1. Key Concepts and Formulas

• Complementary Probability: If $P(A)$ is the probability of an event $A$ occurring, then the probability of its complement $A'$ (event $A$ not occurring) is given by $P(A') = 1 - P(A)$. This is extremely useful for problems involving "at least once" scenarios, as $P(\text{at least once}) = 1 - P(\text{none})$.
• Probability of Independent Events: If multiple events are independent (the outcome of one does not affect the others), the probability that all of them occur is the product of their individual probabilities. For example, $P(A \text{ and } B) = P(A) \times P(B)$ if $A$ and $B$ are independent.

2. Step-by-Step Solution

Step 1: Define Probabilities for a Single Shot First, let's establish the probabilities for a single attempt. Let $P(H)$ be the probability of the man hitting the target in a single shot. Given in the problem: $P(H) = \frac{1}{10}$ Next, we need the probability of the man not hitting the target in a single shot (i.e., missing it). This is the complement of hitting. Let $P(M)$ be the probability of missing the target in a single shot. Using the complementary probability concept for a single event: $P(M) = 1 - P(H) = 1 - \frac{1}{10} = \frac{9}{10}$ Explanation: We calculate $P(M)$ because the complementary strategy, which is often the most efficient for "at least once" problems, will involve calculating the probability of missing the target in all shots.

Step 2: Formulate the Event "Hitting the Target at Least Once" using Complementary Probability Let $n$ be the total number of shots taken. We are interested in the event $E$: "the man hits the target at least once in these $n$ shots." Directly calculating $P(E)$ would involve summing the probabilities of hitting exactly 1 time, exactly 2 times, ..., up to exactly $n$ times, which can be tedious.

Instead, we consider the complementary event, $E'$. The complement of "hitting at least once" is "not hitting even once." This means "missing the target in all $n$ shots." So, $E'$ is the event that the man misses the target in every single one of the $n$ shots.

According to the complementary probability rule: $P(E) = 1 - P(E')$ Explanation: This step sets up the core strategy. By converting the problem into its complementary form, we significantly simplify the calculation.

Step 3: Calculate the Probability of the Complementary Event ($P(E')$) The event $E'$ is "missing the target in all $n$ shots." We assume that each shot is an independent event (the outcome of one shot does not influence subsequent shots). This is a standard assumption in such problems unless explicitly stated otherwise. Since the shots are independent, the probability of missing all $n$ shots is the product of the probabilities of missing each individual shot: $P(E') = P(\text{miss 1st shot}) \times P(\text{miss 2nd shot}) \times \dots \times P(\text{miss } n\text{th shot})$ Since $P(M) = \frac{9}{10}$ for each shot: $P(E') = \left( \frac{9}{10} \right) \times \left( \frac{9}{10} \right) \times \dots \times \left( \frac{9}{10} \right) \quad (n \text{ times})$ $P(E') = \left( \frac{9}{10} \right)^n$ Explanation: The crucial assumption of independence allows us to simply multiply the individual probabilities. If the events were dependent, a more complex conditional probability approach would be needed.

Step 4: Set Up the Inequality The problem states that the probability of hitting the target at least once ($P(E)$) must be greater than $\frac{1}{4}$. So, we need to find the least integer $n$ such that: $P(E) > \frac{1}{4}$ Substitute the expression for $P(E)$ from Step 2 and $P(E')$ from Step 3: $1 - \left( \frac{9}{10} \right)^n > \frac{1}{4}$ Explanation: This step directly translates the problem's condition into a mathematical inequality involving $n$, which we can now solve.

Step 5: Solve the Inequality for $n$ We need to find the least positive integer $n$ that satisfies: $1 - \left( \frac{9}{10} \right)^n > \frac{1}{4}$ To isolate the term with $n$, let's rearrange the inequality: Subtract $\frac{1}{4}$ from both sides and add $\left( \frac{9}{10} \right)^n$ to both sides: $1 - \frac{1}{4} > \left( \frac{9}{10} \right)^n$ $\frac{3}{4} > \left( \frac{9}{10} \right)^n$ Now, let's express the fractions as decimals for easier comparison: $0.75 > (0.9)^n$ We need to find the smallest positive integer $n$ for which this inequality holds. We can test values of $n$ starting from $n=1$:

• For $n=1$: $(0.9)^1 = 0.9$ Is $0.75 > 0.9$? No, this is false. ($0.75$ is not greater than $0.9$)

• For $n=2$: $(0.9)^2 = 0.81$ Is $0.75 > 0.81$? No, this is false. ($0.75$ is not greater than $0.81$)

• For $n=3$: $(0.9)^3 = 0.9 \times 0.9 \times 0.9 = 0.81 \times 0.9 = 0.729$ Is $0.75 > 0.729$? Yes, this is true! ($0.75$ is greater than $0.729$)

Since the base $(0.9)$ is less than 1, the term $(0.9)^n$ decreases as $n$ increases. This means that if the inequality holds for $n=3$, it will also hold for all integer values of $n$ greater than 3. Therefore, the least number of shots required is $n=3$.

Explanation: We rearranged the inequality into a simpler form. For exponential inequalities involving integer exponents, especially without a calculator, testing small integer values sequentially is the most practical and quickest method. The property of the exponential function (decreasing for a base between 0 and 1) helps confirm that the first integer satisfying the condition is indeed the least.

3. Common Mistakes & Tips

• Forgetting Complementary Probability: Many students try to sum probabilities of "hitting exactly once," "hitting exactly twice," etc. Using $1 - P(\text{none})$ is almost always simpler for "at least once" problems.
• Inequality Direction: Pay close attention to the direction of the inequality sign. Remember that multiplying or dividing both sides by a negative number reverses the inequality sign (though not applicable in this specific solution).
• Trial and Error Efficiency: For exponential inequalities with integer exponents, don't try to use logarithms unless absolutely necessary (e.g., if the numbers are very large or if calculator use is allowed). Simple trial and error with small integer values is faster and less prone to error.

4. Summary

This problem effectively demonstrates the power of the complementary probability concept. By transforming the complex event "hitting the target at least once" into its simpler complement "missing the target every time," we could easily set up and solve an inequality. We found that by taking 3 shots, the probability of hitting the target at least once (which is $1 - (9/10)^3 = 1 - 0.729 = 0.271$) becomes greater than $1/4$ (or $0.25$). Thus, the least number of shots required is 3. This approach is a fundamental problem-solving technique in probability.

5. Final Answer

The final answer is $\boxed{3}$.