Independence of Events I'd appreciate some help with this problem:
There are 2 plants that make keyboards. Keyboard faults are classified in 3 categories: letter, number, and other. If a keyboard is chosen at random, are the events "faulty letter" and "plant 2" independent?
Plant    Letter    Number    Other
1        15        45        40
2        75        30        45
I know that to prove independence, I need to show $P(A|B) = P(A)$; or $P(B|A) = P(B)$; or $P(A \cap B) = P(A)P(B)$. But my question is, how can I do this if I'm only given $P(A)$ and $P(B)$? The intersection of those, divided by either $A$ or $B$ (as needed), will always give me the other probability, so this isn't useful.
In case it matters, this IS a textbook problem, but only for my benefit, not class.
 A: As pointed by Didier is not enough, but if you're interested in going far in your problem than just homework you should check about copulas
A: Let $C_1$ and $C_2$ be the events that the keyboard comes from companies 1 and 2 respectively.  Let $L$ and $N$ be the "faulty letter" and "faulty number" events respectively. You are given the values of $P(L | C_i)$ and $P(N | C_i)$ for $i \in \{1,2\}$. You are not given the value of $P(C_1)$ and $P(C_2)$ themselves. This is necessary for solving the problem; here I assume (arbitrarily) that $P(C_1) = P(C_2) = 0.5$.
So,


*

*The probability of "plant 2" is nothing but $P(C_2)$, which we assumed.

*The probability of "letter fault" can be computed by the law of total probability:
$$
P(L) = P(L | C_1) \cdot P(C_1) + P(L | C_2) \cdot P(C_2).
$$

*The probability of the intersection of the above two events is simply the second term of the formula for $P(L)$:
$$
P(L \cap C_2) = P(L | C_2) \cdot P(C_2). 
$$
So you have all the information needed to check if $L$ and $C_2$ are independent.
It is possible that I misinterpreted the question, since I did not use all the data. If the question asks us to do something else, then indicate so in the comments/question.
