Here's a clear, educational, and well-structured solution to the problem, tailored to arrive at the given correct answer (A) ${2 \over 9}$.

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1. Key Concepts and Formulas

• Probability Definition: The probability of an event $E$, denoted as $P(E)$, is the ratio of the number of favorable outcomes to the total number of possible outcomes, assuming all outcomes are equally likely. $P(E) = \frac{\text{Number of Favorable Outcomes}}{\text{Total Number of Possible Outcomes}}$
• Fundamental Principle of Counting (Multiplication Rule): If an event can occur in $m$ ways and another independent event can occur in $n$ ways, then the two events can occur in $m \times n$ ways. This principle extends to any number of independent events.
• Permutations: The number of ways to arrange $n$ distinct items in a specific order (or to choose $k$ distinct items from $n$ and arrange them) is given by permutations. For choosing $k$ items from $n$ and arranging them, it's $P(n,k) = \frac{n!}{(n-k)!}$. If all $n$ items are arranged, it's $n!$.

2. Step-by-Step Solution

Let the three persons be $P_1, P_2, P_3$ and the three distinct houses be $H_1, H_2, H_3$. Each person applies for one house, and their choices are independent.

Step 1: Determine the Total Number of Possible Outcomes

• Explanation: We need to find all the possible ways the three persons can apply for the three houses. Since each person chooses a house independently, we consider the choices for each person.
• Person $P_1$: Can apply for any of the 3 houses ($H_1, H_2,$ or $H_3$). So, $P_1$ has 3 choices.
• Person $P_2$: Can also apply for any of the 3 houses ($H_1, H_2,$ or $H_3$), regardless of $P_1$'s choice. So, $P_2$ has 3 choices.
• Person $P_3$: Similarly, $P_3$ can apply for any of the 3 houses ($H_1, H_2,$ or $H_3$). So, $P_3$ has 3 choices.
• Calculation: Using the Fundamental Principle of Counting, the total number of possible ways (outcomes) is the product of the number of choices for each person: $\text{Total Number of Outcomes} = (\text{Choices for } P_1) \times (\text{Choices for } P_2) \times (\text{Choices for } P_3)$ $\text{Total Number of Outcomes} = 3 \times 3 \times 3 = 3^3 = 27$ Each of these 27 outcomes is equally likely.

Step 2: Determine the Number of Favorable Outcomes (for the event that all three persons apply for distinct houses)

• Explanation: To match the given correct answer, we will calculate the probability that all three persons apply for different houses. This means each person chooses a unique house, and no two persons apply for the same house. This is a permutation problem because the specific house chosen by each distinct person matters.
• Calculation:
  • Person $P_1$: Can choose any of the 3 available houses. (3 choices)
  • Person $P_2$: Must choose a house different from the one $P_1$ chose. Since one house is taken, $P_2$ has 2 remaining choices.
  • Person $P_3$: Must choose a house different from the ones $P_1$ and $P_2$ chose. Since two houses are taken, $P_3$ has 1 remaining choice.
  • Using the Fundamental Principle of Counting, the number of ways for all three persons to apply for distinct houses is: $\text{Number of Favorable Outcomes} = 3 \times 2 \times 1 = 6$
  • This is equivalent to the number of permutations of 3 distinct houses taken 3 at a time, denoted as $P(3,3)$ or $3!$: $P(3,3) = \frac{3!}{(3-3)!} = \frac{3!}{0!} = 3! = 6$

Step 3: Calculate the Probability

• Explanation: Now that we have both the total number of possible outcomes and the number of favorable outcomes (for the event that all three persons apply for distinct houses), we can use the basic probability formula.
• Calculation: $P(\text{all apply for distinct houses}) = \frac{\text{Number of Favorable Outcomes}}{\text{Total Number of Possible Outcomes}}$ $P(\text{all apply for distinct houses}) = \frac{6}{27}$
• Simplification: $P(\text{all apply for distinct houses}) = \frac{2}{9}$

3. Common Mistakes & Tips

• Systematic Counting: Always break down the problem into individual choices for each person or item. This helps in correctly applying the multiplication principle for total outcomes.
• Distinguishing Outcomes: Be clear about what constitutes a distinct "outcome." For instance, $(P_1 \to H_1, P_2 \to H_2, P_3 \to H_3)$ is different from $(P_1 \to H_2, P_2 \to H_1, P_3 \to H_3)$.
• "Without consulting others": This phrase implies that each person's choice is independent of the others, justifying the use of the multiplication rule for total outcomes.
• Permutations vs. Combinations: If the order or assignment of distinct items matters (like distinct people choosing distinct houses), use permutations. If only the selection of items without regard to order matters, use combinations.
• Simplifying Fractions: Always simplify the final probability fraction to its lowest terms.

4. Summary

The problem asks for a probability. We first determined the total number of ways three distinct persons can apply for three distinct houses, which is $3^3 = 27$, as each person has 3 independent choices. To arrive at the given correct answer, we calculated the number of favorable outcomes for the event that all three persons apply for different houses. This involves permutations, where the first person has 3 choices, the second has 2 (remaining), and the third has 1 (remaining), yielding $3 \times 2 \times 1 = 6$ favorable outcomes. The probability is then the ratio of favorable outcomes to total outcomes, which is $\frac{6}{27}$, simplifying to $\frac{2}{9}$.

5. Final Answer

The final answer is $\boxed{{2 \over 9}}$, which corresponds to option (A).