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: AA, aa, Aa?

Step 1Let's use some numbers first and then generalize. I'll set A as probability 34 and a as probability 14.The probability of drawing the three different draws are (and I'm assuming order matters and so P(Aa) and P(aA) are different):P(∀)=34×34=916P(aa)=14×14=116P(Aa)=34×14=316Step 2We can now generalize using p and 1−p:P(∀)=p×p=p2P(aa)=(1−p)×(1−p)=(1−p)2=1−2p+p2P(Aa)=p×(1−p)=p−p2If order doesn't matter, then we multiply the P(Aa) results by 2 to account for P(aA), giving:P(Aa)=2×34×14=616.and P(Aa)=2×p×(1−p)=2p−2p2

"The probability of event A occurring is p...." solved and explained

Jacob Stein

Answered question

2022-02-14

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: AA, aa, Aa?

Answer & Explanation

bijteny85

Beginner2022-02-15Added 10 answers

Step 1
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 (A a) and P (a A)

are different):

P (\forall) = \frac{3}{4} \times \frac{3}{4} = \frac{9}{16}

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

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

Step 2
We can now generalize using p and

1 - p

P (\forall) = p \times p = p^{2}

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

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

If order doesn't matter, then we multiply the P(Aa) results by 2 to account for P(aA), giving:

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

.
and

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

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