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

Answer 1

Answer: (C)

p = p (correct answer), q = p (wrong answer)
$\Rightarrow \pi = \frac{1}{3} ,q = \frac{2}{3}, n = 5$
By using Binomial distribution
Required probability
$ = ^{5}C_{4} \left(\frac{1}{3}\right)^{4}\cdot \frac{2}{3} + ^{5}C_{5} \left(\frac{1}{3}\right)^{5} $
$= 5\cdot \frac{2}{3^{5}} + \frac{1}{3^{5}} = \frac{11}{3^{5}} $