To solve this problem, we need to calculate the probability that a student randomly guesses more than 4 questions correctly on a 10-question multiple choice quiz, where each question has 4 possible choices. Since the student is guessing, the probability of guessing any single question correctly is $\frac{1}{4}$, and the probability of guessing incorrectly is $\frac{3}{4}$.

This situation follows a binomial distribution because each question has two possible outcomes: correct or incorrect. The binomial probability formula is:

$P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}$

Where:

• $n = 10$ (number of trials, i.e., questions),
• $k$ is the number of correct answers (this varies),
• $p = \frac{1}{4}$ (the probability of success, i.e., getting a question right).

We are asked to find the probability that the student gets more than 4 questions correct, i.e., $P(X > 4)$. This is equivalent to calculating the sum of the probabilities for getting 5, 6, 7, 8, 9, and 10 correct answers:

$P(X > 4) = P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10)$

I will now calculate this probability.The probability that the student gets more than 4 questions correct by randomly guessing is approximately 7.81%.

Would you like further details on how this was calculated or any other clarification?

Here are some related questions for practice:

1. What is the probability of getting exactly 4 questions correct?
2. What is the probability of getting all 10 questions correct?
3. What is the expected number of correct answers in this quiz?
4. How does the probability change if the quiz has 20 questions?
5. What is the variance of the number of correct answers in this scenario?

Tip: In binomial distributions, the expected value is calculated as $E(X) = np$, where $n$ is the number of trials and $p$ is the probability of success.