A computer has printer (P), disk (D), terminal (T), and magnetic (M) outputs. 55% of all output is on D, 25% are on P, 15% are on T, and 5% are on M. The error rate for D is 1/2000, for P it is 2/1000, for T it is 1/1000, and for M it is 1/500. An experiment is conducted where a character is output and we observe whether:
- What device made the output
- If the character is correct
What is the sample space? And what is the probability that the character was written on the disk, given A={it was incorrect}.
From my understanding this question requires the use of Bayes rule. However I'm not exactly sure how to apply it.
Using the obvious abbreviations, we want $\Pr(D|A)$, the conditional probability that the device used was the disk, given that the result was incorrect.
We will use the definition of conditional probability, which is close to Bayes's Rule, but not exactly the same. By definition, we have $$\Pr(D|A)=\frac{\Pr(D\cap A)}{\Pr(A)}.$$ Now we compute the probabilities on the right.
First we find the harder one, $\Pr(A)$. An error can happen in $4$ disjoint ways: (i) We used P, and got an error; (ii) We used D and got an error; We used T and got an error; (iv) We used M and got an error.
The probability we used P is, we are told, $\frac{25}{100}$. (Check this, the order of the letters changes during your post.) Given that we used P, the probability that $A$ happened (we had an error) is $\frac{2}{1000}$. Thus the probability of (i) is $\frac{25}{100}\cdot\frac{2}{1000}$.
The probabilities of (ii), (iii), and (iv) are computed in a similar way. For the probability of $A$, add up.
Finally, we need $\Pr(D\cap A)$. We know how to compute this, it is the probability of (ii).