Question

In a multiple-choice examination, there are \(10\) questions with one correct option out of \(4\) options for each question. A student gets \(4\) marks for each correct answer and \(1\) mark is deducted for each incorrect answer. The probability that a student, guessing randomly on every question, scores \(30\) marks in this exam is...

• A. \(\frac{45}{4^{10}}\)
• B. \(\frac{135}{4^{10}}\)
• C. \(\frac{405}{4^{10}}\)
• D. \(\frac{810}{4^{10}}\)

Hint:

Find the number of correct answers from \(4k-(10-k)=30\), then use the binomial formula.

Answer 1

The Correct Option is C

Step 1: Plan:
Count the favourable answer patterns and multiply by the probability of each.

Step 2: Steps:
Eight correct and two wrong gives $32 - 2 = 30$ marks. The number of ways to pick which two are wrong is $\frac{10\times9}{2} = 45$.
Each pattern has probability $\frac{1}{4^8}\cdot\frac{3^2}{4^2} = \frac{9}{4^{10}}$. So the total is $45\times\frac{9}{4^{10}} = \frac{405}{4^{10}}$.

Final Answer:
The probability is $\frac{405}{4^{10}}$, option (C). \[ \boxed{\frac{405}{4^{10}}} \]