If three letters can be posted to any one of the 5 different addresses, then the probability that th

Q: Why do we use ${}^5C_2$ instead of ${}^5P_2$?

Great question! We use ${}^5C_2$ because we are simply selecting a 'set' of two addresses. The order in which we pick the two addresses doesn't matter—whether we pick address A then B, or B then A, we end up with the same pair of destinations to work with.

Q: Why do we subtract 2 in the distribution step ($2^3 - 2$)?

When we calculate $2^3$, we are counting every possible way to put 3 letters into 2 boxes. This includes the scenario where all 3 letters end up in the first box (leaving the second empty) and the scenario where all 3 end up in the second box (leaving the first empty). Since the problem demands 'exactly two' addresses, we must exclude these two 'all-in-one' cases.

Q: Can we solve this using the Inclusion-Exclusion Principle?

Absolutely! You could define sets $A_i$ as the event that address $i$ is empty. However, for a small problem like this, the 'Selection then Distribution' method is much more intuitive and less prone to calculation errors. Stick to the method that feels most logical to you!

Q: Does it matter if the letters are distinct or identical?

It matters immensely! The problem states 'three letters,' and in probability, unless specified as identical, we treat items as distinct. If the letters were identical, the number of ways to distribute them would change significantly (using stars and bars), but here, the distinct nature of the letters is what gives us the $2^3$ distribution factor.

Q: Is there a way to check if my answer is reasonable?

Always! A probability must be between 0 and 1. Our result, $\frac{12}{25} = 0.48$, is a very healthy number. If you had gotten something like 1.2 or a negative number, you would know immediately that a counting error occurred. Always perform a sanity check!

Question

If three letters can be posted to any one of the 5 different addresses, then the probability that the three letters are posted to exactly two addresses is:

Options:

(A) $\frac{18}{25}$

(B) $\frac{12}{25}$

(C) $\frac{6}{25}$

(D) $\frac{4}{25}$

Sigma Percentile · Accepted Answer

### Analyzing the Universe

In any probability problem, we must first define our 'Universe'—the sample space. If each of our three letters has the freedom to choose any of the five addresses, how many total scenarios exist?

Since each letter is independent, the first letter has 5 choices, the second has 5, and the third has 5. Mathematically, we represent this as:

$$n(S) = 5 	imes 5 	imes 5 = 125$$

This is the total number of ways the universe could unfold if we let chaos reign. This is our denominator, the bedrock of our probability.

### The Strategy of Selection

Now, we impose our constraint: we want exactly two addresses to be used. We cannot just jump into the distribution; we must first curate our environment.

We need to choose which two addresses out of the five will be the 'lucky' ones. This is a classic selection problem. We use the combination formula, ${}^nC_r$, because the order in which we select the two addresses is irrelevant.

Choosing address $A$ and $B$ is the same as choosing $B$ and $A$. Thus, we calculate:

$${}^5C_2 = \frac{5 	imes 4}{2 	imes 1} = 10$$

We have 10 distinct pairs of addresses that could potentially host our letters.

### The Distribution Dance

With our two chosen addresses—let's call them $X$ and $Y$—we now distribute the three letters. Each letter has 2 choices: it can go to $X$ or it can go to $Y$.

The total number of ways to distribute the letters into these two specific boxes is:

$$2^3 = 8$$

But wait! We must be careful. Does this count include cases where we fail our mission? Yes.

If all three letters go to $X$, then $Y$ is empty. If all three go to $Y$, then $X$ is empty. In both cases, we have used only one address, not two.

To satisfy the condition of 'exactly two,' we must subtract these two 'all-in-one' scenarios. Therefore, the number of valid ways to distribute the letters into our two chosen addresses is:

$$8 - 2 = 6$$

### The Grand Synthesis

We are almost there. We have 10 ways to pick our pair of addresses, and for each pair, we have 6 ways to distribute the letters such that both are used.

By the Fundamental Counting Principle, we multiply these together to find our total favorable outcomes:

$$n(E) = 10 	imes 6 = 60$$

Finally, we calculate the probability $P(E)$ by dividing our favorable outcomes by the total sample space:

$$P(E) = \frac{60}{125}$$

By simplifying this fraction, dividing both numerator and denominator by 5, we arrive at our elegant final answer:

**$$\frac{12}{25}$$**

### Reflecting on the Elegance

Look at what we have achieved. We broke a complex constraint into a selection phase and a distribution phase. We respected the distinct nature of the letters and the addresses.

We accounted for the 'trap' of using only one address by subtracting the invalid cases. This is the essence of JEE-level thinking: breaking the intimidating into the manageable. Keep this logical structure in your toolkit, and no combinatorial problem will ever stand in your way again.

Animated Solution for Mathematics - Probability: If three letters can be posted to any one of the 5 different addresses, then the probability that the three letters are posted to exactly two addresses is:

Select Answer:

Visualized Solution

Visualizing the Setup

Total Possible Outcomes

Calculating $n (S)$

Defining Favorable Outcomes

Step A: Selecting $2$ Addresses

Calculating Selection Ways

Step B: Distributing Letters

Removing Invalid Cases

Total Favorable Outcomes $n (E)$

Calculating Final Probability

Simplifying the Result

The Sigma Insight: Classical Definition of Probability

Analyzing the Universe

The Strategy of Selection

The Distribution Dance

The Grand Synthesis

Reflecting on the Elegance

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