Question

A multiple choice test consists of \(5\) questions, each question having \(4\) responses. There is only one correct response and the remaining \(3\) are incorrect responses. If a candidate attempts all the \(5\) questions, then the probability that he answers at least \(3\) questions incorrectly is

• A. \(\dfrac{675}{1024}\)
• B. \(\dfrac{459}{512}\)
• C. \(\dfrac{81}{128}\)
• D. \(\dfrac{135}{512}\)

Hint:

For MCQ guessing problems, use binomial distribution where the probability of success or failure remains the same for each question. \[ P(X=r)={}^nC_rp^rq^{n-r} \]

Answer 1

The Correct Option is B

Step 1: Set up the binomial model.
Each question has probability $ p = \dfrac{3}{4} $ of being answered incorrectly (3 wrong out of 4 options). Let $ X $ = number of incorrect answers. Then $ X \sim B\left(5, \dfrac{3}{4}\right) $. We want $ P(X \geq 3) $.
Step 2: Break down $ P(X \geq 3) $.
\[ P(X \geq 3) = P(X=3) + P(X=4) + P(X=5) \]
Step 3: Compute each term using the binomial formula $ P(X=r) = \binom{n}{r}p^r q^{n-r} $.
\[ P(X=3) = \binom{5}{3}\left(\frac{3}{4}\right)^3\left(\frac{1}{4}\right)^2 = 10 \cdot \frac{27}{64} \cdot \frac{1}{16} = \frac{270}{1024} \] \[ P(X=4) = \binom{5}{4}\left(\frac{3}{4}\right)^4\left(\frac{1}{4}\right)^1 = 5 \cdot \frac{81}{256} \cdot \frac{1}{4} = \frac{405}{1024} \] \[ P(X=5) = \binom{5}{5}\left(\frac{3}{4}\right)^5 = \frac{243}{1024} \]
Step 4: Add the three terms.
\[ P(X \geq 3) = \frac{270 + 405 + 243}{1024} = \frac{918}{1024} = \frac{459}{512} \]
Step 5: Verify the simplification.
$ \gcd(918, 1024) = 2 $, so $ \dfrac{918}{1024} = \dfrac{459}{512} $. This matches option 2.
Step 6: State the answer.
\[ \boxed{\frac{459}{512}} \]