One hundred identical coins, each with probability, p, of showing up heads are tossed once. If 0

Q: Wait, if we have 100 coins, shouldn't the probability be $0.5$ for the peak? Why is the answer not $1/2$?

That is a brilliant intuition, but let's refine it! The peak of a binomial distribution is at $np$ only when $np$ is an integer. If $p=0.5$, the peak is at $50$. However, for $P(X=50)$ to equal $P(X=51)$, the distribution must be perfectly symmetric around the midpoint of $50$ and $51$. This only happens if $p=0.5$ in specific scenarios. Here, the condition forces $p$ to be slightly different from $0.5$ to 'shift' the distribution so that the heights at $50$ and $51$ align. Always trust the algebra over the intuition when the parameters are constrained!

Q: Do I really need to expand $100!$ to solve this? That sounds like a nightmare.

Never! If you find yourself trying to calculate $100!$, you are walking into a trap. In JEE problems, we never calculate large factorials. We use the property $n! = n \times (n-1)!$ to cancel terms. The beauty of this problem lies in the elegant cancellation of the factorials, not in the arithmetic. Keep your eyes on the ratios, not the raw numbers.

Q: Why can't we just use the normal approximation for the binomial distribution?

You could, but it would be an approximation, not an exact solution. The JEE Advanced exam tests your ability to handle exact algebraic relationships. Using the normal approximation would give you an estimate, but the problem asks for the exact value of $p$. Stick to the exact Binomial PMF to get the precise answer.

Question

One hundred identical coins, each with probability, $p$, of showing up heads are tossed once. If $0 < p < 1$ and the probability of heads showing on $50$ coins is equal to that of heads showing on $51$ coins, then the value of $p$ is

Options:

(A) $1/2$

(B) $49/101$

(C) $50/101$

(D) $51/101$

Sigma Percentile · Accepted Answer

### The Dance of Probability: Unlocking the Binomial Mystery

My dear student, welcome to the arena. Today, we are not just solving a probability problem; we are stepping into the shoes of a statistician observing the chaotic yet beautiful dance of one hundred coins.

Imagine standing in a vast hall, tossing one hundred identical coins into the air. Each coin has a probability $p$ of landing heads-up. We are looking for the moment where the universe balances perfectly—where the likelihood of seeing exactly $50$ heads is identical to the likelihood of seeing $51$ heads.

Let us peel back the layers of this problem together.

### Phase 1: Defining the Battlefield

When we deal with a fixed number of independent trials—in our case, $n=100$ coin tosses—we are firmly in the territory of the Binomial Distribution. This is our primary weapon.

The probability of achieving exactly $k$ successes (heads) is given by the elegant formula:

$$P(X=k) = \binom{n}{k} p^k (1-p)^{n-k}$$

Here, $\binom{n}{k}$ represents the number of ways to choose $k$ coins out of $n$ to be heads. The term $p^k$ accounts for the $k$ heads, and $(1-p)^{n-k}$ accounts for the remaining $n-k$ tails. It is a perfect balance of success and failure.

Our problem gives us a specific condition: the probability of $50$ heads is equal to the probability of $51$ heads. So, we write:

$$P(X=50) = P(X=51)$$

### Phase 2: The Algebraic Dance

Now, let us substitute our values into the formula. We have $n=100$. For $k=50$, we have:

$$P(X=50) = \binom{100}{50} p^{50} (1-p)^{50}$$

And for $k=51$, we have:

$$P(X=51) = \binom{100}{51} p^{51} (1-p)^{49}$$

Equating these two, we get:

$$\binom{100}{50} p^{50} (1-p)^{50} = \binom{100}{51} p^{51} (1-p)^{49}$$

I know what you are thinking: "This looks like a mess of exponents and combinations." But take a breath. This is where the magic of mathematics reveals itself; we do not need to calculate these values, we only need to simplify.

### Phase 3: The Art of Cancellation

Look at the powers of $p$ and $(1-p)$. On the left, we have $p^{50}$ and $(1-p)^{50}$. On the right, we have $p^{51}$ and $(1-p)^{49}$.

If we divide both sides by $p^{50}$, the left side loses its $p$ term, and the right side is left with just $p^1$. Similarly, if we divide both sides by $(1-p)^{49}$, the right side loses its $(1-p)$ term, and the left side is left with $(1-p)^1$.

Our equation transforms into something much more manageable:

$$\binom{100}{50} (1-p) = \binom{100}{51} p$$

Now, let us expand the combinations using the definition $\binom{n}{r} = \frac{n!}{r!(n-r)!}$:

$$\frac{100!}{50! 50!} (1-p) = \frac{100!}{51! 49!} p$$

See how the $100!$ on both sides vanishes? It is as if it was never there. We are left with:

$$\frac{1-p}{50! 50!} = \frac{p}{51! 49!}$$

### Phase 4: The Grand Finale

We are almost there. We have $50!$ and $51!$ in the denominators. Remember that $51! = 51 	imes 50!$ and $50! = 50 	imes 49!$.

Let us substitute these into our equation:

$$\frac{1-p}{50! 	imes (50 	imes 49!)} = \frac{p}{(51 	imes 50!) 	imes 49!}$$

Cancel out the $50!$ and $49!$ from both sides. The equation collapses into a beautiful, simple linear form:

$$\frac{1-p}{50} = \frac{p}{51}$$

Cross-multiplying gives us:

$$51(1-p) = 50p$$

$$51 - 51p = 50p$$

Moving the $-51p$ to the other side, we find:

$$51 = 101p$$

$$p = \frac{51}{101}$$

And there it is. The value of $p$ that makes the universe balance perfectly is **$p = \frac{51}{101}$**. You see, my friend, the complexity was just a veil. By staying calm and trusting the process of cancellation, we turned a daunting probability problem into a simple algebraic victory.

Animated Solution for Mathematics - Probability: One hundred identical coins, each with probability, $p$ , of showing up heads are tossed once. If $0 < p < 1$ and the probability of heads showing on $50$ coins is equal to that of heads showing on $51$ coins, then the value of $p$ is

Select Answer:

Visualized Solution

The Coin Toss Experiment

The Given Condition

Binomial Probability Formula

Substituting the Values

Simplifying the Exponents

Expanding the Combinations

Cancelling Common Terms

Simplifying the Factorials

Cross-Multiplication

Final Calculation for $p$

The Sigma Insight: Binomial Distribution

The Dance of Probability

Phase 1

Phase 2

Phase 3

Phase 4

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