# Bayes' Theorem - Probability Pants problem

I'm having an issue with a question regarding Bayes' Theorem. Here is the question:

An online clothing store carries three brands of jeans. 40% of sales are brand A, 20% are brand B and the remainder are brand C. 20% of brand A jeans cost over 100, 40% of brand B jeans cost over $100 and 90% of brand C jeans cost over 100. Given that a pair of jeans is purchased for over 100, what is the probability that they are brand A? My work is as follows: Using Bayes Theorem I classified two events$A$: Picking brand A$B$: Jeans cost over 100 So I would need to find$P(A|B)$$$P(A) = 2/5$$ $$P(B|A) = 1/5$$ $$P(\neg A) = 3/5$$ $$P(B|\neg A) = 11/25$$ When I do calculations with Bayes' theorem, I get; $$\frac{2/25}{2/25 + (11/25)\times(3/5)} = 10/43$$ But the answer is$2/13$. Now, when I first did the question, I forgot to multiply$P(B|\neg A)$by$P(\neg A)$in the denominator, and I got the right answer. Is there a reason I should leave$P(\neg A)$out? Or did I approach the problem in the wrong way completely? • Where do you get$\mathrm{P}(\mathsf{B}|!\mathsf{A})$from ? (And does this need a 'homework' tag?) – Scortchi - Reinstate Monica Dec 6 '12 at 1:04 • I would suggest that you give some thought to your choice of notation. Given that A, B, and C are already used as name brands and you have decided to use A to mean Brand A is chosen, you are just setting yourself for confusion by choosing B to denote Jeans cost over \$100 instead of B being chosen. Couldn't you have thought of something else? Say O for over \$100 the way that both answers did? – Dilip Sarwate Dec 6 '12 at 3:33 ## 2 Answers This would be correct:$\mathrm{P}(\mathsf{O}|¬\mathsf{A})=\mathrm{P}(\mathsf{O}|\mathsf{B})\mathrm{P}(\mathsf{B}|¬\mathsf{A})+\mathrm{P}(\mathsf{O}|\mathsf{C})\mathrm{P}(\mathsf{C}|¬\mathsf{A})=\frac{11}{15}$Not this:$\mathrm{P}(\mathsf{O}|¬\mathsf{A})=\mathrm{P}(\mathsf{O}|\mathsf{B})\mathrm{P}(\mathsf{B})+\mathrm{P}(\mathsf{O}|\mathsf{C})\mathrm{P}(\mathsf{C})=\frac{11}{25}$It is easier to say that you have 3 disjoint events:$A$: picking up brand A$B$: picking up brand B$C$: picking up brand C and you want to compute$P(A|O)$with$O$: jeans cost over 100 so $$P(A|O) = \frac{P(O|A)P(A)}{P(O)} = \frac{P(O|A)P(A)}{P(O|A)P(A) + P(O|B)P(B) + P(O|C)P(C)} = \\ \frac{20/100 \cdot 40/100}{20/100 \cdot 40/100 + 40/100 \cdot 20/100 + 90/100 \cdot 40/100} = 8/52 = 2/13$$ In your case$P(B|\neg A)$cannot be computed by$40/100 \cdot 20/100 + 90/100 \cdot 40/100 = 11/25\$.

• That seems like a more clear cut way to do the question. Thank-you. – user1209379 Dec 6 '12 at 2:22