Question 415228
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There are 26 ways to pick the letter, then 9 ways to pick the first digit (because you must exclude zero), then 10 ways to pick each of the second and third digits, and finally 5 ways to pick the last digit (because you must exclude the 5 odd digits and you must include zero as an even digit).  Hence:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 26\ \times\ 9\ \times\ 10\ \times\ 10\ \times\ 5]


You can do your own arithmetic to get the denominator of your probability fraction.  Presuming there is only one correct access code, the numerator of your probability fraction would be 1.  This is a reasonable presumption because of the wording of the problem: "...selecting <i><b>the</b></i> correct access code..."  Be that as it may the numerator would change to however many valid access codes exist if in fact there were more than one valid code)


John
*[tex \LARGE e^{i\pi} + 1 = 0]
My calculator said it, I believe it, that settles it
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