The probability of event A occurring is p. If A doesn't occur, that is event a. What is the probability of the following occurring: A A, aa, Aa?

See below:

Explanation:

Let's use some numbers first and then generalize. I'll set $A$ as probability $\frac{3}{4}$ and $a$ as probability $\frac{1}{4}$.

The probability of drawing the three different draws are (and I'm assuming order matters and so $P \left(A a\right) \mathmr{and} P \left(a A\right)$ are different):

$P \left(A A\right) = \frac{3}{4} \times \frac{3}{4} = \frac{9}{16}$

$P \left(a a\right) = \frac{1}{4} \times \frac{1}{4} = \frac{1}{16}$

$P \left(A a\right) = \frac{3}{4} \times \frac{1}{4} = \frac{3}{16}$

We can now generalize using $p$ and $1 - p$:

$P \left(A A\right) = p \times p = {p}^{2}$

$P \left(a a\right) = \left(1 - p\right) \times \left(1 - p\right) = {\left(1 - p\right)}^{2} = 1 - 2 p + {p}^{2}$

$P \left(A a\right) = p \times \left(1 - p\right) = p - {p}^{2}$

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If order doesn't matter, then we multiply the $P \left(A a\right)$ results by 2 to account for $P \left(a A\right)$, giving:

$P \left(A a\right) = 2 \times \frac{3}{4} \times \frac{1}{4} = \frac{6}{16}$

and

$P \left(A a\right) = 2 \times p \times \left(1 - p\right) = 2 p - 2 {p}^{2}$