UPSC MainsMANAGEMENT-PAPER-II20212 Marks
Q6.

Probability Calculation: A, B, and C Hitting Target

A’ can hit a target 4 times in 5 shots, ‘B' 3 times in 4 shots and 'C' twice in 3 shots. Calculate the probability : (A) ‘A’, ‘B’, ‘C’ all may hit the target. (B) ‘B’ and ‘C’ may hit the target, but ‘A’ may not hit it. (C) ‘C’ and ‘A’ may hit the target, but ‘B’ may not hit it.

How to Approach

This question tests the application of basic probability principles. The approach involves understanding independent events and calculating probabilities of combined events (intersection of events). We need to calculate the probability of each person hitting the target and then use these individual probabilities to determine the probabilities of the specified combinations. The key is to remember that the events are independent, meaning one person's success or failure doesn't affect the others. The solution will involve multiplying probabilities for events occurring together.

Model Answer

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Introduction

Probability, a fundamental concept in statistics and mathematics, quantifies the likelihood of an event occurring. It is widely used in various fields, including risk assessment, decision-making, and predictive modeling. In this problem, we are dealing with independent events – the success or failure of each person hitting the target does not influence the others. Calculating the probability of combined events requires understanding the principles of multiplication of probabilities for independent events. This question assesses the candidate’s ability to apply these principles to a practical scenario.

Understanding the Problem

Let A, B, and C represent the events that persons A, B, and C hit the target, respectively. We are given the following probabilities:

  • P(A) = 4/5 = 0.8
  • P(B) = 3/4 = 0.75
  • P(C) = 2/3 ≈ 0.667

We need to calculate the probabilities of three different scenarios.

(A) Probability that A, B, and C all hit the target

Since the events are independent, the probability of all three hitting the target is the product of their individual probabilities:

P(A ∩ B ∩ C) = P(A) * P(B) * P(C) = (4/5) * (3/4) * (2/3) = 24/60 = 2/5 = 0.4

(B) Probability that B and C hit the target, but A does not

First, we need to find the probability that A does *not* hit the target: P(A') = 1 - P(A) = 1 - (4/5) = 1/5 = 0.2

Now, we calculate the probability of B and C hitting the target, and A missing the target:

P(A' ∩ B ∩ C) = P(A') * P(B) * P(C) = (1/5) * (3/4) * (2/3) = 6/60 = 1/10 = 0.1

(C) Probability that C and A hit the target, but B does not

First, we need to find the probability that B does *not* hit the target: P(B') = 1 - P(B) = 1 - (3/4) = 1/4 = 0.25

Now, we calculate the probability of C and A hitting the target, and B missing the target:

P(C ∩ A ∩ B') = P(C) * P(A) * P(B') = (2/3) * (4/5) * (1/4) = 8/60 = 2/15 ≈ 0.133

Summary of Results

Scenario Probability
A, B, and C all hit 0.4
B and C hit, A misses 0.1
C and A hit, B misses 2/15 (≈ 0.133)

Conclusion

In conclusion, we have successfully calculated the probabilities for each of the given scenarios using the fundamental principles of probability. The key takeaway is the application of the multiplication rule for independent events. Understanding these concepts is crucial for analyzing and predicting outcomes in various real-world situations, from quality control to risk management. The ability to accurately assess probabilities is a valuable skill in many fields.

Answer Length

This is a comprehensive model answer for learning purposes and may exceed the word limit. In the exam, always adhere to the prescribed word count.

Additional Resources

Key Definitions

Independent Events
Events are considered independent if the outcome of one event does not affect the outcome of another. Mathematically, P(A ∩ B) = P(A) * P(B) if A and B are independent.
Conditional Probability
The probability of an event occurring given that another event has already occurred. It is denoted as P(A|B), read as "the probability of A given B".

Key Statistics

According to a report by the National Safety Council (2023), the probability of being involved in a car crash is approximately 1 in 366.

Source: National Safety Council, 2023

In actuarial science, life insurance premiums are calculated based on the probability of death at different ages, derived from mortality tables (data as of 2022).

Source: Society of Actuaries, 2022

Examples

Coin Toss

Flipping a fair coin twice are independent events. The outcome of the first flip (heads or tails) does not influence the outcome of the second flip.

Frequently Asked Questions

What if the events were not independent?

If the events were dependent, we would need to use conditional probability, P(A|B), which represents the probability of event A occurring given that event B has already occurred. The formula would then be P(A ∩ B) = P(A|B) * P(B).

Topics Covered

StatisticsMathematicsProbabilityEventsConditional Probability