Note that you’re assuming that the probabilities of scoring are statistically independent.

The easiest way to do these problems is to calculate the probability that *nobody* scores, and then subtract that from 1.

With two, the probability that nobody scores is:

(1 − 0.25)(1 − 0.25) = 0.75^{2} = 0.5625

Therefore, the probability that at least one scores is:

1 − 0.5625 = 0.4375 = **43.75%**

With three players, the probability that nobody scores is:

(1 − 0.25)(1 − 0.25)(1 − 0.25) = 0.75^{3} = 0.421875

and the probability that at least one scores is:

1 − 0.421875 = 0.578125 = **57.8125%**

With four,

1 – 0.75^{4} = **68.359375%**

And so on.

If you want to extend the addition rule to more than two events, here’s how it works:

P(A *or* B) = P(A) + P(B) – P(AB)

P(A *or* B *or* C) = P(A) + P(B) + P© – P(AB) – P(AC) – P(BC) + P(ABC)

P(A *or* B *or* C *or* D) = P(A) + P(B) + P© + P(D) – P(AB) – P(AC) – P(AD)

– P(BC) – P(BD) – P(CD) + P(ABC) + P(ABD) + P(ACD) + P(BCD) – P(ABCD)

In general, it’s:

P(At least one of *n* things) = Σ(probabilities taken *one* at a time)

– Σ(probabilities taken *two* at a time)

- Σ(probabilities taken
*three* at a time)

– Σ(probabilities taken *four* at a time)

± Σ(probabilities taken *n* at a time)