UPSC NDA II 2018 Mathematics Solved Paper

Concept:We use the binomial probability formula to find the chance of exactly $r$ successes in $n$ independent trials, where each trial has success probability $p$ and failure probability $q = 1-p$.Explanation:Here, $n = 8$ coins are tossed simultaneously.Getting a head on a single toss is considered a success.For a fair coin, $p = \frac{1}{2}$ and $q = 1 - \frac{1}{2} = \frac{1}{2}$.We need the probability of at least 6 heads, i.e., $x = 6$, $x = 7$, or $x = 8$ heads.The probability is therefore: $P(x \ge 6) = P(x=6) + P(x=7) + P(x=8)$.Using the binomial formula $P(x=r) = \binom{n}{r} p^{r} q^{n-r}$:$P(x=6) = \binom{8}{6}\left(\frac{1}{2}\right)^6\left(\frac{1}{2}\right)^{2} = \binom{8}{6}\left(\frac{1}{2}\right)^8$$P(x=7) = \binom{8}{7}\left(\frac{1}{2}\right)^7\left(\frac{1}{2}\right)^{1} = \binom{8}{7}\left(\frac{1}{2}\right)^8$$P(x=8) = \binom{8}{8}\left(\frac{1}{2}\right)^8\left(\frac{1}{2}\right)^{0} = \binom{8}{8}\left(\frac{1}{2}\right)^8$Factor out $\left(\frac{1}{2}\right)^8$: $\left(\frac{1}{2}\right)^8 \left[\binom{8}{6} + \binom{8}{7} + \binom{8}{8}\right]$.Compute the combinations: $\binom{8}{6} = 28$, $\binom{8}{7} = 8$, $\binom{8}{8} = 1$. Sum = $28 + 8 + 1 = 37$.Thus, the required probability is $\frac{37}{2^8} = \frac{37}{256}$.Answer:$\frac{37}{256}$ (Option C).