The probability that a missile hits a target successfully is 0.75. In order to destroy the target co

Q: Wait, why do we use the complement rule? Can't we just sum the probabilities from 3 to $n$?

That is a great question! While you technically *could* sum from 3 to $n$, you don't know the value of $n$ yet. Calculating a sum where the upper limit is unknown is algebraically impossible. By using the complement rule, we shift our focus to the 'failure' cases ($X=0, 1, 2$), which are fixed and independent of $n$. It turns an impossible problem into a solvable one.

Q: Why are we testing values for $n$ instead of solving the inequality algebraically?

In this problem, the variable $n$ appears both as a base and in the exponent (e.g., $4^n$). This creates a transcendental inequality. There is no standard algebraic method to isolate $n$ in such expressions. In JEE Advanced, when you encounter such inequalities, testing integer values is a standard and expected mathematical strategy.

Q: Why is $P(X < 3)$ equal to $P(X=0) + P(X=1) + P(X=2)$?

Remember that the Binomial Distribution is a discrete probability distribution. The variable $X$ can only take integer values: $0, 1, 2, \dots, n$. Therefore, the event '$X < 3$' is strictly composed of the mutually exclusive events $X=0$, $X=1$, and $X=2$. Summing these individual probabilities gives you the total probability of the event '$X < 3$'.

Question

The probability that a missile hits a target successfully is 0.75. In order to destroy the target completely, at least three successful hits are required. Then the minimum number of missiles that have to be fired so that the probability of completely destroying the target is NOT less than 0.95, is ____.

Sigma Percentile · Accepted Answer

### Analyzing the Strategic Battlefield

Imagine you are a defense strategist tasked with ensuring a target is destroyed. You have a missile system with a success probability of $p = 0.75$, or $\frac{3}{4}$.

To guarantee the target is obliterated, you need at least three successful hits. The question is: how many missiles must you fire to ensure a success probability of at least $0.95$?

We model this using the **Binomial Distribution**, where $X \sim B(n, 0.75)$.

### The Power of the Complement

When we face the condition $P(X \ge 3) \ge 0.95$, our intuition might suggest calculating the probability of 3 hits, 4 hits, 5 hits, and so on. However, since we do not know $n$, this approach leads to an algebraic loop.

Instead, we use the **Complement Rule**. The total probability space sums to $1$, so the probability of getting 3 or more hits is simply $1$ minus the probability of getting fewer than 3 hits.

Mathematically, we have:
$$P(X \ge 3) = 1 - P(X < 3)$$

Our condition becomes $1 - P(X < 3) \ge 0.95$, which simplifies to:
$$P(X < 3) \le 0.05$$

Now, we only need to evaluate the cases where we fail to achieve the required hits: $X=0$, $X=1$, and $X=2$.

### The Algebraic Siege

We apply the **Binomial Probability formula**:
$$P(X=r) = \binom{n}{r} p^r q^{n-r}$$

For $r=0, 1, 2$, with $p = \frac{3}{4}$ and $q = \frac{1}{4}$, the inequality becomes:
$$\binom{n}{0} \left(\frac{1}{4}\right)^n + \binom{n}{1} \left(\frac{3}{4}\right)^1 \left(\frac{1}{4}\right)^{n-1} + \binom{n}{2} \left(\frac{3}{4}\right)^2 \left(\frac{1}{4}\right)^{n-2} \le 0.05$$

Factoring out the common term $\frac{1}{4^n}$, the expression simplifies to:
$$\frac{1}{4^n} \left[ \binom{n}{0} + \binom{n}{1}(3) + \binom{n}{2}(3^2) \right] \le \frac{1}{20}$$

Expanding the combinations where $\binom{n}{0}=1$, $\binom{n}{1}=n$, and $\binom{n}{2}=\frac{n(n-1)}{2}$, we obtain:
$$\frac{1}{4^n} \left[ 1 + 3n + \frac{9n(n-1)}{2} \right] \le \frac{1}{20}$$

### The Final Breakthrough

Simplifying the numerator to $9n^2 - 3n + 2$ and adjusting the denominator, we arrive at the inequality:
$$\frac{9n^2 - 3n + 2}{2 \cdot 4^n} \le \frac{1}{20} \Rightarrow \frac{9n^2 - 3n + 2}{4^n} \le \frac{1}{10}$$

Now, we test integer values for $n$:

For $n=5$:
$$\frac{9(25) - 3(5) + 2}{1024} = \frac{212}{1024} \approx 0.207$$
Since $0.207 > 0.1$, we require more firepower.

For $n=6$:
$$\frac{9(36) - 3(6) + 2}{4096} = \frac{308}{4096} \approx 0.075$$
Since $0.075 \le 0.1$, the condition is satisfied.

We have found our answer: it takes **6 missiles** to cross the threshold of reliability.

Enter Numerical Value:

Visualized Solution

Define Binomial Distribution

Setup the Given Condition

Apply the Complement Rule

Rearrange the Inequality

Binomial Probability Formula

Substitute Values of $p$ and $q$

Factor Out Common Term

Expand the Combinations

Take Common Denominator

Simplify to Final Inequality

Test for $n = 5$

Test for $n = 6$

Conclusion

The Sigma Insight: Binomial Distribution

Analyzing the Strategic Battlefield

The Power of the Complement

The Algebraic Siege

The Final Breakthrough

Similar Questions

The minimum number of times one has to toss a fair coin so that the probability of observing at least one head is at least 90% is :

Minimum number of times a fair coin must be tossed so that the probability of getting at least one head is more than 99% is :

A student appeared in an examination consisting of 8 true-false type questions. The student guesses the answers with equal probability. The smallest value of $n$ , so that the probability of guessing at least ' $n$ ' correct answers is less than $1/2$ , is

The probability that an event $A$ happens in one trial of an experiment is $0.4$ . Three independent trials of the experiment are performed. The probability that the event $A$ happens at least once is

A multiple choice examination has 5 questions. Each question has three alternative answers of which exactly one is correct. The probability that a student will get 4 or more correct answers just by guessing is

A pair of dice is thrown 5 times. For each throw, a total of 5 is considered a success. If the probability of at least 4 successes is $\frac{k}{3 ^{11}}$ , then $k$ is equal to

The sum and product of the mean and variance of a binomial distribution are 82.5 and 1350 respectively. They the number of trials in the binomial distribution is:

If the mean and the variance of a binomial variate $X$ are 2 and 1 respectively, then the probability that $X$ takes a value greater than one is equal to .........

Suppose the probability for $A$ to win a game against $B$ is 0.4. If $A$ has an option of playing either a "best of 3 games" or a "best of 5 games" match against $B$ , which option should be choose so that the probability of his winning the match is higher? (No game ends in a draw).

For an initial screening of an admission test, a candidate is given fifty problems to solve. If the probability that the candidate can solve any problem is 4/5, then the probability that he is unable to solve less than two problems is :