Question

On a multiple-choice exam with three possible answers for each of the five questions, what is the probability that a student would get four or more correct answers just by guessing?

Short answer

The probability of a student guessing and getting at least 4 correct answers is \(11/243\), or approximately 0.0452675.

Step-by-step solution

Understanding the binomial probability formula

The probability of getting exactly k successes in n independent Bernoulli trials with probability p of success on each trial can be calculated using the binomial probability formula: P(X = k) = \(\binom{n}{k}\) p^k (1-p)^(n-k) In this case, each question on the exam represents an independent Bernoulli trial with success probability p = 1/3 (since there are three possible answers, and only one correct answer). We are looking for the probability of getting at least four correct answers, so we will need to find P(X = 4) + P(X = 5).

Calculating the binomial coefficients

We will first need to compute the binomial coefficients \(\binom{n}{k}\) for the cases X = 4 and X = 5. These coefficients represent the number of ways we can choose k successes from n trials: \(\binom{n}{k} = \frac{n!}{k!(n-k)!}\) For X = 4: \(\binom{5}{4} = \frac{5!}{4!(5-4)!} = 5\) For X = 5: \(\binom{5}{5} = \frac{5!}{5!(5-5)!} = 1\)

Computing the individual probabilities

Now, we will compute the probability of 4 and 5 correct answers using the binomial probability formula from Step 1: For X = 4: P(X = 4) = \(\binom{5}{4}\) (1/3)^4 (2/3)^1 = 5 * (1/81) * (2/3) = 10/243 For X = 5: P(X = 5) = \(\binom{5}{5}\) (1/3)^5 (2/3)^0 = 1 * (1/243) * 1 = 1/243

Computing the total probability

Finally, we can add the probabilities of getting 4 or 5 correct answers just by guessing: P(X >= 4) = P(X = 4) + P(X = 5) = 10/243 + 1/243 = 11/243 So, the probability of guessing and getting at least 4 correct answers is 11/243, or approximately 0.0452675.