My question is about Independent events of the following question in which tree diagram is said to make. How to do this? Suppose identical tags are placed on both the left ear and the right
ear of a fox. The fox is then let loose for a period of time.
Consider the two events $C_1 = \{\text{left ear tag is lost}\}$ and $C_2 = \{\text{right ear tag is lost}\}$. Let $\pi = P(C_1) = P(C_2)$, and assume $C_1$ and $C_2$
are independent events. Derive an expression (involving $\pi$) for the
probability that exactly one tag is lost given that at most one is
lost.

Let $A$ be the event where exactly one tag is lost and $B$ be the event where at most one tag is lost. Then we are looking for
$$
P(A\mid B).
$$
By the definition of conditional probability,
$$
P(A\mid B) = \frac{P(A\cap B)}{P(B)}.
$$
If exactly one tag is lost, then it's true that at most one tag is lost, so $A\subset B$. Hence $P(A\cap B) = P(A)$. Hence we have to compute
$$
P(A\mid B) = \frac{P(A)}{P(B)}.
$$
The probability $P(A)$ that exactly one event of $C_1,C_2$ happens is given by
\begin{align*}
P(A) &= P(C_1) + P(C_2) - 2\cdot P(C_1\cap C_2) \\
&= \pi + \pi - 2\pi^2 = 2\pi(1-\pi).
\end{align*}
We used the identity $P(C_1\cap C_2) = P(C_1)\cdot P(C_2)$ because the events $C_1,C_2$ are independent.
The probability $P(B)$ that at most one tag is lost is $1-P(B^c)$, where $P(B^c)$ is the probability $P(C_1\cap C_2)$ that both tags are lost. Hence,
$$
P(B) = 1 - P(B^c) = 1 - P(C_1\cap C_2) = 1 - \pi^2 = (1-\pi)(1+\pi).
$$
Putting it all together,
$$
P(A\mid B) = \frac{2\pi(1-\pi)}{(1-\pi)(1+\pi)} = \color{blue}{\frac{2\pi}{1+\pi}}.
$$
