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:
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:
Updated On: Mar 25, 2026
- $\frac{12}{25}$
- $\frac{18}{25}$
- $\frac{4}{25}$
- $\frac{6}{25}$
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The Correct Option is A
Approach Solution - 1
To solve this problem, we need to find the probability that exactly two addresses are used when posting three letters to any one of the five different addresses.
- First, calculate the total number of ways to post three letters to any of the five addresses.
Since each letter can be posted to any one of the five addresses independently, the total number of combinations is given by: \(5^3 = 125\). - Next, we determine the number of favorable outcomes where exactly two addresses are used.
- Choose 2 addresses out of the 5 available for the letters. This can be done in: \(\binom{5}{2} = 10\) ways.
- Once the addresses are chosen, we need to distribute the 3 letters such that both addresses receive at least one letter. The distribution where exactly two addresses are used can be split into two cases:
- Case 1: One address receives 1 letter, and the other receives 2 letters.
- Case 2: The same structure as Case 1, but the addresses are swapped.
- For Case 1: Choose 1 letter to go to the first address (3 ways), and the remaining 2 letters go to the second address (1 way). Hence, there are: \(\binom{3}{1} \cdot 1 = 3\) ways.
- Since either address can receive any set of letters, another 3 ways arise by exchanging letters. Therefore, total ways = \(3 + 3 = 6\).
- For each pair of addresses, there are 6 ways to distribute the letters. Thus, the total number of favorable outcomes is: \(10 \cdot 6 = 60\).
- Finally, the probability is the ratio of favorable outcomes to the total number of outcomes: \(\frac{60}{125} = \frac{12}{25}\).
Therefore, the probability that the three letters are posted to exactly two addresses is \(\frac{12}{25}\).
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Approach Solution -2
\[ \text{Total methods} = 5^3 \]
\[ \text{Favorable} = ^3C_2 \times (2^3 - 2) = 60 \]
\[ \text{Probability} = \frac{60}{125} = \frac{12}{25} \]
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