Following questions like this one, I have some very simple probability questions for which I guess that Bayes theorem is the tool to answer. I think that Bayes theorem is the tool because each question involves a posterior probability and an update of the current information after an event. My first question is: please, tell me if I am wrong and I am not supposed to use Bayes.

  1. Laura has a few necklaces (5 gold and 4 silver) and a few rings (4 gold and 3 silver) in the safe. Today she chose a piece of jewelry at random and her husband said to her “gold ("$g$") always really suits you”. What is the probability that it was a necklace ("$n$")?


  1. A box contains 4 red balls and 5 blue balls. One ball is taken out and then, without putting the one just taken out back in the box, another one is taken out. The second ball taken out is red ("$r_{2}$"). What is the probability that the first one was also red ("$r_{1}$")?


  1. There are 2 boxes: box $A$ contains 10 red balls and 10 blue balls. Box $B$ contains 5 red, 5 green, and 5 blue balls. One box is chosen at random and one ball is taken out. The ball taken out is red ("$r$"). What is the probability that this comes from the box $A$ ("$A$")?


Second question: however, it seems that the last answer is wrong and that $P(A|r)=\frac{3}{5}\ne\frac{7}{12}$. How would you explain it?


1 Answer 1


You should use the Bayes theorem, and all looks good. Your numerator in 3 is correct; as for the denominator, remember, you marginalise as follows;

$$P(r)= P(r|A)P(A)+ P(r|B)P(B) $$

This gives you:

$$P(A|r)=\frac{P(r|A)P(A)}{P(r|A)P(A)+ P(r|B)P(B)}=\frac{\frac{1}{2}\frac{1}{2}}{\frac{5}{12}}=\frac{3}{5}$$


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