Problem 50 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

The probability of a man hitting a target is \(\frac{1}{10} .\) The least number of shots required, so that the probability of his hitting the target at least once is greater than \(\frac{1}{4}\), is \(\quad\) [NA Sep. 04, 2020 (I)]

Step-by-Step Solution

Answer

The least number of shots required is 3.

Step 1: Understanding the Problem

We need to find the least number of shots such that the probability of hitting the target at least once is greater than \( \frac{1}{4} \). The probability of hitting the target with one shot is \( \frac{1}{10} \).

Step 2: Complementary Probability Method

To calculate the probability of hitting the target at least once with \( n \) shots, it is easier first to calculate the probability of missing all shots. The probability of missing with one shot is \( 1 - \frac{1}{10} = \frac{9}{10} \). Therefore, the probability of missing all \( n \) shots is \( \left( \frac{9}{10} \right)^n \).

Step 3: Setting up the Inequality

We need the probability of hitting at least once to be greater than \( \frac{1}{4} \), so first calculate the complementary probability (missing all shots) and subtract from 1: \[1 - \left( \frac{9}{10} \right)^n > \frac{1}{4}\]Rearrange the inequality to solve for \( n \):\[\left( \frac{9}{10} \right)^n < \frac{3}{4}\]

Step 4: Solving the Inequality

Take the natural logarithm of both sides to solve for \( n \):\[\ln\left( \left( \frac{9}{10} \right)^n \right) < \ln\left( \frac{3}{4} \right)\]Which simplifies to:\[n \cdot \ln\left( \frac{9}{10} \right) < \ln\left( \frac{3}{4} \right)\]Divide by \( \ln\left( \frac{9}{10} \right) \):\[n > \frac{\ln\left( \frac{3}{4} \right)}{\ln\left( \frac{9}{10} \right)}\]

Step 5: Calculating the Minimum Number of Shots

Calculate the value:\[\ln\left( \frac{3}{4} \right) \approx -0.28768 \quad \text{and} \quad \ln\left( \frac{9}{10} \right) \approx -0.10536\]\[n > \frac{-0.28768}{-0.10536} \approx 2.73\]Since \( n \) must be an integer, round up to find that the least number of shots needed is 3.

Key Concepts

Complementary Probability

Complementary probability is a useful concept in finding the likelihood of an event happening by first understanding the probability of the opposite event. In the given problem, we want to know the probability that a man hits the target at least once. However, calculating this directly can be complex. Instead, we calculate the opposite event — missing the target with every shot. If each shot has a probability of hitting the target of \( \frac{1}{10} \), then missing the target is \( 1 - \frac{1}{10} = \frac{9}{10} \).

• If he takes \( n \) shots and misses all of them, the probability is \( \left( \frac{9}{10} \right)^n \).
• Therefore, the complementary probability, the probability of hitting the target at least once, is \( 1 - \left( \frac{9}{10} \right)^n \).

This approach simplifies the calculation, allowing us to easily set up and solve the problem using inequalities.

Inequality Solving

Inequalities are used in mathematics to denote that one quantity is greater or lesser than another. In our probability problem, we set up an inequality to determine the minimum number of shots needed. After establishing the complementary probability, we express the problem as:\[1 - \left( \frac{9}{10} \right)^n > \frac{1}{4}\] This inequality means the probability of hitting at least once should exceed \( \frac{1}{4} \). Manipulating this requires rearranging terms and applying mathematical operations

• First, subtract \(1\) on both sides, which is an operation directly affecting the inequality sign.
• This simplifies to \( \left( \frac{9}{10} \right)^n < \frac{3}{4} \).

Solving inequalities, especially in cases involving exponential terms, often requires the use of logarithms, as they can simplify complex exponential expressions into linear ones.

Natural Logarithm

Logarithms, especially natural logarithms (denoted as \( \ln \)), are crucial for simplifying exponential equations. In our problem, the inequality \( \left( \frac{9}{10} \right)^n < \frac{3}{4} \) can be difficult to solve directly when \( n \) is the exponent. By taking natural logarithms of both sides, the exponent \( n \) is brought down:\[\ln\left( \left( \frac{9}{10} \right)^n \right) < \ln\left( \frac{3}{4} \right)\] This becomes:\[n \cdot \ln\left( \frac{9}{10} \right) < \ln\left( \frac{3}{4} \right)\]Dividing both sides by \( \ln\left( \frac{9}{10} \right) \), which is negative, flips the inequality sign, resulting in:\[n > \frac{\ln\left( \frac{3}{4} \right)}{\ln\left( \frac{9}{10} \right)}\]This step permits us to solve for \( n \), determining how many shots are required to achieve the desired probability.

Rounding Up Integers

When we solve the inequality \( n > \frac{\ln\left( \frac{3}{4} \right)}{\ln\left( \frac{9}{10} \right)} \), we find \( n \approx 2.73 \). However, \( n \) represents the number of shots, which must be a whole number since you can't shoot a fraction of a shot.Rounding up is essential here because even a small fraction often means the minimum necessary to achieve the probability goal is not quite reached by rounding down. So, although \( 2.73 \) rounded normally would be \( 3 \), applying the context-specific rule of rounding up solidifies our requirement ensuring any fraction of chance remaining is covered by agreeing to a full shot.Thus, we arrive at the conclusion of needing at least 3 shots to have a probability greater than \( \frac{1}{4} \) of hitting the target at least once. Rounding up in such scenarios ensures that all conditions are satisfied clearly and effectively.