Math Problem Statement

A multiple-choice test consists of 30 questions with possible answers of a, b, c, d. Estimate the probability that with random guessing, the number of correct answers is at least 12.

Solution

This problem can be solved using a binomial distribution, which is appropriate here since each question has two possible outcomes (correct or incorrect), and each guess is independent.

Let's define the variables:

  • Total number of questions n=30n = 30
  • Probability of guessing a question correctly p=14=0.25p = \frac{1}{4} = 0.25
  • We want to find the probability of getting at least 12 correct answers, P(X12)P(X \geq 12), where XX is the number of correct answers.

Since calculating the exact binomial probabilities for each value from 12 to 30 would be complex, we can use a normal approximation to the binomial distribution.

Step 1: Verify Normal Approximation Validity

For a binomial distribution, the normal approximation is valid when both np5np \geq 5 and n(1p)5n(1 - p) \geq 5.

  • np=30×0.25=7.5np = 30 \times 0.25 = 7.5
  • n(1p)=30×0.75=22.5n(1 - p) = 30 \times 0.75 = 22.5

Both conditions are satisfied, so the normal approximation is appropriate.

Step 2: Mean and Standard Deviation of XX

The mean μ\mu and standard deviation σ\sigma for a binomial distribution are: μ=np=7.5\mu = np = 7.5 σ=np(1p)=30×0.25×0.752.37\sigma = \sqrt{np(1 - p)} = \sqrt{30 \times 0.25 \times 0.75} \approx 2.37

Step 3: Calculate P(X12)P(X \geq 12) using Normal Approximation

Using the continuity correction, we approximate P(X12)P(X \geq 12) as P(X11.5)P(X \geq 11.5).

Now, we convert this to a standard normal variable ZZ: Z=11.57.52.371.69Z = \frac{11.5 - 7.5}{2.37} \approx 1.69

Using a standard normal distribution table, we find: P(Z1.69)0.0455P(Z \geq 1.69) \approx 0.0455

Final Answer

The probability of guessing at least 12 answers correctly by random guessing is approximately 0.0455, or 4.55%.

Would you like further explanation on the normal approximation or more details on binomial probability?

Related Questions

  1. How would the probability change if the test had 50 questions instead of 30?
  2. What is the probability of getting exactly 10 questions correct?
  3. How does the probability change if there are 5 answer choices instead of 4?
  4. How would the probability be affected if the passing mark was set to 15 correct answers?
  5. What is the expected number of correct answers on a test of 30 questions with random guessing?

Tip

Using a normal approximation for binomial probabilities is helpful when dealing with large sample sizes, as it simplifies complex binomial calculations.

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Math Problem Analysis

Mathematical Concepts

Probability
Binomial Distribution
Normal Approximation

Formulas

Binomial probability: P(X = k) = C(n, k) * p^k * (1 - p)^(n - k)
Mean: μ = np
Standard deviation: σ = √(np(1 - p))
Z-score: Z = (X - μ) / σ

Theorems

Central Limit Theorem (for normal approximation)

Suitable Grade Level

Grades 11-12