Question 197961
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All Wednesdays:


There are seven different days in the week, so the probability that one person is born on a given day is *[tex \LARGE \frac{1}{7}].  Since each of the three birthdays is an independent event, the total probability is the product of the three individual probabilities:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(\frac{1}{7}\right)\left(\frac{1}{7}\right)\left(\frac{1}{7}\right)]


Different day of the week:


The probability that the first guy is born on any one of the 7 days in the week is certainty, or 1.  The probability that the second guy was born on a different day is the number of days unused so far, or 6 divided by the number of possible days, or 7.  The probability that the third guy was born on yet a different day is calculated the same way, this time 5 divided by 7.  Again the total probability is the product:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 1\cdot\left(\frac{6}{7}\right)\left(\frac{5}{7}\right)]


None on Saturday:


6 out of 7 ways <i>not</i> to be born on Saturday, so:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(\frac{6}{7}\right)\left(\frac{6}{7}\right)\left(\frac{6}{7}\right)]


By the way, you get to do your own arithmetic.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
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