Math Problem Statement

A multiple-choice test consists of 9 questions. Each question has answer choices of a, b, c, and d, and only one of the choices is correct. If a student randomly guesses on each question, what is the probability that she gets fewer than 2 of them correct?

Carry your intermediate computations to at least four decimal places, and round your answer to two decimal places.

Solution

To solve this problem, we need to determine the probability that a student guesses fewer than 2 correct answers out of 9 questions on a multiple-choice test. Each question has 4 choices, and only one choice is correct. Since the student is guessing, the probability of getting a question correct is 14\frac{1}{4}, and the probability of getting a question wrong is 34\frac{3}{4}.

This is a binomial probability problem, where the number of trials n=9n = 9, the probability of success (getting a question correct) p=14p = \frac{1}{4}, and we want to find the probability that the student gets fewer than 2 correct answers.

We need to find the sum of the probabilities of getting 0 correct answers and 1 correct answer.

Step 1: Probability of getting 0 correct answers

The probability of getting exactly k=0k = 0 correct answers is given by the binomial probability formula:

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

For k=0k = 0:

P(X=0)=(90)(14)0(34)9P(X = 0) = \binom{9}{0} \left(\frac{1}{4}\right)^0 \left(\frac{3}{4}\right)^9

P(X=0)=1×1×(34)9(0.75)90.1001P(X = 0) = 1 \times 1 \times \left(\frac{3}{4}\right)^9 \approx (0.75)^9 \approx 0.1001

Step 2: Probability of getting 1 correct answer

For k=1k = 1:

P(X=1)=(91)(14)1(34)8P(X = 1) = \binom{9}{1} \left(\frac{1}{4}\right)^1 \left(\frac{3}{4}\right)^8

P(X=1)=9×14×(34)89×0.25×(0.75)89×0.25×0.10010.2252P(X = 1) = 9 \times \frac{1}{4} \times \left(\frac{3}{4}\right)^8 \approx 9 \times 0.25 \times (0.75)^8 \approx 9 \times 0.25 \times 0.1001 \approx 0.2252

Step 3: Sum the probabilities

Now, we add the probabilities of getting 0 or 1 correct answer:

P(X<2)=P(X=0)+P(X=1)0.1001+0.22520.3253P(X < 2) = P(X = 0) + P(X = 1) \approx 0.1001 + 0.2252 \approx 0.3253

Final Answer

Rounding to two decimal places, the probability that the student gets fewer than 2 correct answers is approximately 0.33\boxed{0.33}.

Would you like more details or have any questions about this problem?

Here are some related questions you might find useful:

  1. What is the probability of getting exactly 3 correct answers?
  2. How does the probability change if there are 5 answer choices instead of 4?
  3. What is the expected number of correct answers if a student guesses on all 9 questions?
  4. How would the probability change if the test had 15 questions instead of 9?
  5. What is the variance of the number of correct answers?
  6. How would you calculate the probability of getting at least 7 correct answers?
  7. Can you find the cumulative distribution function (CDF) for this problem?
  8. What is the probability that the student gets all questions wrong?

Tip: When dealing with binomial probability problems, it's useful to remember that the sum of probabilities for all possible outcomes (from 0 to n correct answers) should always equal 1.