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                  Misc  Hardware Software
Within Warranty 15.00%   14.00%  29.00%
After Warranty   10.00%  26.00%  6.00%

I've come across a seemingly simplistic question in a textbook (regarding discrete random variables) and I'm wondering if there is more to it than simply reading the appropriate value from the table.

The above table concerns type of complaints made at a computer shop. The question reads:

i) If a complaint has been made, what is the probability the complaint was about Hardware, given that the complaint was filed within the warranty period?

Is the answer really as simple as 14.00%?

If not, can anybody explain or point me in the right direction to solve this?

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Hint:

  • You might start by finding the probability that a complaint is filed within the warranty period
  • Then you can find the conditional probability the complaint was about Hardware, given that the complaint was filed within the warranty period
  • which is the probability that the complaint was about Hardware and was filed within the warranty period (the $14\%$ in the table), divided by the probability that the complaint was filed within the warranty period
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  • $\begingroup$ So P(A+B)/P(B), where P(A+B) = 14%. And P(B) is equal to (15+14+29) = 58%. 14/58 = 0.24. That's 0.24%? $\endgroup$ May 29 '20 at 12:44
  • $\begingroup$ @user3012926 Does $0.24\%$ seem plausible? It seems a little small to me. $\endgroup$
    – Henry
    May 29 '20 at 12:46
  • $\begingroup$ Yes, you're right - it's 0.14/0.58 = 0.241 i.e. 24.1%. $\endgroup$ May 29 '20 at 13:00
  • $\begingroup$ Are the events described in the question considered dependent or independent by the way? $\endgroup$ May 29 '20 at 13:01
  • $\begingroup$ @user3012926 Since the six percentages add up to $100\%$ you might reasonably assume they are supposed to be mutually disjoint $\endgroup$
    – Henry
    May 29 '20 at 13:42
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The trick here is modifying our universe, if the question just asked about Hardware AND during warranty, then the table info would be correct, but the problem gives us a clue because we already know the complaint was done during the period of warrante, now our universe gets smaler, befora it was the whole table, but know it is just the first row. To solve this I just do a $\frac{Succes cases}{total cases}$ which in this case would be> $$P(Hardware|During Waranty)=\frac{0.14}{0.14+0.15+0.29}$$ $$P(Hardware|During Waranty)=\frac{0.14}{0.58}=0.24=24\%$$

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