examination hall. You have eight true-false questions in front of you. You have no idea what the answers are, so you decide to guess.

It feels like a game of chance, doesn't it? But in the world of mathematics, this 'guessing' is a perfectly structured experiment. This is the Binomial Distribution, and today, we are going to master it.

### Phase 1: The Anatomy of the Guess

Let us define our random variable $X$ as the number of correct answers. Since each question is a binary choice—either you are right or you are wrong—and you are guessing with equal probability, the probability of success is $p = 1/2$.

With $n = 8$ questions, we are dealing with a classic Binomial Distribution: $X \sim B(8, 1/2)$. The probability of getting exactly $k$ correct answers is given by the elegant formula:

$$P(X = k) = \binom{8}{k} \left( \frac{1}{2} ight)^8$$

This formula is the engine of our problem. It tells us exactly how likely it is to get any specific score, from a complete disaster ($0$ correct) to a perfect score ($8$ correct).

### Phase 2: The Mirror of Symmetry

Here is where the intuition kicks in. Because our probability of success is exactly $1/2$, our distribution is not skewed; it is perfectly symmetric.

The mean, or the expected value, is $\mu = np = 8 	imes \frac{1}{2} = 4$. If you were to plot this on a graph, the bar at $x=4$ would be the tallest, and the bars to the left would perfectly mirror the bars to the right.

This means that the probability of getting $3$ correct is the same as getting $5$ correct, $2$ is the same as $6$, and so on. Mathematically, we express this as:

$$P(X = k) = P(X = 8 - k)$$

This symmetry is our greatest weapon.

### Phase 3: The Algebraic Dance

We know that the sum of all probabilities in any distribution must equal $1$. So, $\sum_{k=0}^{8} P(X = k) = 1$.

To solve our problem, we need to partition this sum. Let us break it into three distinct regions: the left tail ($X \le 3$), the middle peak ($X = 4$), and the right tail ($X \ge 5$).

Our equation becomes:

$$P(X \le 3) + P(X = 4) + P(X \ge 5) = 1$$

Now, apply the symmetry we just discussed. The left tail $P(X \le 3)$ is identical to the right tail $P(X \ge 5)$. We can rewrite our equation as:

$$2 P(X \ge 5) + P(X = 4) = 1$$

This is the moment of clarity. We have reduced a complex summation into a simple algebraic relationship.

### Phase 4: The Final Verdict

We are looking for the smallest $n$ such that $P(X \ge n) < 1/2$. Let us test $n = 5$.

From our equation, we can isolate $P(X \ge 5)$:

$$P(X \ge 5) = \frac{1 - P(X = 4)}{2}$$

Since $P(X = 4)$ is the probability of getting exactly $4$ correct, it is definitely a positive value. Therefore, $1 - P(X = 4)$ is strictly less than $1$. When we divide this by $2$, the result $P(X \ge 5)$ must be strictly less than $1/2$. This satisfies our condition!

But what about $n = 4$? If we look at $P(X \ge 4)$, we are including the middle term:

$$P(X \ge 4) = P(X = 4) + P(X \ge 5)$$

Using our previous derivation, this is equal to $1 - P(X \ge 5)$. Since we know $P(X \ge 5) < 1/2$, subtracting it from $1$ gives us a value greater than $1/2$.

Thus, $P(X \ge 4) > 1/2$. The condition fails for $n=4$. Therefore, the smallest integer $n$ that makes the probability less than $1/2$ is **$5$**.

### Conclusion

Mathematics is not about memorizing formulas; it is about seeing the structure beneath the numbers. By visualizing the symmetry of the Binomial Distribution, we turned a potentially tedious calculation into a logical deduction.

You didn't need to calculate $\binom{8}{5}$ or $\binom{8}{6}$. You simply needed to understand the balance of the distribution. Keep this perspective in your toolkit, and you will find that even the most intimidating JEE problems have a beautiful, simple core waiting to be discovered.

Select Answer:

Visualized Solution

Defining the Random Variable $X$

Identifying the Binomial Distribution

Visualizing the Symmetry

The Total Probability Equation

Applying Symmetry to the Sum

Analyzing $P (X \geq 5)$

Bounding $P (X \geq 5)$

Checking the Case for $n = 4$

Bounding $P (X \geq 4)$

The Final Conclusion

The Sigma Insight: Binomial Distribution

The Art of the Guess

Phase 1

Phase 2

Phase 3

Phase 4

Conclusion

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