UPSC MainsMANAGEMENT-PAPER-II20165 Marks
Q4.

Question 4

A multiple-choice quiz has 200 questions, each with four possible answers, of which only one is the correct answer. What is the probability that sheer guesswork yields from 25 to 30 correct answers for 80 of the 200 questions about which the student has no knowledge?

How to Approach

This question requires applying the principles of binomial probability distribution. The core idea is to calculate the probability of getting a specific range of successes (25-30 correct answers) in a fixed number of trials (80 questions) where each trial has only two possible outcomes (correct or incorrect). We need to understand the formula for binomial probability, calculate the probability for each success value (25, 26,...30), and then sum these probabilities to get the final answer. The question tests analytical and quantitative skills.

Model Answer

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Introduction

Probability theory is a fundamental branch of mathematics dealing with the likelihood of events occurring. In scenarios involving a large number of independent trials, such as a multiple-choice quiz, the binomial distribution provides a powerful tool for calculating probabilities. This question presents a practical application of the binomial distribution, asking us to determine the probability of achieving a specific range of correct answers through random guessing. Understanding this concept is crucial in various fields, including statistics, risk assessment, and decision-making. The problem focuses on 80 questions out of 200, where the student resorts to pure guesswork.

Understanding the Binomial Distribution

The binomial distribution is a probability distribution that describes the number of successes in a fixed number of independent trials. The formula for calculating the probability of exactly k successes in n trials is:

P(X = k) = (n choose k) * pk * (1-p)(n-k)

Where:

  • P(X = k) is the probability of exactly k successes
  • n is the number of trials
  • k is the number of successes
  • p is the probability of success on a single trial
  • (n choose k) is the binomial coefficient, calculated as n! / (k! * (n-k)!)

Applying the Formula to the Question

In this case:

  • n = 80 (the number of questions the student guesses on)
  • p = 0.25 (the probability of getting a correct answer by guessing, as there are four options)
  • We want to find P(25 ≤ X ≤ 30), which means we need to calculate P(X = 25) + P(X = 26) + ... + P(X = 30)

Calculating Individual Probabilities

We need to calculate each P(X = k) for k = 25 to 30 using the formula above. This involves calculating the binomial coefficient and then the probability. Due to the computational complexity, this is best done using statistical software or a calculator with binomial distribution functions. Let's illustrate with P(X=25):

P(X = 25) = (80 choose 25) * (0.25)25 * (0.75)55

Similarly, we calculate P(X = 26), P(X = 27), P(X = 28), P(X = 29), and P(X = 30).

Summing the Probabilities

Once we have calculated each individual probability, we sum them up to get the final probability:

P(25 ≤ X ≤ 30) = P(X = 25) + P(X = 26) + P(X = 27) + P(X = 28) + P(X = 29) + P(X = 30)

Approximation using Normal Distribution

Since n is large (n=80), we can approximate the binomial distribution with a normal distribution. The mean (μ) and standard deviation (σ) of the binomial distribution are:

  • μ = n * p = 80 * 0.25 = 20
  • σ = √(n * p * (1-p)) = √(80 * 0.25 * 0.75) = √15 ≈ 3.87

We can then use the normal distribution to approximate the probability. We need to apply a continuity correction. We want P(25 ≤ X ≤ 30), which becomes P(24.5 < X < 30.5) in the normal approximation.

We calculate the Z-scores for 24.5 and 30.5:

  • Z1 = (24.5 - 20) / 3.87 ≈ 1.19
  • Z2 = (30.5 - 20) / 3.87 ≈ 2.72

Then, P(24.5 < X < 30.5) = P(Z < 2.72) - P(Z < 1.19). Using a standard normal distribution table, P(Z < 2.72) ≈ 0.9967 and P(Z < 1.19) ≈ 0.8830. Therefore, P(24.5 < X < 30.5) ≈ 0.9967 - 0.8830 ≈ 0.1137.

Therefore, the probability that sheer guesswork yields from 25 to 30 correct answers for 80 of the 200 questions is approximately 0.1137 or 11.37%.

Conclusion

In conclusion, using the binomial distribution (or its normal approximation), we've determined the probability of achieving between 25 and 30 correct answers through random guessing on 80 questions. The approximate probability of 11.37% suggests that while not highly likely, it's certainly a plausible outcome given the number of trials. This illustrates the power of probability distributions in analyzing random events and quantifying uncertainty. Further refinement of the calculation would require computational tools for precise binomial coefficient calculations.

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

Binomial Distribution
A probability distribution that describes the number of successes in a fixed number of independent trials, each with the same probability of success.
Z-score
A measure of how many standard deviations an element is from the mean. It is used to standardize data and compare values from different distributions.

Key Statistics

The average score on the UPSC Civil Services Examination (Prelims) is around 30-35% (as of 2023 data).

Source: Various UPSC analysis websites and coaching centers

The probability of getting a perfect score (all questions correct) by random guessing on a 200-question multiple-choice test with four options per question is approximately 1.3 x 10<sup>-60</sup>.

Source: Calculated based on binomial probability formula (knowledge cutoff 2024)

Examples

Coin Toss

Flipping a fair coin 10 times and counting the number of heads is an example of a binomial experiment. Each flip is a trial, and getting heads is a success.

Frequently Asked Questions

What is the difference between a binomial and a normal distribution?

The binomial distribution is discrete (deals with whole numbers), while the normal distribution is continuous. The normal distribution can approximate the binomial distribution when the number of trials is large.