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.

  • 2
    $\begingroup$ Knowing P(A) and P(B) but not P(A and B) (or P(A or B), by the way) is not enough to determine whether A and B are independent or not. $\endgroup$
    – Did
    Sep 11 '11 at 15:31
  • $\begingroup$ For this question, you can calculate all the required probabilities (and so decide whether the events are independent). Can you tell us where you are stuck? $\endgroup$
    – Srivatsan
    Sep 11 '11 at 15:44
  • $\begingroup$ I calculate probability of faulty letter = P(A)=(15+75)/(15+45+40+75+30+45) and probability from plant 2 = P(B) = (75+30+45)/(15+45+40+75+30+45) Then, P(A|B) = P(A Intersect B)/P(B) would always give me P(A), and always give me "true" since P(A|B) would = P(A). Can someone please explain where my logic went wrong? $\endgroup$
    – Jeremy
    Sep 11 '11 at 15:56
  • $\begingroup$ Jeremy, I do not agree with your "faulty letter" calculation, or I do not understand what those probability numbers mean. $\endgroup$
    – Srivatsan
    Sep 11 '11 at 16:01

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

  • $\begingroup$ Frank, Since you seem to be knowledgeable about copulas, you could explain which aspects of theory and/or applications they provide, that the usual conditioning notion does not. $\endgroup$
    – Did
    Sep 11 '11 at 15:48

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 $P(C_1) = P(C_2) = 0.5$.


  • 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.


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