$\text{Kashvi, Kalyani and Sara of class}12\text{are given a problem in Accounts whose respective probabilities of solving are}\frac{2}{5},\frac{1}{4}\text{and}\frac{1}{\mathrm{ 6}}.\text{They were asked}$

$\text{to solve it independently.}$ ```$\text{The probability that either only Kalyani or Kashvi or Sara solves it is:}$

The correct answer is 9/20.

We are given the individual probabilities of solving the problem:

P(Kashvi solves) = $\frac{2}{5}$, so P(Kashvi does NOT solve) = $1-\frac{2}{5}=\frac{3}{5}$
P(Kalyani solves) = $\frac{1}{4}$, so P(Kalyani does NOT solve) = $1-\frac{1}{4}=\frac{3}{4}$
P(Sara solves) = $\frac{1}{6}$, so P(Sara does NOT solve) = $1-\frac{1}{6}=\frac{5}{6}$

Since they solve independently, we need the probability that exactly one of them solves it. This means one person solves it while the other two do not. There are three mutually exclusive cases:

Case 1: Only Kashvi solves it (Kalyani and Sara do not)

$P\left(\text{only Kashvi}\right)=\frac{2}{5}\times \frac{3}{4}\times \frac{5}{6}=\frac{30}{120}=\frac{1}{4}$

Case 2: Only Kalyani solves it (Kashvi and Sara do not)

$P\left(\text{only Kalyani}\right)=\frac{3}{5}\times \frac{1}{4}\times \frac{5}{6}=\frac{15}{120}=\frac{1}{8}$

Case 3: Only Sara solves it (Kashvi and Kalyani do not)

$P\left(\text{only Sara}\right)=\frac{3}{5}\times \frac{3}{4}\times \frac{1}{6}=\frac{9}{120}=\frac{3}{40}$

Since these three cases are mutually exclusive, we add the probabilities. To add them, we use a common denominator of 40:

$\frac{1}{4}=\frac{10}{40},\frac{1}{8}=\frac{5}{40},\frac{3}{40}=\frac{3}{40}$

$P\left(\text{exactly one solves}\right)=\frac{10}{40}+\frac{5}{40}+\frac{3}{40}=\frac{18}{40}=\frac{9}{20}$

Therefore, the probability that either only Kashvi, or only Kalyani, or only Sara solves the problem is 9/20.