JEE Main 2020 - 5 Sep (Evening)

LEVELJEE Main

Animated Solution for Mathematics - Probability: In a bombing attack, there is 50% chance that a bomb will hit the target. Atleast two independent hits are required to destroy the target completely. Then the minimum number of bombs, that must be dropped to ensure that there is at least 99% chance of completely destroying the target, is

Q: Wait, why do we use the complement rule? Can't we just calculate $P(X \ge 2)$?

You absolutely could, but it would be a test of your patience rather than your physics! Calculating $P(X \ge 2)$ means summing $P(X=2) + P(X=3) + \dots + P(X=n)$. Since $n$ is unknown, this is algebraically messy. The complement rule, $P(X \ge 2) = 1 - P(X < 2)$, reduces the problem to just two simple cases: $X=0$ and $X=1$. Always look for the path of least resistance in probability problems!

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

Because these events are mutually exclusive. You cannot have exactly zero hits and exactly one hit at the same time. When events are mutually exclusive, the probability of their union is simply the sum of their individual probabilities. It is a fundamental property of probability theory.

Q: Why can't we just solve $2^n \ge 100(n+1)$ using logarithms?

That is a great question! While you could use logarithms, you would end up with a transcendental equation involving $n$ and $\log(n)$, which cannot be solved for $n$ using basic algebraic methods. In JEE problems, when you encounter an exponential inequality like this, testing integer values is often the intended and most efficient strategy.

Enter Numerical Value:

Progressive Hints

Hint 1

Hint 1 (The Spark): Instead of calculating the probability of getting at least two hits directly, which involves summing up many scenarios, consider the complement. What is the probability that the target survives? The target survives if we get zero hits or exactly one hit.

Hint 2

Hint 2 (The Arsenal): Utilize the Binomial Distribution formula

P (X = r) =^{n} C_{r} p^{r} q^{n - r}

, where

p = 0.5

is the probability of a hit and

q = 0.5

is the probability of a miss. You need to calculate

P (X = 0)

and

P (X = 1)

and sum them to find the probability of failure.

Hint 3

Hint 3 (The Execution): Set up the inequality

1 - P (X < 2) \geq 0.99

, which simplifies to

P (X < 2) \leq 0.01

. Substitute your expressions for

P (X = 0)

and

P (X = 1)

to get the inequality

\frac{n + 1}{2 ^{n}} \leq 0.01

, or

2^{n} \geq 100 (n + 1)

. Now, test integer values for

n

to find the minimum.

Visualized Solution

The Bombing Mission

Target requires at least $2$ hits to be destroyed.
Probability of a single bomb hitting is $50%$ .

Defining Probabilities

Probability of hit: $p = \frac{1}{2}$
Probability of miss: $q = 1 - p = \frac{1}{2}$
Let $n$ be the total number of bombs dropped.

The $99%$ Guarantee

Let $X$ be the number of successful hits.
Target is destroyed if $X \geq 2$ .
Required condition: $P (X \geq 2) \geq 0.99$

The Complement Rule

Calculating $P (X \geq 2)$ directly involves many terms: $P (X = 2) + P (X = 3) + \dots + P (X = n)$ .
It is easier to use the complement rule: $P (X \geq 2) = 1 - P (X < 2)$ .

The Failure Scenarios

The target survives if $X < 2$ .
This happens in two cases: exactly $0$ hits or exactly $1$ hit.
$P (X < 2) = P (X = 0) + P (X = 1)$

Binomial Distribution Formula

We use the Binomial Probability formula.
$P (X = r) =^{n} C_{r} p^{r} q^{n - r}$

Calculating $P (X = 0)$ and $P (X = 1)$

For $X = 0$ : $P (X = 0) =^{n} C_{0} (\frac{1}{2})^{0} (\frac{1}{2})^{n} = \frac{1}{2 ^{n}}$
For $X = 1$ : $P (X = 1) =^{n} C_{1} (\frac{1}{2})^{1} (\frac{1}{2})^{n - 1} = \frac{n}{2 ^{n}}$

Substituting into the Main Condition

Total failure probability: $P (X < 2) = \frac{1}{2 ^{n}} + \frac{n}{2 ^{n}} = \frac{n + 1}{2 ^{n}}$
Substitute back: $1 - (\frac{n + 1}{2 ^{n}}) \geq 0.99$

Rearranging the Inequality

Rearrange: $1 - 0.99 \geq \frac{n + 1}{2 ^{n}}$
Simplify: $0.01 \geq \frac{n + 1}{2 ^{n}}$
Final form: $2^{n} \geq 100 (n + 1)$

Testing $n = 10$

We need to find the minimum integer $n$ that satisfies $2^{n} \geq 100 (n + 1)$ .
Let's test $n = 10$ :
LHS: $2^{10} = 1024$
RHS: $100 (10 + 1) = 1100$
$1024 \geq 1100$ is False.

Testing $n = 11$

Let's test the next integer, $n = 11$ :
LHS: $2^{11} = 2048$
RHS: $100 (11 + 1) = 1200$
$2048 \geq 1200$ is True.

Final Answer

The minimum number of bombs required is $11$ .
Key Takeaway: Exponential growth ( $2^{n}$ ) eventually overtakes linear growth ( $100 n$ ).

00:00 / 00:00

The Sigma Insight: Random Variables and Probability Distributions

Solution Diagram

Analyzing the Strategic Objective

To ensure the destruction of the target, we require at least two direct hits. Each bomb dropped has a probability of success

p = \frac{1}{2}

and a probability of failure

q = 1 - p = \frac{1}{2}

We are dropping

n

bombs, and we define

X

as the random variable representing the number of successful hits. Our goal is to find the minimum integer

n

such that the probability of success is at least

99%

P (X \geq 2) \geq 0.99

The Power of the Complement

Calculating

P (X \geq 2)

directly involves summing the probabilities of all outcomes from

2

n

. Instead, we utilize the complement rule to simplify the calculation.

