Calculate probability via Bayes' theorem Amongst three group of products, only the first group contains one third of irregular products while other 2 contain only correct products.. Calculate the probability that the randomly taken product is from the group that contains incorrect products if we know the product is correct. I thought of using Bayes' theorem bit somehow my results then out illogical. Thanks.

Using Bayes's theorem is, indeed, the way to go. You have the following events:

* The object is from the $i$-th group. Let us mark that as $G_i$ ($G_1$ is the event "product comes from first group"). Also, in this numbering, the first group contains irregular products while the other two do not.

* The object is correct. Let us mark that as $C$.



Now, what you want to calculate is $$P(G_1|C).$$ By Bayes' theorem, you know that that equals $$\frac{P(C|G_1) \cdot P(G_1)}{P(C)}.$$

You already know what $P(C|G_1)$ is, and you know what $P(G_1)$ is. For $P(C)$, you have two options:

* Try to calculate the probability of $\
eg C$.
* Resort to the law of total probability, $P(C) = \sum_iP(C|H_i)P(H_i)$ for some appropriate hypotheses $H_i$.