Let A = the event that John is selected and B = the event that Dani is selected.

Given that $P\left(A\right)$ = 1/3 and $P\left(B\right)$ = 1/5

We know that $\bar{\text{A}}$ is the event that A does not occur and $\bar{\text{B}}$ is the event that B does not occur

Probability that only one of them is selected

$= \text{P}\left[\left(A \cap \bar{\text{B}}\right)\cup \left(\text{B} \cap \bar{\text{A}}\right)\right]$[ Reference : Algebra of Events]

$= \text{P}\left(A \cap \bar{\text{B}}\right)+ \text{P}\left(\text{B} \cap \bar{\text{A}}\right)$[ Reference :Mutually Exclusive Events and Addition Theorem of Probability)

$= \text{P(A)P(}\bar{\text{B}}) + \text{P(B)P(}\bar{\text{A}})$[ Here A and B are and refer theorem on independent events]

$=\text{P(A)}\left[1 - \text{P(B)}\right] + \text{P(B)}\left[1 - \text{P(A)}\right]$

$= \dfrac{1}{3}\left(1 - \dfrac{1}{5}\right) + \dfrac{1}{5}\left(1 - \dfrac{1}{3}\right) = \dfrac{1}{3} \times \dfrac{4}{5} + \dfrac{1}{5}\times \dfrac{2}{3} = \dfrac{4}{15} + \dfrac{2}{15} = \dfrac{2}{5}$