If two balls are chosen at random and one of them is found to be of standard quality. What is the probability that the other one is defective? A football factory has two machines, A and B. Machine A manufactures 70% of the balls and machine B manufactures 30%. 80% of the balls from machine A and 90% of the balls from machine B are rated of standard quality.
Q1. A ball is chosen at random and is found to be of standard quality. What is the probability that it has come from machine A? The answer here is 0.67 and can be found using Bayes' rule.
Q2. If two balls manufactured using the same machine are chosen at random and one of them is found to be of standard quality, what is the probability that the other ball is defective?
My attempt:
I have used Bayes theorem to answer the question.
Here, D = event of having defective item, S = event of having  Standard quality item.
Pr(D|S) = Pr(D).Pr(S|D)/ Pr(S) 
Pr(S) = 0.7*0.8 + 0.3*0.9 = 0.83
Pr(D) = 0.7*0.2 + 0.3*0.1 = 0.17
Since they are independent events. They have no common intersection between them, Pr(S|D) = Pr(S)
So, In my opinion, the answer is 0.17.
But, when I discussed this with my friends, one of them highlighted a point saying that "how do you know the exact first pick will be standard or defective?" In his opinion, I need to dig more. So it can happen in 4 ways. They are given below: 
Event 1 could be One Standard, other is defective;
Event 2 could be One Defective, other is Standard; 
Event 3 could be Both Defective;
Event 4 could be both Standard.
I have used the conditional probability approach to solve that. 
Now let's say X = event of other one is defective, Y = event of one of them is Standard
Pr(one of them being Standard) = 0.83*0.83 + 0.83*0.17 = 0.9711 (sum of event 1,2,4) ---> Pr(Y)
Pr(other one defective | one of them Standard) = Pr(X intersets Y) / P(Y)
Pr(X intersets Y) = 0.83*0.17 + 0.17*0.83 = 0.2822 (Sum of event 1,2)
therefore the answer is 0.2822/0.9711 = 0.2906
I have applied Bayes, got the same answer in this case.
which one was the correct approach? and if you could contrast between them, it will help me a lot to understand it better. TYIA
 A: The second part of the question is ill-posed in two ways.  First, it is not clear whether the two balls were drawn from the same randomly selected machine.  Second, it is not clear whether the standard quality ball was determined by random selection and the second ball's quality is unknown, or if the quality of both balls is known to another party and you are told that at least one ball is of standard quality.  To illustrate why these things matter, consider the following scenarios:


*

*The machines that make the balls dump them into a large bin, mixing up all of the balls; however, balls are stamped with a serial number that identifies which machine that made them.  You randomly select one ball from the bin, inspect it, and find it is of standard quality.  You then draw a second ball from the bin.  What is the probability it is defective?  The answer is $$1 - ((0.8)(0.7) + (0.9)(0.3)) = 0.17.$$

*Same as Scenario 1, except this time, a factory worker randomly select two balls from the bin and inspects them.  Then he tells you that one of the balls is standard quality.  What is the probability the other ball is defective?  The answer is $$\frac{2(0.17)(0.83)}{1 - (0.17)^2} = \frac{34}{117} \approx 0.290598.$$

*Now instead suppose the machines put the balls into separate bins.  The factory worker, with equal probability, selects one of the two bins, and randomly draws one ball for you to inspect; it is standard.  If he draws a second ball from the same bin (thus the same machine), what is the probability it is defective?  The answer is computed as follows.  Let $X_i = 1$ if ball $i$ is standard, and $0$ if defective.  Then the posterior probability of the worker having selected Bin $A$ given the $X_1 = 1$ is $$\Pr[A \mid X_1 = 1] = \frac{\Pr[X_1 = 1 \mid A]\Pr[A]}{\Pr[X_1 = 1]} = \frac{(0.8)(0.5)}{(0.8)(0.5)+(0.9)(0.5)} = \frac{8}{17} \approx 0.470588.$$  Consequently, the probability the second ball drawn from the same bin is defective is $$\begin{align*}
\Pr[X_2 = 0 \mid X_1 = 1] &= \Pr[X_2 = 0 \mid A]\Pr[A \mid X_1 = 1] + \Pr[X_2 = 0 \mid B]\Pr[B \mid X_1 = 1] \\
&= (0.2)(8/17) + (0.1)(9/17) = \frac{5}{34} \approx 0.147059.
\end{align*}$$

*As in Scenario 3, we have separate bins, but now this time, the factory worker selects a bin with probability proportional to the number of balls it contains; that is to say, because machine $A$ makes $70\%$ of the balls, he chooses bin $A$ with $70\%$ probability rather than $50\%$.  Everything else is the same.  Then the answer becomes $$(0.2)(56/83) + (0.1)(27/83) = \frac{139}{830} \approx 0.16747.$$

*Now let's consider Scenario 3 except the factory worker has drawn two balls from the same bin, inspects them, and informs you that one of them is standard.  What is the probability the other is defective?  By now you should clearly realize that this probability will not be the same as any of the ones we have calculated so far.  I will not do the computation; I leave it as a straightforward exercise for the reader.

*Consider Scenario 5 except now the probability of selecting a bin is as in Scenario 4.  Again, the computation will not be performed here and is left as an exercise.
As you can see, we must be absolutely clear about how the balls are drawn in order to perform the requisite computation.  The way the question is posed does not adequately describe the sampling scheme.

Now that the question has been rephrased, we can reasonably interpret it to match Scenario 3 or Scenario 4.  But which one is again not clear because it is not explained how two balls are randomly selected from the same machine.
