Events A, B and C are independent, $\mathbb P (A) = 0.1$; $\mathbb P (B) = 0.4$ and $\mathbb P (C) = 0.9.$
- Find the probability of an event $D = (A + B) (A + C) (B + C).$
- Find the probability of event D if it is known that event A has already occurred.
At first glance, the task seemed easy and I used the multiplication and addition formulas for the probabilities:
I) $\mathbb P(A+B)= \mathbb P(A) + \mathbb P(B)$
II) $P(A\cdot\ B) = \mathbb P(A) \cdot\ \mathbb P(\frac{B}{A}) $
- $D=(0,1+0,4)*(0,1+0,7)*(0,4+0,7)=0,44$
- $D=(1+0,4)*(1+0,7)*(0,4+0,7)=2,618$
but the answer came out wrong, this is on the first point. The second point, I took probability A as a unit and also used the formulas for adding and multiplying the probabilities, but the answer is also incorrect. Why? What is the highlight of this task?
Correct answer:
- 0.4180.
- 0.940.
Consider the following Venn's Diagram
(1) what they are requesting you is to calculate the probability of the event
$$D=(A \cup B)\cap(A \cup C)\cap(B \cup C)$$
that is the probability of the purple area:
$$0.1\times0.4+0.1\times 0.9+0.4\times0.9-2\times0.1\times0.4\times0.9=0.418$$
(2) Can you try to proceed by yourself? (the previous diagram can be VERY useful)
after a small reasoning you get
$$\mathbb{P}(D|A)=\frac{\mathbb{P}(AD)}{\mathbb{P}(A)}=\frac{\mathbb{P}(AB)+\mathbb{P}(AC)-\mathbb{P}(ABC)}{\mathbb{P}(A)}=$$
$$=\frac{0.1\times0.4+0.1\times0.9-0.1\times0.4\times0.9}{0.1}=0.94$$
this because $A\cap D$ is the following area
that means its probability is $P(AB)+P(AC)-P(ABC)$