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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.

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    Jeremy, I do not agree with your "faulty letter" calculation, or I do not understand what those probability numbers mean.2011-09-11

2 Answers 2

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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.

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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

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    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.2011-09-11