UPSC NDA I 2024 Mathematics Paper

To solve for the value of $n$, we need to use the given probabilities and the fact that the coin is unbiased, meaning the probability of getting heads (H) and tails $(T)$ are both $\;\frac{1}{2}$.The probability of getting at least one tail in $n$ coin tosses is $p$. The complementary event, i.e., getting no tails (all heads), happens with a probability of $\left(\;\frac{1}{2}\right)^n$.Therefore, we can write the probability of getting at least one tail as:$p=1- \left(\;\frac{1}{2}\right)^n$Now, for the probability of getting at least two tails $(q)$, we consider the complementary events, which are getting zero tails or exactly one tail. The probability of getting exactly one tail in $n$ coin tosses can be calculated using the binomial distribution:$\binom{n}{1} \left(\;\frac{1}{2}\right)^1 \left(\;\frac{1}{2}\right)^{n-1} = n \left(\;\frac{1}{2}\right)^n$Therefore, the probability of getting at least two tails is:$q=1- \left(\;\frac{1}{2}\right)^n - n \left(\;\frac{1}{2}\right)^n = 1- \left(\;\frac{1}{2}\right)^n(1+n)$We are given that $p-q=\;\frac{5}{32}$. Substituting the expressions for $p$ and $q$, we get:$\left(1- \left(\;\frac{1}{2}\right)^n\right) - \left(1- \left(\;\frac{1}{2}\right)^n(1+n)\right) =\;\frac{5}{32}$Simplifying, we obtain:$1- \left(\;\frac{1}{2}\right)^n -1+ \left(\;\frac{1}{2}\right)^n(1+n) =\;\frac{5}{32}$$\left(\;\frac{1}{2}\right)^n(1+n)- \left(\;\frac{1}{2}\right)^n =\;\frac{5}{32}$$\left(\;\frac{1}{2}\right)^n n =\;\frac{5}{32}$ Solving for $n$, we look at reasonable values of $n$ that make the equation true. Let's explore the given options:For $n=4$ :$\left(\;\frac{1}{2}\right)^4 \times 4 =\;\frac{1}{16} \times 4 =\;\frac{4}{16} =\;\frac{1}{4}$This does not match $\;\frac{5}{32}$.For $n=5$ :$\left(\;\frac{1}{2}\right)^5 \times 5 =\;\frac{1}{32} \times 5 =\;\frac{5}{32}$This matches the given equation $\;\frac{5}{32}$.So, the correct value of $n$ is 5 .