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?

  • $\begingroup$ Where do you get $\mathrm{P}(\mathsf{B}|!\mathsf{A})$ from ? (And does this need a 'homework' tag?) $\endgroup$ – Scortchi - Reinstate Monica Dec 6 '12 at 1:04
  • 2
    $\begingroup$ 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? $\endgroup$ – Dilip Sarwate Dec 6 '12 at 3:33

This would be correct:


Not this:



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


$$P(A|O) = \frac{P(O|A)P(A)}{P(O)} = \frac{P(O|A)P(A)}{P(O|A)P(A) + P(O|B)P(B) + P(O|C)P(C)} = \\ \frac{20/100 \cdot 40/100}{20/100 \cdot 40/100 + 40/100 \cdot 20/100 + 90/100 \cdot 40/100} = 8/52 = 2/13$$

In your case $P(B|\neg A)$ cannot be computed by $40/100 \cdot 20/100 + 90/100 \cdot 40/100 = 11/25$.

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

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