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

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.

| cite | improve this answer | |

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

| cite | improve this answer | |
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.