Math Problem Statement

multiple-choice quiz has 200 questions, each with 4 possible answers of which only 1 is correct. What is the probability that sheer guesswork yields from 25 to 30 correct answers for the 80 of the 200 problems about which the student has no knowledge?

Solution

To solve this problem, we use the binomial probability distribution, as the situation involves repeated independent trials (guessing on each question) with two outcomes: correct or incorrect.

Key Information:

  1. Number of trials (n): 80 questions
  2. Probability of success (p): 14\frac{1}{4} (since only one of four answers is correct)
  3. Number of successes (X): We need the probability that XX, the number of correct guesses, falls between 25 and 30 inclusive.
    • P(25X30)P(25 \leq X \leq 30)

The probability mass function for the binomial distribution is:

P(X=k)=(nk)pk(1p)nkP(X = k) = \binom{n}{k} p^k (1-p)^{n-k}

Where:

  • (nk)\binom{n}{k} is the binomial coefficient n!k!(nk)!\frac{n!}{k!(n-k)!}
  • pp is the probability of success
  • nn is the number of trials
  • kk is the number of successes

Since this involves a range of values, we compute:

P(25X30)=P(X=25)+P(X=26)+P(X=27)+P(X=28)+P(X=29)+P(X=30)P(25 \leq X \leq 30) = P(X = 25) + P(X = 26) + P(X = 27) + P(X = 28) + P(X = 29) + P(X = 30)

Exact Computation

If exact computation is required, we calculate each term P(X=k)P(X = k) using the formula. However, this can be tedious.

Approximation with Normal Distribution

For large nn, the binomial distribution can be approximated by the normal distribution:

XN(μ,σ2)X \sim N(\mu, \sigma^2)

Where:

  • μ=np=8014=20\mu = n \cdot p = 80 \cdot \frac{1}{4} = 20
  • σ=np(1p)=801434=153.87\sigma = \sqrt{n \cdot p \cdot (1-p)} = \sqrt{80 \cdot \frac{1}{4} \cdot \frac{3}{4}} = \sqrt{15} \approx 3.87

The probability P(25X30)P(25 \leq X \leq 30) can be approximated as:

P(25X30)P(25203.87Z30203.87)P(25 \leq X \leq 30) \approx P\left(\frac{25 - 20}{3.87} \leq Z \leq \frac{30 - 20}{3.87}\right)

P(25X30)P(1.29Z2.58)P(25 \leq X \leq 30) \approx P(1.29 \leq Z \leq 2.58)

Using a standard normal distribution table:

  • P(Z2.58)0.9951P(Z \leq 2.58) \approx 0.9951
  • P(Z1.29)0.9015P(Z \leq 1.29) \approx 0.9015

P(25X30)0.99510.9015=0.0936P(25 \leq X \leq 30) \approx 0.9951 - 0.9015 = 0.0936

Final Answer:

The probability that the student guesses correctly on 25 to 30 questions out of 80 is approximately 0.0936 or 9.36%.

Would you like detailed steps on calculating exact binomial probabilities, or have any other questions? Here are some related questions for further exploration:

  1. How is the binomial distribution derived and what are its assumptions?
  2. What is the difference between exact and approximate methods for binomial probabilities?
  3. How do you calculate P(Xk)P(X \leq k) for a binomial distribution?
  4. How does the normal approximation work and why is it used?
  5. What factors affect the shape of the binomial distribution?

Tip: Use software tools like Python,

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Math Problem Analysis

Mathematical Concepts

Probability
Binomial Distribution
Normal Approximation

Formulas

Binomial Probability Formula: P(X = k) = C(n, k) * p^k * (1-p)^(n-k)
Normal Approximation Formula: Z = (X - μ) / σ

Theorems

Central Limit Theorem
Properties of Binomial Distribution

Suitable Grade Level

Grades 11-12 or introductory college level