UPSC NDA I 2024 Mathematics Paper

To determine the value of $p+q$, let's first analyze the given probabilities for an unbiased coin being tossed $n$ times.The probability of getting at least one tail when an unbiased coin is tossed $n$ times is given by:$p=1-$ Probability of getting all headsSince the probability of getting all heads (which is $\text{H H H . . . H}$ ) on $n$ coin tosses is $\left(\;\frac{1}{2}\right)^n$, we have:$p=1-\left(\;\frac{1}{2}\right)^n$The probability of getting at least two tails is:$q=1-$ Probability of zero tails - Probability of one tailThe probability of zero tails is $\left(\;\frac{1}{2}\right)^n$, and the probability of exactly one tail (which can occur in any of the $n$ positions) is $n \cdot \left(\;\frac{1}{2}\right) \cdot \left(\;\frac{1}{2}\right)^{n-1}=n \cdot \left(\;\frac{1}{2}\right)^n$. Therefore,$q=1-\left(\;\frac{1}{2}\right)^n-n \cdot \left(\;\frac{1}{2}\right)^n$We are given that:$p-q=\;\frac{5}{32}$Substituting the expressions for $p$ and $q$, we get:$\left(1-\left(\;\frac{1}{2}\right)^n\right)-\left(1-\left(\;\frac{1}{2}\right)^n-n \cdot \left(\;\frac{1}{2}\right)^n\right)=\;\frac{5}{32}$Simplifying the equation above:$1-\left(\;\frac{1}{2}\right)^n-1+\left(\;\frac{1}{2}\right)^n+n \cdot \left(\;\frac{1}{2}\right)^n=\;\frac{5}{32}$$n \cdot \left(\;\frac{1}{2}\right)^n=\;\frac{5}{32}$ We need to solve for $n$ :Let's say $n=5$, then:$5 \cdot \left(\;\frac{1}{2}\right)^5=\;\frac{5}{32}$Which simplifies to:$\;\frac{5}{32}=\;\frac{5}{32}$This shows that our assumption $n=5$ is correct. Now, substituting $n=5$ back into the expressions for $p$ and $q$ :$p=1-\left(\;\frac{1}{2}\right)^5=1-\;\frac{1}{32}=\;\frac{31}{32}$$q=1-\left(\;\frac{1}{2}\right)^5-5 \cdot \left(\;\frac{1}{2}\right)^5=1-\;\frac{1}{32}-\;\frac{5}{32}=1-\;\frac{6}{32}=\;\frac{26}{32}$Adding $p$ and $q$ :$p+q=\;\frac{31}{32}+\;\frac{26}{32}=\;\frac{57}{32}$