We can solve this problem by using the binomial probability.
In this case, there are two possible outcomes for each trial (a player has scored a goal and a player has not scored a goal), and the probability of a success is equal for each trial (each occurrence has the same probability that a particular player has scored a goal).
Since we assumed that an opponent is evenly matched, we will state that there is a probability of $0.5$ that a team A (where a desired player plays) scores a goal.
So, the probability that a player scores a randomly selected goal is :
$p = 0.5 \cdot 0.15=0.075$
Moreover, the probability that a player does NOT score a randomly selected goal is :
$q=1-p=1-0.075=0.925$
Also, we will assume that exactly $6$ goals will occur (since we do not have any data on how the distribution of goals is, apart from expected number of goals).
Moreover, the number of trials is fixed ($6$ goals) and the trials are independent from each other.
To determine the probability of $x$ successes in $n$ trials, we will use the Binomial Formula :
$ P(X=x) = \dfrac{n!}{x! \cdot (n-x)!} \cdot p^x \cdot q^{n-x}$
where $n$ represents sample size, $x$ is the number of successes, $n-x$ represents the number of failures. Variable $p$ represents the probability of a success, and $q$ denotes the probability of failure.
The probability that a player will score EXACTLY one goal is :
$ P(X=1) = \dfrac{6!}{1! \cdot (6-1)!} \cdot 0.075^1 \cdot 0.925^{6-1} \approx 0.30473$
Furthermore, we will determine the probability that a player scores two or more goals by using the Complement Rule :
$ P(X \geq 2) =1-P(X=0)-P(X=1)$
Hence, we will determine the probability that a player does not score a singular goal :
$ P(X=0) = \dfrac{6!}{0! \cdot (6-0)!} \cdot 0.075^0 \cdot 0.925^{6-0} \approx 0.6264$
Now, we can determine the probability that a player scores two or more goals by using the Complement Rule :
$ P(X \geq 2) =1-0.6264-0.30473 = 0.06887$
To conclude, there is an approximately $30.473 \%$ chance that a player scores EXACTLY one goal and a $6.887 \%$ chance that a player scores two or more goals.