Problem 20 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

If the probability of hitting a target by a shooter, in any shot, is \(\frac{1}{3}\), then the minimum number of independent shots at the target required by him so that the probability of hitting the target at least once is greater than \(\frac{5}{6}\), is: [Jan. 10, 2019 (II)] (a) 3 (b) 6 (c) 5 (d) 4

Step-by-Step Solution

Answer

The minimum number is 5 shots.

Step 1: Understand the problem

We are given the probability that a shooter hits a target in one shot as \( \frac{1}{3} \). We need to find the minimum number of shots needed such that the probability of hitting the target at least once is greater than \( \frac{5}{6} \).

Step 2: Define probability of not hitting the target

The probability of not hitting the target in one shot is \( 1 - \frac{1}{3} = \frac{2}{3} \).

Step 3: Calculate probability of not hitting the target in n shots

The probability of not hitting the target after \( n \) shots, assuming each shot is independent, is \( \left( \frac{2}{3} \right)^n \).

Step 4: Express the probability of hitting the target at least once

The probability of hitting the target at least once in \( n \) shots is \( 1 - \left( \frac{2}{3} \right)^n \).

Step 5: Set up the inequality

We want this probability to be greater than \( \frac{5}{6} \), so we set up the inequality: \( 1 - \left( \frac{2}{3} \right)^n > \frac{5}{6} \).

Step 6: Solve the inequality

Rearranging the inequality gives \( \left( \frac{2}{3} \right)^n < \frac{1}{6} \). We need to find the smallest integer \( n \) that satisfies this.

Step 7: Test values for n

- For \( n = 3 \), \( \left( \frac{2}{3} \right)^3 = \frac{8}{27} \approx 0.296 > \frac{1}{6} \).- For \( n = 4 \), \( \left( \frac{2}{3} \right)^4 = \frac{16}{81} \approx 0.198 > \frac{1}{6} \).- For \( n = 5 \), \( \left( \frac{2}{3} \right)^5 = \frac{32}{243} \approx 0.131 < \frac{1}{6} \).

Step 8: Conclusion

The minimum number of shots required is \( n = 5 \).

Key Concepts

Independent Events

When we talk about independent events in probability, it means that each event occurs independently of others. For instance, in the shooting scenario, each shot the shooter takes does not affect the outcome of the next shot. This independence is crucial because it allows us to multiply the probabilities of distinct events to get a combined probability. In the problem, each time the shooter fires, there is a probability of \( \frac{1}{3} \) to hit and \( \frac{2}{3} \) not to hit the target. It is important to understand that no matter how many shots have been fired, the probability of hitting the target in the next shot remains \( \frac{1}{3} \). This illustrates the independent nature of each event. Once you grasp this, understanding how probabilities combine over multiple trials becomes simpler.

Probability Inequality

Probability inequalities help to determine conditions under which certain outcomes are likely to occur. In this exercise, we seek the minimum number of shots required so that the probability of hitting the target at least once is greater than a specific threshold, \( \frac{5}{6} \).The inequality is set up as \[ 1 - \left( \frac{2}{3} \right)^n > \frac{5}{6} \] which expresses: the probability of hitting the target at least once (over \( n \) shots) must be greater than \( \frac{5}{6} \). Rearranging this inequality gives us \( \left( \frac{2}{3} \right)^n < \frac{1}{6} \). Solving this helps us find the minimum \( n \). - It tells us how good our chances are after multiple attempts.- It determines when we exceed the confidence level and solve the task.By understanding and solving this inequality, we align towards reaching the desired outcome with minimum trials.

Cumulative Probability

Cumulative probability is all about adding probabilities progressively—"cumulative" means increasing. In the context of this exercise, it tells us the likelihood of hitting the target at least once when multiple shots are combined.Here, we don't just want the probability of a single successful shot, but of success at least once in \( n \) attempts. This can happen with a few hits out of several shots, and it sharply increases as more shots are taken: - The cumulative probability cobbles together all chances from 1 up to \( n \) shots succeeding any number of times but at least once.This cumulative approach results in our goal inequality: \[ 1 - \left( \frac{2}{3} \right)^n \] where \( \left( \frac{2}{3} \right)^n \) shows the probability of missing every single time with \( n \) shots, hence becoming effective for combined attempts.

Probability Distribution

Probability distribution provides an overview of all potential outcomes and their likelihoods. It's like checking every possible result and seeing how probable each is. In our shooting example, we have two primary single-shot outcomes—hit or miss—with respective probabilities \( \frac{1}{3} \) for a hit and \( \frac{2}{3} \) for a miss.The distribution here focuses on what happens over multiple shots:- For \( n \) shots, different combination scenarios may play out based on miss and hit results.Looking at these possible combinations helps to define the total likelihood across multiple trials, and each possible outcome from zero to several successful hits. Probability distribution informs how these probabilities stack up when observing potential outcomes jointly. Understanding this helps anyone participating in the exercise to easily gauge any step and what the chances stand to appear as more attempts are made.