How do I find $P(A), P(B), P(C)$

Question

Would someone help to solve and explain this problem?

Total exported devices $2425000$ per year. The top exported device: Device A ($725790$), device B ($537390$), and device C ($159877$).

1. Calculate $P(A), P(B)$, and $P(C)$.
2. Are the events $A$ and $B$ mutually exclusive? Why? and compute $P(A ∩B)$.
3. Find the probability exported device from $A$ or $C$?
4. What is the probability to export other than one of these three?

Here is my solution, would check it please:

Q1/

$P(A)=725790/(725790+537390+159877)=0.51002173$

$P(B)=537390/(725790+537390+159877)=0.37763069$

$P(C)=159877/(725790+537390+159877)=0.11234757$

Q2

Yes, they are mutually exclusive, that the same device cannot be exported twice, therefore; $P(A∩C)=0$.

Q3

$P(A∪C)=P(A)+ P(C)= 0.51002173+ 0.37763069= 0.88765242$

Q4

$(2425000-1423057)/425000= 0.4132$

Answer 1

Thanks for providing your work. Here are some Comments.

You are partly on the right track, but in (1) denominators should be 2425000. The way you have it $P(A)+P(B)+P(C)=1$ and that contradicts what you have in (4), which may be OK.

You may be right about (2), but not for the reason given. If $A$ and $C$ are exported independently, then $P(A \cap C) = P(A)P(C).$ If they are mutually exclusive then $P(A \cap C) = 0.$ I hope there is more information about this than you have given. If not, you need to make and state assumptions.

Then in (3) you have $P(A \cup C) = P(A)+P(C)-P(A\cap C).$ In addition to the uncertainty about $P(A \cap C)$, you have the wrong values of $P(A)$ and $P(C)$ from (1).

Answer 2

1.

$P(A)= \frac{725790}{2425000}=0.299294845$

$P(B)= \frac{537390}{2425000}=0.221604124$

$P(C)= \frac{159877}{2425000}=0.06592866$

2.

If two events are mutually exclusive then $P(Event_1) ∩ P(Event_2) = 0 $. And

Yes, they are mutually exclusive, that the same device cannot be exported twice, therefore; P(A∩C)=0.

is wrong statement. Firstly, Device A and Device B are not same devices. In this context "mutually exclusive" means Devices A and B cannot be exported at the same time. Which means they are dependent.

3.

Answer is wrong if two events are independent.

4.

(Total minus sum of A, B and C) divided by Total.