Question

A multiple choice examination has 5 questions. Each question has three alternative answers of which exactly one is correct. The probability that a student will get 4 or more correct answers just by guessing is

• A. \( \frac{13}{3^5} \)
• B. \( \frac{11}{3^5} \)
• C. \( \frac{10}{3^5} \)
• D. \( \frac{17}{3^5} \)

Hint:

When dealing with probabilities of getting a certain number of successes in a fixed number of independent trials, use the binomial distribution formula: \[ P(X = k) = \binom{n}{k} p^k q^{n-k} \]

Answer 1

The Correct Option is B

Let the number of trials \( n = 5 \).
Probability of success (correct answer by guessing) \( p = \frac{1}{3} \),
Probability of failure \( q = \frac{2}{3} \).
We want: \( P(X \geq 4) = P(X = 4) + P(X = 5) \) where \( X \sim \text{Binomial}(n=5, p=1/3) \).
Step 1: Calculate \( P(X = 4) \)
\[ P(X = 4) = \binom{5}{4} \left(\frac{1}{3}\right)^4 \left(\frac{2}{3}\right)^1 = 5 \cdot \frac{1}{81} \cdot \frac{2}{3} = \frac{10}{243} \] Step 2: Calculate \( P(X = 5) \)
\[ P(X = 5) = \binom{5}{5} \left(\frac{1}{3}\right)^5 = 1 \cdot \frac{1}{243} = \frac{1}{243} \] Step 3: Add the probabilities
\[ P(X \geq 4) = \frac{10}{243} + \frac{1}{243} = \frac{11}{243} \]