The probability of a man hitting a target is 1/10. The least number of shots required, so that the p

Q: Why do we use the complement rule $1 - P(\text{none})$ instead of just adding up probabilities?

Great question! Calculating 'at least one' directly would require summing the probability of exactly 1 hit, exactly 2 hits, ..., up to $n$ hits. That is a massive amount of work! The complement rule is a shortcut that collapses all those complex scenarios into one simple subtraction: $1 - P(\text{zero hits})$. It is the elegant way to navigate probability.

Q: Can I use logarithms to solve $(0.9)^n < 0.75$?

Absolutely! You can take the natural log on both sides: $n \ln(0.9) < \ln(0.75)$. Just remember that since $\ln(0.9)$ is negative, the inequality sign will flip when you divide by it. However, for small integers like this, manual testing is often faster and less prone to calculation errors.

Q: Why is the answer 3? If I hit with 0.9 probability, shouldn't I hit it almost immediately?

Be careful! The probability of hitting is $1/10$ (or 0.1), not 0.9. The 0.9 is the probability of *missing*. Because the chance of hitting is quite low, it takes a few attempts to make it statistically likely that you'll land at least one shot.

Q: Does the order of shots matter here?

In this specific problem, no. We are looking for the total number of trials $n$ to achieve at least one success. Since the shots are independent, the sequence doesn't change the probability of the final outcome.

Q: What if the question asked for 'at least two hits'? Would the complement rule still work?

The complement rule is most powerful for 'at least one'. For 'at least two', the complement would be 'zero hits OR exactly one hit'. You would calculate $1 - [P(0) + P(1)]$. It still works, but it's slightly more involved!

Question

The probability of a man hitting a target is $1/10$. The least number of shots required, so that the probability of his hitting the target at least once is greater than $1/4$, is

Sigma Percentile · Accepted Answer

### The Philosophy of the Complement Imagine you are standing on a shooting range. You have a target, and your probability of hitting it is $p = \frac{1}{10}$. You want to know how many shots you need to fire to be reasonably sure—specifically, with a probability greater than $\frac{1}{4}$—that you will hit the target at least once. Most students immediately try to calculate the probability of hitting once, then twice, then three times, and so on. But stop! That is the long, winding road. In probability, when you see the phrase 'at least one', your brain should immediately trigger the **Complement Rule**. The complement of 'hitting at least once' is simply 'never hitting at all'. ### Setting the Stage If the probability of success is $p = \frac{1}{10}$, then the probability of failure, which we call $q$, is simply $q = 1 - p = \frac{9}{10}$. Now, if you take $n$ shots, what is the probability that you miss every single time? Since each shot is an independent event, we multiply the probabilities: $$P( ext{zero hits}) = q^n = \left(\frac{9}{10} ight)^n$$ Our goal is to ensure that the probability of hitting at least once is greater than $\frac{1}{4}$. Mathematically, we write this as: $$1 - P( ext{zero hits}) > \frac{1}{4}$$ Substituting our value for $P( ext{zero hits})$, we get: $$1 - \left(\frac{9}{10} ight)^n > \frac{1}{4}$$ ### The Algebraic Dance Now, let us rearrange this inequality to isolate the term with $n$. Subtracting 1 from both sides and multiplying by $-1$ (remembering to flip the inequality sign!), we arrive at: $$\left(\frac{9}{10} ight)^n < \frac{3}{4}$$ In decimal form, this is $(0.9)^n < 0.75$. This is the heart of our problem. We are looking for the smallest integer $n$ that makes this statement true. ### The Final Countdown Let us test our values of $n$ one by one: 1. For $n=1$: $(0.9)^1 = 0.9$. Is $0.9 < 0.75$? No, that is clearly false. 2. For $n=2$: $(0.9)^2 = 0.81$. Is $0.81 < 0.75$? Still false, though we are getting closer. 3. For $n=3$: $(0.9)^3 = 0.729$. Is $0.729 < 0.75$? **Yes!** At $n=3$, the inequality finally holds. This means that after three shots, the probability of having missed every single time drops below 0.25, which mathematically guarantees that the probability of having hit at least once has climbed above 0.25. ### Why This Matters This problem isn't just about shooting targets. It is about understanding how uncertainty behaves. Whether you are calculating the reliability of a circuit board or the success rate of a chemical reaction, the logic remains the same: sometimes, it is easier to calculate the probability of total failure and subtract it from the certainty of the universe (which is 1) to find the probability of success. Keep practicing this mindset. When you see 'at least one', think 'one minus none'. It is a small shift in perspective that will save you hours of calculation in the JEE Advanced exam.

Animated Solution for Mathematics - Probability: The probability of a man hitting a target is $1/10$ . The least number of shots required, so that the probability of his hitting the target at least once is greater than $1/4$ , is

Enter Numerical Value:

Visualized Solution

Identify Success Probability $p$

Calculate Failure Probability $q$

Define the 'At Least One' Logic

Formulate the Inequality

Rearrange the Inequality

Convert to Decimal Form

Test for $n = 1$

Test for $n = 2$

Test for $n = 3$

Final Conclusion

The Sigma Insight: Random Variables and Probability Distributions

The Philosophy of the Complement

Setting the Stage

The Algebraic Dance

The Final Countdown

Why This Matters

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