Poisson distribution and soccer scoring

Question

A soccer player scores at least one goal in roughly half of her games. How would you estimate the percentage of games where she scores exactly three goals?

$\textbf{My Attempt:}$

I try modeling this process by a Poisson distribution with parameter $\lambda=1/2.$ Then the probability that she will score exactly three goals is given by $$\mathbb P(X=3)=e^{-1/2}\frac{1}{6\cdot 2^3}=\frac{1}{48e^{1/2}}.$$

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Is this the correct way to model to model this process?

Any feedback is much appreciated. Thank you for your time.

Answer 1

If $Y \sim \mathsf{Pois}(\lambda)$ and $.5 = P(Y \ge 1) = 1 - P(Y=0) = 1 - e^{-\lambda},$ then $\lambda = 0.6931.$ Hence $P(Y = 3) = 0.0278.$

Computations in R:

lam = -log(.5); lam
[1] 0.6931472
dpois(0, lam)
[1] 0.5
dpois(3, lam)
[1] 0.02775205