Summary
The probability can meaningfully differ from $\frac 23$. When the biological child is a boy or girl with equal chances, it can be $\frac 34$, $\frac 23$, or even tends to $\frac 12$ or $1$. In fact, $\frac 23$ is only obtained for a lottery-like situation.
The probability is a decreasing function of the odds of a specific event, which enables us to easily analysis different assumptions.
A key modeling point is that wining in a championship race is not similar to wining in a lottery where buying two tickets double the wining chance.
Events
$W $: The adopted girl of the family is the winner
$F$: The family of the adopted girl holds the championship
$G_1$: The family of the adopted girl has a biological boy (has only one girl)
$G_2$: The family of the adopted girl has a biological girl (has two girls).
Analysis
Given the problem statement:
A family has a child and adopts a girl as well. If some years later one of the children become the champion of women's swimming championship, what is the probability that this girl is the adopted child?
we want to compute the probability $\color{blue}{p}$ of that the adopted girl of the family wins the race ($W$) given her family holds the championship ($F$). This can be written as the following conditional probability:
$$\color{blue}{p}= P(W|F).$$
This probability can be computed using each of the following formulas:
$$ P(W|F)=\frac{P(W \cap F)}{P(F)}=\frac{P(A)+P(B)}{ P(A)+P(B)+P(C)} \tag{1}$$
$$ P(W|F)=P(W|F \cap G_1)P(G_1|F)+P(W|F \cap G_2)P(G_2|F) \tag{2}$$
where the new auxiliary events $A$, $B$, and $C$ are nothing but the three disjoints possible situations in front of us:
$\; \; \;A=W \cap F\cap G_1 $: The biological child is a boy, and the adopted girl is the winner
$\; \; \; B=W \cap F\cap G_2$: The biological child is a girl, and the adopted girl is the winner
$\; \; \; C=W' \cap F\cap G_2$: The biological child is a girl, and the biological girl is the winner.
To use ($1$) or ($2$), the probability terms in the right side should be determined under a suitable assumption. Some possible assumptions are considered in the following.
Assumption 1: The three events $A$, $B$, and $C$ occur with equal probabilities.
In this case, $P(A)=P(B)=P(C)$, and from ($1$) one can easily get $$p=\color{blue}{\frac23}.$$
The above is a common assumption used in elementary probability courses and text books to simplify problems, especially when students are not familiar with conditional probabilities and only able to compute the probabilities by counting methods.
Maybe the above method was intended by the examiner because to answer a 4-choice question like the one in the OP, around a minute to 90 seconds is typically given.
Assumption 2: It is equally likely for the family to have a biological boy or girl. When the family holds the championship and has two girls, the winner can be the biological girl or the adopted girl equally likely.
In this case, by defining $\alpha >0 $ (which is the odds of the event $G_2$ given $F$)
$$\alpha=\frac{P(G_2|F)}{P(G_1|F)}=\frac{P(F|G_2)}{P(F|G_1)}, \tag{3}$$
one can use the formula $(2)$ as follows:
$$\color{blue}{p}= 1 \times \frac {1}{1+\alpha} +\frac12 \times \frac {\alpha}{1+\alpha},$$
to get the general formula:
$$\fbox{$\color{blue}{p}=\frac {2+\alpha}{2(1+\alpha)}.$} \tag{4}$$
This gives $\color{blue}{\frac34}$ for $\alpha=1$, $\color{blue}{\frac23}$ for $\alpha=2$, tends to $\color{blue}{\frac 12}$ as $\alpha \to +\infty$, and tends to $\color{blue}{1}$ as $\alpha \to 0$.
Let us briefly discuses some cases on $\alpha$. Other reasonings may be given by readers.
- Case 1: $0< \alpha <1$. For this case, we have $P(F|G_2)<P(F|G_1)$, which may seem unacceptable at first, but it is possible for a toxic family environment where the performances of the girls decrease because of jealousy, recognition of unfairness, or other factors.
- Case 2: $\alpha \ge 1$. For this case, we have $P(F|G_2) \ge P(F|G_1)$, which seems more sensible if birds in their little nests agree. In fact, a positive environment may help both girls perform better. In addition to intra-familial factors, the value of $\alpha$ may also depend on extra-familial factors, such as the number of other competitors.
- Case 3: $\alpha = 2$. In this case, we have a lottery-like situation, as wining in the championship race is assumed to be similar to wining in a lottery with unique bills where buying two tickets double the wining chance [1]. It may be worth noting that $\alpha=2$ implies that $P(A)=P(B)=P(C)$, which follows from $$\small{P(A)=P(W |F\cap G_1)P(F|G_1)P(G_1)=\frac 12 P(F|G_1)\\=P(B)=P(W |F\cap G_1)P(F|G_2)P(G_2)= \frac 14 P(F|G_2).}$$ This means that Assumption 1 and the lottery-like situation are equivalent under Assumption 2. Indeed, $$\alpha=2 \iff p=\color{blue}{\frac23} \iff P(A)=P(B)=P(C)$$ as the function in ($4$) is strictly decreasing in $\alpha$.
Remark 1: Almost all the given answers to the OP's question consider the lottery-like situation ($\alpha=2$), see e.g., Mike's answer [2] and Xander's second answer [3] (Xander's first answer [4] estimates the probability of that an adopted girl wins the race). David's answer [5] is an exception. With the assumption he made, for $N\ge 0$, we have $$1 \le \alpha=\frac{\frac{2}{N+2}}{\frac{1}{N+1}}=\frac{2(N+1)}{N+2}<2,$$ which from $(4)$ gives $\frac{2N+3)}{3N+4} > \frac 23$ (but it rapidly tends to the lottery-like situation as $N$ increases). The above is an instance for which $\alpha$ depends on the number of the other competitors (an extra-familial factor).
Remark 2: Indeed, one can consider more realistic assumptions to get a better estimate based on some available data. For example, in reality, the probability of the baby being a boy or a girl is not the same. Moreover, the adopted girl or the biological girl may have a more favorable genetic profile or better mental or psychological well-being, which implies that the wining chances of the girls are not the same. A similar analysis can be used to handle these extensions to get:
$$\color{blue}{p}= \frac {1+\theta}{1+\alpha} \tag{5}$$
where $\kappa=P(G_2)$, $\theta=P(U|F \cap G_2)$, and $\alpha=\frac{P(G_2|F)}{P(G_1|F)}=\frac{\kappa}{1-\kappa}\frac{P(F|G_2)}{P(F|G_2)}$. $\kappa$ can be estimated by the probability that a baby is a girl [6]. $\theta$ can be estimated by $\frac{p_A}{p_A+p_B}$, where $p_A$ and $p_B$ are the success probabilities of the adopted and biological girls, respectively.
