Question

A student appeared in an examination consisting of 8 true - false type questions. The student guesses the answers with equal probability. The smallest value of n, so that the probability of guessing at least 'n' correct answers is less than $\frac{1}{2}$, is :

• A. 3
• B. 4
• C. 5
• D. 6

Hint:

For a symmetric binomial distribution ($p=1/2$), the median is at the mean Np = 8(1/2) = 4. The probability $P(X \ge k)$ will be greater than 0.5 if k is less than or equal to the mean, and less than 0.5 if k is greater than the mean.

Answer 1

The Correct Option is C

This is a binomial probability problem.
Number of trials (questions), N = 8.
For each question, there are two outcomes: correct or incorrect.
The probability of guessing a correct answer, $p = 1/2$.
The probability of guessing an incorrect answer, $q = 1 - p = 1/2$.
Let X be the number of correct answers. The probability of getting exactly k correct answers is given by the binomial probability formula:
$P(X=k) = \binom{N}{k} p^k q^{N-k} = \binom{8}{k} (\frac{1}{2})^k (\frac{1}{2})^{8-k} = \binom{8}{k} (\frac{1}{2})^8$.
We want to find the smallest value of n such that the probability of guessing at least 'n' correct answers is less than $\frac{1}{2}$.
$P(X \ge n)<\frac{1}{2}$.
$P(X \ge n) = P(X=n) + P(X=n+1) + ....... + P(X=8)$.
Let's calculate the probabilities for different values of k:
$P(X=k) = \frac{\binom{8}{k}}{2^8} = \frac{\binom{8}{k}}{256}$.
$\binom{8}{0}=1, \binom{8}{1}=8, \binom{8}{2}=28, \binom{8}{3}=56, \binom{8}{4}=70, \binom{8}{5}=56, \binom{8}{6}=28, \binom{8}{7}=8, \binom{8}{8}=1$.
Total probability is $\sum_{k=0}^8 P(X=k) = \frac{1+8+28+56+70+56+28+8+1}{256} = \frac{256}{256} = 1$.
Now let's check the condition $P(X \ge n)<1/2 = 0.5$.
For n=1: $P(X \ge 1) = 1 - P(X=0) = 1 - 1/256 = 255/256>0.5$.
For n=2: $P(X \ge 2) = 1 - (P(X=0)+P(X=1)) = 1 - 9/256>0.5$.
For n=3: $P(X \ge 3) = 1 - (P(X=0)+P(X=1)+P(X=2)) = 1 - 37/256>0.5$.
For n=4: $P(X \ge 4) = P(X=4)+...+P(X=8) = \frac{70+56+28+8+1}{256} = \frac{163}{256} \approx 0.63>0.5$.
For n=5: $P(X \ge 5) = P(X=5)+P(X=6)+P(X=7)+P(X=8) = \frac{56+28+8+1}{256} = \frac{93}{256} \approx 0.36<0.5$.
The smallest value of n for which the condition holds is n=5.