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You can not add them up.
Ya, so... What to do?
Must be different for me and you then. Mine requests a + b in a / b.
That method is faulty. If there were 5 people before D, you would end up with 5/5 and no chance for D.
I would say yes, because the answer you gave me works
Is that the exact question?
Consider that the 4th person chose dish X.
Although I have the exact answer by experimental methods I am still working on a good solid method that would please an olympic dude.
New Problem There are 5 kinds of dishes i, ii, iii, iv, and v. Four Customers A, B, C and D choose a dish at random. What is the probability that person D chooses a previously unordered dish?
Yes, if you are signed in as a member.
The Featured Solution: https://brilliant.org/i/b6ozMJ/