Jeremy's Solution:

For five flips, the answer is 5/16.
There are 2 choices for each of five coin flips, thus the number of possible paths I could take is 2 to the fifth power, or 32. Of all these paths, the only ones that will take me to school are the ones that involve a total gain of 1 unit East. The only way to do this is a path with 3 Easts and 2 Wests. thus, 2 of the Easts and 2 of the Wests cancel out. so to figure out the number of possible paths to school, lone must figure out the number of ways to arrange the letters EEEWW. This is the same as inserting two Wests into 5 empty slots. There are 10 ways, as follows:

W W _ _ _
W _ W _ _
W _ _ W _
W _ _ _ W
_ W W _ _
_ W _ W _
_ W _ _ W
_ _ W W _
_ _ _ W W

Or, if we assume that the 3 E's are different and the 2 W's are different, one could say that there are 5 choices for the first letter, 4 choices for the second letter, 3 choices for the third letter, 2 choices for the fourth letter, and 1 choice for the last letter. This is equal to 5 x 4 x 3 x 2 x 1, or 120. But the 3 E's are in fact the same, as are the W's, thus we must divide by 12, because we have counted each arrangement of the 3 E's 3! or 6 times, and each arrangement of the 2 W's 2! or 2 times, and we again arrive at 10. thus the answer is 10/32, or 5/16.

For the question where there are 10 flips, the answer is zero, because I can only end up an even number away from my staring point (and school is an odd number away). similarly, there is a problem for any even number.