# What is the probability that a customer will not use a credit card? Pays in cash or with a credit card?

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.

• The reasoning is well described, and correct. Commented Sep 3, 2015 at 0:32
• 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.
– lulu
Commented Sep 3, 2015 at 0:46
• @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. Commented Sep 3, 2015 at 0:59
• Ah, the joys of machine interaction. Glad it worked!
– lulu
Commented Sep 3, 2015 at 1:02