TITLE: What is the probability that both tubes are good?
QUESTION [1 upvotes]: This a conditional probability where we calculate: 
Probability that both are good knowing that one of them is good= prob that both are good/ probability that one of them is good
Which gives me: (1/3) / (6/10)=5/9
But the correct answer is: 5/13
How should I proceed?

REPLY [1 votes]: You presumably calculated the probability both are good as $$\frac{6}{10}\times\frac{5}{9}=\frac{1}{3}$$
Similarly the probability both are bad is $$\frac{4}{10}\times\frac{3}{9}=\frac{2}{15}$$
And so the probability at least one is good  $$1-\frac{2}{15} = \frac{13}{15}$$
which you could alternatively calculate as $\frac{6}{10}+\frac{4}{10}\times\frac{6}{9}$ (since if the first is bad then the second might be good) or as $\frac{6}{10}\times\frac{4}{9}+\frac{4}{10}\times\frac{6}{9}+\frac{6}{10}\times\frac{5}{9}$ (looking at all the possibilities for both: good & bad, or bad & good, or good & good)
so the answer to the question is $$\frac{\frac13}{\,\frac{13}{15}\,}=\frac{5}{13}$$

REPLY [1 votes]: $P(A\mid B)$ is not $P(A)/P(B)$ in general, but it is when $A\subseteq B$, so all good.
$P(A)=\frac13$, you're right.
To compute $P(B)$, the easiest is to go through $P(\overline B)=\frac{6}{45}$. That's where your mistake is.
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