A pair of dice is thrown 5 times. For each throw, a total of 5 is considered a success. If the proba

Q: Wait, why can't we just multiply the probabilities? Why do we need the Binomial Distribution?

That is a fantastic question! Simple multiplication only works if you are looking for a specific sequence of events (e.g., Success, Success, Failure, Success, Success). But here, the successes can happen in any order! The Binomial Distribution accounts for all those different combinations (like SSSSF, SSFSS, etc.) automatically using the $\binom{n}{r}$ term. It is the difference between looking at one specific path and looking at the entire forest of possibilities.

Q: I got $41/9^5$, but the question asks for $k/3^{11}$. Did I do something wrong?

Not at all! You are actually on the right track. The question is testing your ability to manipulate exponents. Remember that $9 = 3^2$, so $9^5 = (3^2)^5 = 3^{10}$. Once you have $41/3^{10}$, you just need to multiply the numerator and denominator by 3 to get the denominator to $3^{11}$. It is a common JEE trick to make you adjust your final answer to match a specific format.

Q: Why do we include $P(X=5)$? The question says 'at least 4'.

In probability, 'at least 4' is an inclusive phrase. It means 4 or more. Since we are only throwing the dice 5 times, the only possibilities that satisfy 'at least 4' are exactly 4 successes and exactly 5 successes. If you only calculated $P(X=4)$, you would be missing the case where you get lucky and succeed all 5 times!

Question

A pair of dice is thrown 5 times. For each throw, a total of 5 is considered a success. If the probability of at least 4 successes is $\frac{k}{3^{11}}$, then $k$ is equal to

Options:

(A) 82

(B) 75

(C) 164

(D) 123

Sigma Percentile · Accepted Answer

### The Geometry of Chance: A Journey into Binomial Probability

Welcome, future engineer! Today, we are not just solving a probability problem; we are stepping into the shoes of a statistician analyzing a series of independent events.

Imagine you are standing in a casino, holding two dice. You are about to throw them five times. The goal is to get a sum of 5 as often as possible. This is the essence of the **Binomial Distribution**—a powerful framework that turns chaos into predictable patterns.

### Phase 1: The Grid of Possibilities

First, let us ground ourselves in the physical reality of the experiment. When you throw two dice, each die has 6 faces. The total number of unique outcomes is $6 	imes 6 = 36$.

Visualize a $6 	imes 6$ grid. Each cell in this grid represents one possible outcome of the pair. Now, we define our 'success'. A success occurs when the sum of the two dice is exactly 5.

Let us hunt for these pairs: $(1, 4)$, $(2, 3)$, $(3, 2)$, and $(4, 1)$. There are exactly 4 favorable outcomes. Thus, the probability of success in a single throw is:

$$p = \frac{4}{36} = \frac{1}{9}$$

Consequently, the probability of failure is:

$$q = 1 - p = \frac{8}{9}$$

### Phase 2: The Binomial Engine

Now, we move to the core of the problem. We are throwing the dice 5 times. This is a classic Binomial Distribution scenario because the trials are independent, and the probability of success remains constant.

We want the probability of 'at least 4 successes'. This means we need to calculate the probability of exactly 4 successes, $P(X=4)$, and exactly 5 successes, $P(X=5)$, and add them together.

Using the Binomial formula $P(X = r) = \binom{n}{r} p^r q^{n-r}$, let us calculate $P(X=4)$:

$$P(X = 4) = \binom{5}{4} \left(\frac{1}{9}ight)^4 \left(\frac{8}{9}ight)^1 = 5 	imes \frac{1}{9^4} 	imes \frac{8}{9} = \frac{40}{9^5}$$

Next, let us calculate $P(X=5)$:

$$P(X = 5) = \binom{5}{5} \left(\frac{1}{9}ight)^5 \left(\frac{8}{9}ight)^0 = 1 	imes \frac{1}{9^5} 	imes 1 = \frac{1}{9^5}$$

### Phase 3: The Final Bridge

Adding these together, we get the total probability:

$$P(X \ge 4) = \frac{40}{9^5} + \frac{1}{9^5} = \frac{41}{9^5}$$

Here is where the JEE examiner tests your attention to detail. The question demands the answer in the form $\frac{k}{3^{11}}$.

We know that $9 = 3^2$, so $9^5 = (3^2)^5 = 3^{10}$. Our expression is $\frac{41}{3^{10}}$. To reach the target denominator of $3^{11}$, we multiply both the numerator and the denominator by 3:

$$\frac{41 	imes 3}{3^{10} 	imes 3} = \frac{123}{3^{11}}$$

Comparing this to $\frac{k}{3^{11}}$, we find that **$k = 123$**. You have successfully navigated the trap!

Animated Solution for Mathematics - Probability: A pair of dice is thrown 5 times. For each throw, a total of 5 is considered a success. If the probability of at least 4 successes is $\frac{k}{3 ^{11}}$ , then $k$ is equal to

Select Answer:

Visualized Solution

Sample Space of Two Dice

Condition for Success

Probability of Success and Failure

Binomial Distribution Setup

Probability of Exactly 4 Successes

Probability of Exactly 5 Successes

Total Probability $P (X \geq 4)$

Matching the Given Format

Final Conclusion

The Sigma Insight: Binomial Distribution

The Geometry of Chance

Phase 1

Phase 2

Phase 3

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