Topic: probability question?
June 16, 2019 / By Clarette 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?
Now, I know the answer is actually quite suprising: 1/2. But I cannot understand why. Can someone explain this to me, and perhaphs also prove it? Preferably in some way I may be able to understand. (i.d. Don't just make calculations, say what is you are doing) Thank you!
Also, the statement that chance that a student gets his test back for a total N of student should hold in more cases than just 100. Particulary for any 10^n where n is a natural number.
Best Answers: probability question?
Baylee | 1 day ago
The answer is 1/2. This is identical to a problem that was on the radio show Car Talk.
Here is how it works. It all comes down to which homework sheet is taken first between the first person's and the last person's. Let's number them sheet 1 and sheet 100. If sheet 100 is taken first then the last person will not get the correct homework sheet. However, if at some point sheet 1 is taken, then every person afterward, including the last person, will get the correct homework sheet. Since there is a 50% chance of either sheet 1 or 100 being chosen first, the overall probability is one half.
Let me clarify this with an example.
If person one takes sheet 1 then obviously everyone gets the right sheet. Suppose person 1 takes sheet 7. Then people 2 through 6 all get their own. If person 7 takes sheet 1 then everyone after 7 gets the correct sheet. Suppose person 7 takes the sheet of person 12. People 8 to 11 all get the right sheet and when we get to person 12, we are in the same position as before. If person 12 takes sheet 1 then everyone afterward will get the right sheet. Otherwise the problem gets kicked down the road.
When it gets to person 100, there is only one sheet left, so at least one of sheet 1 and sheet 100 must have been chosen, and depending on which was first chosen, person 100 will have the correct one. If sheet 1 is first chosen then sheet 100 will be the one remaining. If sheet 100 is chosen first, it does not matter if sheet 1 is chosen afterward, since the sheet of person 100 has been taken in either case.
👍 230 | 👎 1
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Originally Answered: Probability Question?
6 reds, 4 blacks, 10 whites.
Worst case scenario is to first remove all the whites and all the blacks: 16 socks and still not a pair of each!
You need two more = 18 socks total
which is the 9 pairs sought.
Originally Answered: Probability Question?
In worst case of removing 8 pairs has only red socks and white socks
(find the worst case by sorting max --> min
for this question is 5, 3, 2 )
the worst case of 8 pairs + 1 black pair will make sure that she will has a pair of each color.
so the minimum is 9 pairs
1/100 chance the first student took your sheet
1/100 chance the first student took his own sheet (then you will get yours)
98/100 then first student took another's sheet. Say you are student 1 and the first one is 100, then the one he took is A (between 1 and 100). Then:
everybody until A gets his own paper, then A has A sheets to pick from.
when A picks, there is a 1/A chance he takes your sheet
there is a 1/A chance he takes the sheet of student 100 (then you will get yours)
there is a (A-2)/A chance he takes B's sheet.
This keeps repeating. If the last student is 2, then:
there is a 1/2 chance he takes your sheet
there is a 1/2 chance he takes a prior student's sheet (then you get yours)
So, at whatever point the outcome is determined (the student picks your sheet or his own, in the case of student 100, or a prior student that picked randomly, for any other), it is equal if he takes your sheet or someone else's that makes it final.
And this holds for any number of students (more than 2 :-)
👍 100 | 👎 -6
P(n) be the probability of the last person having his/her own sheet out of a student of n pupil.
P(1) = 1,
since any random pick will get the last and only student his/her sheet.
P(n) = Probability of a random pick in the begining and then followed by atleast 1 win and atmost (n-1) wins, since we cannot exceed the number of students/sheet in the class.
We want atleast 1 win above because we want the last one to surely get his/her sheet and we do not mind if anyone else also gets it.
So, going step by step for n = 2,3,.... we get
P(2) = 1/2 * P(1) = 1/2
A random pick followed by atleast 1 win and atmost 1 win...
P(3) = 1/3 * P(1) + 1/3 * P(2) = 1/3 * 1 + 1/3 * 1/2 = 1/2
A random pick followed by atleast 1 win and atmost 2 wins
=> (A random pick and 1 win) OR (A random pick and 2 wins)
Similarly continuing for n=4 we get,
= 1/4 * P(1) + 1/4 * P(2) + 1/4 * P(3)
= 1/4 * 1 + 1/4 * 1/2 + 1/4 * 1/2 = 1/2 again
P(n) = 1/n * P(1) + 1/n * P(2) + 1/n * P(3) + ..... + 1/n * P(n-1)
P(n) = 1/n + 1/n [ P(2) + P(3) + ..... + P(n-1)]
So for n = 100
P(100) = 1/100 + 1/100 [P(2) + P93) + ...... + P(98)]
P(100) = 1/100 + 1/100 [1/2 * 98]
P(100) = 1/2
No matter what is "n", the answer will always be 1/2
👍 100 | 👎 -13
Originally Answered: Probability question? help?
For the first one, there are 8 possible ways the coins could come down: HHH, HHT, HTH, THH, TTH, THT, HTT, TTT; since 3 of those are "Exactly one heads", we say the probability of seeing "Exactly one heads" is (3/8). Then we do the probability times the number of reps, to get 100,000*(3/8) = 37500.
If you throw 2 dice, there are 3 number of ways to get 4: 1,3; 3,1; 2,2; Out of 36 possible combinations. So the P(Sum=4) = 3/36 = 1/12. And if we're tossing it 1,000 times, E[Sum=4] = n*p, where n is the number of trials and p is the probability of the desired outcome. So in this case n*p = 1000*(1/12) = 83.33 times out of 1000.