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Topic: **First grade homework sheets****Question:**
We have 100 graded homework sheets on the desk. The first student was in hurry and grabbed one sheet randomly (that means he might take anybody’s homework with the equal chance). The following students checked the pile one by one: if one found his own sheet, he took it; otherwise, he would also randomly take one from the remaining pile.
Question: You were the last student to get your homework, what’s the chance you found your own work sheet?

April 22, 2019 / By Adison

Somewhat surprising, the answer is that your chance, as the last student, to find your own work is 1/2. Let P[n] be the probability that that the last student in a class of n students gets his own test back. Then: P[2] = 1/2 P[3] = 1/3 + (1/3)P[2] = 1/3 + 1/6 = 1/2 P[4] = 1/4 + (1/4)(P[3] + P[2]) = 1/4 + 1/4 = 1/2 etc. You can proceed (by induction or any other method), by using the following equation: P[n] = 1/n + (1/n)(P[2] + P[3] + ......+ P[n-1]) to show that P[2] = P[3] = ......= P[n-1] = P[n] = 1/2. So, regardless of whether there are 10, 20, 100 or 1000 students, the probability of the last student to get his own paper back is always 1/2.

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Did you like the answer? We found more questions related to the topic: **First grade homework sheets**

I think that you diddnt put the full riddle in, i think you mean there was a ladder hanging over the side of a boat(thats how the actual riddle is. The answer is the ladder will always be 1ft under the water as the water rises so does the boat! LOOK AT RIDDLE 53 IN THIS LINK, ANSWER AT BOTOM OF THE PAGE ON LINK. http://www.scribd.com/doc/18477353/Riddles-For-Kids-

The first student randomly takes a sheet. There is a 1 in 100 chance that he took his own. The next student takes a sheet. This may be his own, unless the person before him took it. Then he will take one randomly. The chance that the guy before him took his in 1 in 100. The chance that he will have to take one randomly is the chance that the first person took his. Which is one in a 100. The third student will serach the pile of sheets. He may find his own, or he will take one randomly. The chance that the second person took his sheet is 1 on 99. The chance that the first person took the sheet of the second person is one in a hundred. Therefore, the chance that the third person has to take a sheet randomly is 1/100 * 1/99. Etc. The chance that the last student does NOT find his own work sheet is thereby Pnot = 1/100*1/99*1/98*1/97............*1/2 = 1/(100!) = 1.07151029 × 10^-158 The chance that he does is Pdoes = 1-Pnot = 1-1.07151029 × 10^-158 = very close to 1

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7 crossings counting both ways. Parrot across, back, cat across, back with parrot, crackers across, back, parrot across.

http://britton.disted.camosun.bc.ca/jbwo... exact same puzzle, different objects. Look at solution to level 1, substitute parrot for goat, cat for wolf, and cracker for cabbage. take the parrot over. come back alone. take the crackers over. come back but take the parrot with you. leave the parrot, take the cat over. come back alone. take the parrot over. That's seven crossings, if you count coming back.

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