1. Key Concepts and Formulas

• Probability of Success ($p$) and Failure ($q$): For any single trial (like a shot), if the probability of success is $p$, then the probability of failure is $q = 1 - p$.
• Independent Events: Each shot is independent, meaning the outcome of one shot does not affect the outcome of any other shot. This allows us to multiply probabilities for sequences of shots.
• Complementary Probability: The probability of an event happening is $1 - P(\text{Event not happening})$. This is particularly useful for questions involving "at least once," as it's often simpler to calculate the probability of the event never happening. $P(\text{Event happens at least once}) = 1 - P(\text{Event never happens})$

2. Step-by-Step Solution

Step 1: Define Probabilities for a Single Shot Let $p$ be the probability of hitting the target in a single shot (success), and $q$ be the probability of not hitting the target (failure).

• Given probability of hitting the target, $p = \frac{1}{3}$.
• Probability of not hitting the target, $q = 1 - p = 1 - \frac{1}{3} = \frac{2}{3}$.

Step 2: Formulate the Problem Using Complementary Probability We are looking for the minimum number of shots, $n$, such that the probability of hitting the target at least once is greater than $\frac{5}{6}$. Let $X$ be the number of hits in $n$ shots. We want to find $n$ such that $P(X \ge 1) > \frac{5}{6}$. Directly calculating $P(X \ge 1)$ involves summing probabilities of 1 hit, 2 hits, ..., up to $n$ hits. This is complex. Using complementary probability, the event "hitting at least once" is the complement of "hitting zero times" (i.e., missing every shot). $P(X \ge 1) = 1 - P(X = 0)$

Step 3: Calculate the Probability of Zero Hits ($P(X=0)$) The event "hitting zero times" means that out of $n$ shots, all $n$ shots are misses. Since each shot is independent and the probability of missing a single shot is $q = \frac{2}{3}$: $P(X=0) = q \times q \times \dots \times q \quad (\text{n times})$ $P(X=0) = \left(\frac{2}{3}\right)^n$

Step 4: Set Up the Inequality Now, substitute $P(X=0)$ back into our complementary probability expression and apply the given condition: $1 - \left(\frac{2}{3}\right)^n > \frac{5}{6}$

Step 5: Solve the Inequality for the Term with $n$ We need to isolate $\left(\frac{2}{3}\right)^n$:

• Subtract 1 from both sides: $-\left(\frac{2}{3}\right)^n > \frac{5}{6} - 1$ $-\left(\frac{2}{3}\right)^n > -\frac{1}{6}$
• Multiply both sides by -1. Remember to reverse the inequality sign when multiplying or dividing by a negative number: $\left(\frac{2}{3}\right)^n < \frac{1}{6}$

Step 6: Determine the Minimum Value of $n$ We need to find the smallest positive integer $n$ that satisfies the inequality $\left(\frac{2}{3}\right)^n < \frac{1}{6}$. We can do this by testing integer values for $n$ starting from 1.

• For $n=1$: $\left(\frac{2}{3}\right)^1 = \frac{2}{3}$. Is $\frac{2}{3} < \frac{1}{6}$? (i.e., $0.666... < 0.166...$) No, $\frac{2}{3}$ is clearly greater than $\frac{1}{6}$.

• For $n=2$: $\left(\frac{2}{3}\right)^2 = \frac{4}{9}$. Is $\frac{4}{9} < \frac{1}{6}$? (i.e., $0.444... < 0.166...$) No, $\frac{4}{9}$ is greater than $\frac{1}{6}$.

• For $n=3$: $\left(\frac{2}{3}\right)^3 = \frac{8}{27}$. Is $\frac{8}{27} < \frac{1}{6}$? (i.e., $0.296... < 0.166...$) No, $\frac{8}{27}$ is greater than $\frac{1}{6}$.

• For $n=4$: $\left(\frac{2}{3}\right)^4 = \frac{16}{81}$. Is $\frac{16}{81} < \frac{1}{6}$? (i.e., $0.1975... < 0.166...$) By carefully comparing these fractions, we find that this inequality holds true.

Since $n=4$ is the first integer value for which the inequality holds, it is the minimum number of independent shots required.

3. Common Mistakes & Tips

• Don't Forget Complementary Probability: For "at least once" scenarios, using the complementary event is almost always the most efficient approach.
• Inequality Reversal: A common pitfall is forgetting to reverse the inequality sign when multiplying or dividing both sides by a negative number.
• Fraction Comparison: Be precise when comparing fractions. If decimal approximations are used, ensure enough precision to avoid errors. Cross-multiplication ($a/b < c/d \implies ad < bc$) is often the most reliable method.

4. Summary

We began by defining the probabilities of hitting ($p=1/3$) and missing ($q=2/3$) the target. To find the probability of hitting the target at least once, we used the complementary probability approach, calculating $1 - P(\text{no hits})$. This led to the inequality $1 - \left(\frac{2}{3}\right)^n > \frac{5}{6}$, which simplified to $\left(\frac{2}{3}\right)^n < \frac{1}{6}$. By systematically testing integer values for $n$, we found that $n=4$ is the smallest number of shots for which the probability of hitting the target at least once exceeds $\frac{5}{6}$.

The final answer is $\boxed{4}$, which corresponds to option (A).