The target survives if we achieve fewer than two hits, which corresponds to either zero hits or exactly one hit. We express this as:

P (X < 2) = P (X = 0) + P (X = 1)

Our condition

P (X \geq 2) \geq 0.99

is equivalent to

1 - P (X < 2) \geq 0.99

, which simplifies to:

P (X < 2) \leq 0.01

The Mathematical Siege

Using the Binomial Distribution formula,

P (X = r) =^{n} C_{r} p^{r} q^{n - r}

, we calculate the failure scenarios:

For

X = 0

P (X = 0) =^{n} C_{0} (\frac{1}{2})^{0} (\frac{1}{2})^{n} = \frac{1}{2 ^{n}}

For

X = 1

P (X = 1) =^{n} C_{1} (\frac{1}{2})^{1} (\frac{1}{2})^{n - 1} = n \cdot \frac{1}{2 ^{n}} = \frac{n}{2 ^{n}}

Summing these, the total probability of failure is:

P (X < 2) = \frac{1 + n}{2 ^{n}}

The Inequality Showdown

We substitute our result into the inequality

P (X < 2) \leq 0.01

\frac{n + 1}{2 ^{n}} \leq 0.01 \Rightarrow 2^{n} \geq 100 (n + 1)

We now test integer values for

n

to find the threshold where exponential growth overtakes linear growth:

For

n = 10

2^{10} = 1024, 100 (10 + 1) = 1100

Since

1024 < 1100

, ten bombs are insufficient.

For

n = 11

2^{11} = 2048, 100 (11 + 1) = 1200

Since

2048 > 1200

, the condition is satisfied.

The minimum number of bombs required to be

99%

certain of destroying the target is 11.

Common Traps & FAQs

Wait, why do we use the complement rule? Can't we just calculate

P (X \geq 2)

You absolutely could, but it would be a test of your patience rather than your physics! Calculating

P (X \geq 2)

means summing

P (X = 2) + P (X = 3) + \dots + P (X = n)

. Since

n

is unknown, this is algebraically messy. The complement rule,

P (X \geq 2) = 1 - P (X < 2)

, reduces the problem to just two simple cases:

X = 0

and

X = 1

. Always look for the path of least resistance in probability problems!

Why is

P (X < 2)

equal to

P (X = 0) + P (X = 1)

Because these events are mutually exclusive. You cannot have exactly zero hits and exactly one hit at the same time. When events are mutually exclusive, the probability of their union is simply the sum of their individual probabilities. It is a fundamental property of probability theory.

Why can't we just solve

2^{n} \geq 100 (n + 1)

using logarithms?

That is a great question! While you could use logarithms, you would end up with a transcendental equation involving

n

and

lo g (n)

, which cannot be solved for

n

using basic algebraic methods. In JEE problems, when you encounter an exponential inequality like this, testing integer values is often the intended and most efficient strategy.

Enter Numerical Value:

Visualized Solution

The Bombing Mission

Defining Probabilities

The $99%$ Guarantee

The Complement Rule

The Failure Scenarios

Binomial Distribution Formula

Calculating $P (X = 0)$ and $P (X = 1)$

Substituting into the Main Condition

Rearranging the Inequality

Testing $n = 10$

Testing $n = 11$

Final Answer

The Sigma Insight: Random Variables and Probability Distributions

Analyzing the Strategic Objective

The Power of the Complement

The Mathematical Siege

The Inequality Showdown

Similar Questions

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

The minimum number of times a fair coin needs to be tossed, so that the probability of getting at least two heads is at least 0.96, is .........

Four fair dice are thrown independently 27 times. Then the expected number of times, at least two dice show up a three or a five, is

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If the sum and the product of mean and variance of a binomial distribution are 24 and 128 respectively, then the probability of one or two successes is :

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Enter Numerical Value:

Visualized Solution

The Bombing Mission

Defining Probabilities

The .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 99% Guarantee

The Complement Rule

The Failure Scenarios

Binomial Distribution Formula

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P(X=0) and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P(X=1)

Substituting into the Main Condition

Rearranging the Inequality

Testing .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } n=10

Testing .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } n=11

Final Answer

The Sigma Insight: Random Variables and Probability Distributions

Analyzing the Strategic Objective

The Power of the Complement

The Mathematical Siege

The Inequality Showdown

Similar Questions

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The minimum number of times a fair coin needs to be tossed, so that the probability of getting at least two heads is at least 0.96, is .........

Four fair dice are thrown independently 27 times. Then the expected number of times, at least two dice show up a three or a five, is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A fair n(n>1) faces die is rolled repeatedly until a number less than n appears. If the mean of the number of tosses required is 9n​, then n is equal to

If the sum and the product of mean and variance of a binomial distribution are 24 and 128 respectively, then the probability of one or two successes is :

A coin is biased so that a head is twice as likely to occur as a tail. If the coin is tossed 3 times, then the probability of getting two tails and one head is-

In a game, a man wins Rs. 100 if he gets 5 or 6 on a throw of a fair die and loses Rs. 50 for getting any other number on the die. If he decides to throw the die either till he gets a five or a six or to a maximum of three throws, then his expected gain/loss (in rupees) is :

Each of the persons A and B independently tosses three fair coins. The probability that both of them get the same number of heads is :

The $99%$ Guarantee

Calculating $P (X = 0)$ and $P (X = 1)$

Testing $n = 10$

Testing $n = 11$

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

A fair $n (n > 1)$ faces die is rolled repeatedly until a number less than $n$ appears. If the mean of the number of tosses required is $\frac{n}{9}$ , then $n$ is equal to