An anti-aircraft gun can take a maximum of four shots at an enemy plane moving away from it. The pro

Q: Wait, why can't I just add the probabilities $0.4 + 0.3 + 0.2 + 0.1$ to get $1.0$?

That is a classic trap! Adding probabilities is only valid for mutually exclusive events (events that cannot happen at the same time). Here, the shots are independent. If you add them, you are overcounting the scenarios where multiple shots hit the plane. The correct approach is to look at the complement—the scenario where the gun fails completely.

Q: How do I know for sure that the shots are independent?

In probability problems of this type, unless stated otherwise, we assume each shot is an independent trial. This means the outcome of the first shot does not influence the outcome of the second. This independence allows us to multiply the probabilities of individual events to find the probability of their joint occurrence.

Q: What if the question asked for the probability of 'exactly one' hit instead of 'at least one'?

That would be a much more complex problem! You would have to calculate the probability of hitting on the first shot and missing the rest, plus hitting on the second and missing the rest, and so on. 'At least one' is a beautiful shortcut because it covers all successful outcomes in one simple subtraction.

Question

An anti-aircraft gun can take a maximum of four shots at an enemy plane moving away from it. The probabilities of hitting the plane at the first, second, third and fourth shot are 0.4, 0.3, 0.2 and 0.1 respectively. What is the probability that the gun hits the plane?

Sigma Percentile · Accepted Answer

### The Battlefield of Probability

Imagine you are standing on a quiet, dusty field. An enemy plane is flying away, a distant speck in the sky. You are operating an anti-aircraft gun, and you have exactly four chances to take it down.

Each shot you take has a different probability of success: $P(E_1) = 0.4$, $P(E_2) = 0.3$, $P(E_3) = 0.2$, and $P(E_4) = 0.1$. As the plane moves further away, your chances of hitting it dwindle.

Our mission is to determine the probability that you, the gunner, successfully hit the plane at least once.

### The Siren Song of Addition

When students first encounter this problem, the most common instinct is to add the probabilities: $0.4 + 0.3 + 0.2 + 0.1 = 1.0$. It feels intuitive, but this is the siren song of probability—a trap that has claimed many marks in JEE exams.

Addition is only for mutually exclusive events. If you add these, you are assuming that hitting on the first shot and hitting on the second shot are mutually exclusive, which they are not.

You could hit on the first shot, the second, or both. To solve this correctly, we must change our perspective.

### The Power of the Complement

Instead of trying to count every possible way to hit the plane, we look at the only scenario we do not want: total failure. Total failure means you miss every single shot.

If we can calculate the probability of missing all four shots, we can subtract that from the total probability space (which is $1$) to find the probability of at least one hit.

This is the **Complement Rule**:

$$P(	ext{At least one hit}) = 1 - P(	ext{None of the shots hit})$$

### Executing the Strategy

First, we must find the probability of missing each individual shot. If the probability of hitting is $P(E_n)$, the probability of missing is $P(\bar{E}_n) = 1 - P(E_n)$.

The individual miss probabilities are:

$$P(\bar{E}_1) = 1 - 0.4 = 0.6$$

$$P(\bar{E}_2) = 1 - 0.3 = 0.7$$

$$P(\bar{E}_3) = 1 - 0.2 = 0.8$$

$$P(\bar{E}_4) = 1 - 0.1 = 0.9$$

Because each shot is an independent event, the probability of missing all four is the product of these individual miss probabilities:

$$P(	ext{None hit}) = P(\bar{E}_1) 	imes P(\bar{E}_2) 	imes P(\bar{E}_3) 	imes P(\bar{E}_4)$$

Calculating this product:

$$P(	ext{None hit}) = 0.6 	imes 0.7 	imes 0.8 	imes 0.9 = 0.3024$$

This value, $0.3024$, represents the probability that you miss every single time.

### The Final Verdict

Finally, we apply the complement rule to find the probability of success:

$$P(	ext{Hit}) = 1 - 0.3024 = 0.6976$$

The probability of hitting the plane is **$0.6976$**, or roughly **$69.76\%$**.

By focusing on the failure, we found the success. Remember this lesson: when a problem seems to have too many paths to success, look for the one path to failure and subtract it from the whole. That is the mark of a true JEE strategist.

Enter Numerical Value:

Visualized Solution

Visualizing the Scenario

Defining Hit Probabilities

The "At Least One" Strategy

Calculating Miss Probabilities (Part 1)

Calculating Miss Probabilities (Part 2)

Probability of Total Failure

Executing the Multiplication

Computing the Final Result

The Sigma Insight: Addition and Multiplication Theorems

The Battlefield of Probability

The Siren Song of Addition

The Power of the Complement

Executing the Strategy

The Final Verdict

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