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So I'm doing some basic probability problems for homework, and we just recently went over the Inclusion-Exclusion prinicple, which I'm assuming this problem deals with, which is as follows.

Shoppers can pay for their purchases with cash, a credit card, or a debit card. Suppose that the proprietor of a shop determines that 63% of her customers use a credit card, 23% pay with cash, and the rest use a debit card. What is the probability that a person will not use a credit card? What is the probability that a person pays in cash or with a credit card?

Now, it's my understanding that these percentages they give us don't overlap (i.e. a person won't pay part of their bill with cash, and the rest on a card, etc. etc.) So it seems relatively straightforward in the sense that only 14% of people pay with a debit card. So for the first part of the problem, wouldn't the answer simply be to add 14% and 23%? And the second would simply be the addition of 63% and 23%?

I have a feeling I'm misunderstanding something, as I'm not getting the problem right.

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  • $\begingroup$ The reasoning is well described, and correct. $\endgroup$ Commented Sep 3, 2015 at 0:32
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    $\begingroup$ The only thing I can think of is that perhaps you are meant to express your result as a decimal. Thus, instead of $14\%+23\%$ you should have $.14+.23$. Otherwise, what you write sounds entirely correct. $\endgroup$
    – lulu
    Commented Sep 3, 2015 at 0:46
  • $\begingroup$ @lulu that did it! I feel ridiculous but it's funny that the program doesn't tell you anywhere to express it as a percentage. $\endgroup$
    – secondubly
    Commented Sep 3, 2015 at 0:59
  • $\begingroup$ Ah, the joys of machine interaction. Glad it worked! $\endgroup$
    – lulu
    Commented Sep 3, 2015 at 1:02

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Reading the problem and your explanation, I think that you are completely right. If 63% of the people pay with credit, then 100%-63% = 14+23% don't. Then for the other question, 63%+23% would equal the amount of people who pay with credit or cash. So yeah, your reasoning is right.

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  • $\begingroup$ Yeah, I may just ask the professor on Friday if there's something up with the problem, it's just weird $\endgroup$
    – secondubly
    Commented Sep 3, 2015 at 0:38

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