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:

We are told that the student answers 8 true-false questions purely by guessing. For every single question the probability of being correct is $$\tfrac12$$ and the probability of being wrong is also $$\tfrac12$$. Therefore, if we let the random variable $$X$$ denote “number of correct answers”, then $$X$$ follows a binomial distribution with parameters $$n=8$$ and $$p=\tfrac12$$, which we write as $$X\sim\text{Binomial}(8,\tfrac12)$$.

The probability mass function of a binomial random variable is given by the formula

$$P(X=k)=\binom{8}{k}\,p^{\,k}(1-p)^{8-k}.$$

Because $$p=\tfrac12$$, the factor $$p^{\,k}(1-p)^{8-k}=(\tfrac12)^8=\tfrac1{256}$$ is the same for every value of $$k$$. Hence

$$P(X=k)=\frac{\binom{8}{k}}{256}.$$

We need the smallest integer $$n$$ such that the probability of getting at least $$n$$ correct answers is less than $$\tfrac12$$. Symbolically we want

$$P(X\ge n)<\frac12,$$

and $$n$$ must be the least integer for which this inequality holds.

We start testing from the middle of the distribution because a binomial with $$p=\tfrac12$$ is symmetric around its mean $$\mu=8\!\times\!\tfrac12=4$$. First record the individual probabilities for $$k=0$$ through $$k=8$$:

$$\begin{aligned} P(X=0)&=\frac{\binom80}{256}=\frac1{256},\\ P(X=1)&=\frac{\binom81}{256}=\frac8{256},\\ P(X=2)&=\frac{\binom82}{256}=\frac{28}{256},\\ P(X=3)&=\frac{\binom83}{256}=\frac{56}{256},\\ P(X=4)&=\frac{\binom84}{256}=\frac{70}{256},\\ P(X=5)&=\frac{\binom85}{256}=\frac{56}{256},\\ P(X=6)&=\frac{\binom86}{256}=\frac{28}{256},\\ P(X=7)&=\frac{\binom87}{256}=\frac8{256},\\ P(X=8)&=\frac{\binom88}{256}=\frac1{256}. \end{aligned}$$

Now accumulate the probabilities from the top down until the total becomes less than $$\tfrac12$$.

Case $$n=4$$:

$$P(X\ge4)=P(X=4)+P(X=5)+P(X=6)+P(X=7)+P(X=8)$$

$$=\frac{70+56+28+8+1}{256}=\frac{163}{256}\approx0.6367>\frac12.$$

The condition is not satisfied for $$n=4$$.

Case $$n=5$$:

$$P(X\ge5)=P(X=5)+P(X=6)+P(X=7)+P(X=8)$$

$$=\frac{56+28+8+1}{256}=\frac{93}{256}\approx0.3633<\frac12.$$

The inequality $$P(X\ge5)<\tfrac12$$ is now satisfied, so $$n=5$$ works.

To be sure that 5 is indeed the smallest such integer, we already saw that $$n=4$$ fails; therefore $$n=5$$ is minimal.

Hence, the correct answer is Option A.