Question

I just got 71% on my first probability test :(
Here is one I missed. The answers are not released yet so
1) I cannot confirm if we have correctly solved it and.
2) Do no look at this thread if you are still talking the test.

To get a complete collection of stamps, Ann needs to have 5 specific (and different) kinds of stamps, denoted by \(k_1,…,k_5\). She starts out with no stamps. She then receives 5 stamps randomly from a lottery. Each of these 5 stamps has probability 1/5 of being of any particular kind \(k_i\), and the kinds of the different stamps she receives are independent.

Answer 1

Lol im doing probabilty to, and i SUCK :P

Answer 2

Sorry, i dont know this either, i really SUCK at probabilty, but when im done learning i can try my best to help

Answer 3

Questions:
a) What is the probability that Ann has a complete collection after receiving the 5 stamps from the lottery? (this one I got correctly as 24/625)

b) Given that Ann does not have a complete collection after receiving the 5 stamps from the lottery, what is the probability that she is missing exactly one stamp?

for this I tried 96/601 and (if i remember correctly) 24/601. neither were correct. Any thoughts?

Answer 4

I just wanted to give you a medal bye

Answer 5

yay, Professor Zarkon has arrived

Answer 6

Professor Zarkon?

Answer 7

Hola Professor!

Answer 8

Um.... No, T_T Im goin with professor.

Answer 9

At least 71% is a C in MY grade, If you in high-school, i think it might be either C- or D

Answer 10

it's from MIT, so I think they give A's at 80%+, and give some room for bonus points, but still...

Answer 11

i'm doing probability too though i suck horribly at it.

Answer 12

aaaaa; 1*1
aaaab, c, d, e; 4*5
aaabc, bd, be, ce, de; 5*10
aabcd, bce, cde; 3*10
abcde; 1*5

530 different outcomes of (1/5)^5 each maybe?

and 5 ways to pull out abcde?

im sure somethings off there

Answer 13

5^5 different positional ways ....

Answer 14

and there are what, 5! ways to order S5 ?

Answer 15

I may as well show my reasoning. for the first:
\[P(All~~Stamps)=1\cdot\frac45\cdot\frac35\cdot\frac25\cdot\frac15=24/625\]for the second\[P(All~~Stamps-1|Not~~All~~Stamps)\\={P(All~~Stamps-1\cap Not~~All~~Stamps)\over P(Not~~All~~Stamps)}\\={1\cdot\frac45\cdot\frac35\cdot\frac25\cdot\frac45\over1-P(All~~Stamps)}={{4*4!\over5^4}\over\frac{601}{625}}=...\frac{96}{601}~~~?\]wait a minute... was my only error a calculator error? did I get the right answer just now?

Answer 16

I may as well show my reasoning. for the first:
\[P(All~~Stamps)=1\cdot\frac45\cdot\frac35\cdot\frac25\cdot\frac15=24/625\]for the second\[P(All~~Stamps-1|Not~~All~~Stamps)\\={P(All~~Stamps-1\cap Not~~All~~Stamps)\over P(Not~~All~~Stamps)}\\={1\cdot\frac45\cdot\frac35\cdot\frac25\cdot\frac45\over1-P(All~~Stamps)}={{4*4!\over5^4}\over\frac{601}{625}}=\frac{96}{601}\]

Answer 17

pretty sure my problem is in the numerator...

Answer 18

3.84 seems right to me for the first: 120/3125

Answer 19

that looks correct

Answer 20

Brass Monkey

Answer 21

go into more detail on how you worked out your numerator

Answer 22

i thought 1/1 chance for the first being a new coupon, 4/5 the next is new, 3/5 the next is new, 2/5 the next is new, then the last must be one of the prior 4, so 4/5 chance for the last one to already be in the set.... I don't think that covers the intersection of the sets, though

Answer 23

think of it in terms of the multinomial distribution


suppose you got 2 of the first kind and 0 of the last kind then the prob is

\[\frac{5!}{2!1!1!1!0!}\left(\frac{1}{5}\right)^2\left(\frac{1}{5}\right)^1\left(\frac{1}{5}\right)^1\left(\frac{1}{5}\right)^1\left(\frac{1}{5}\right)^0\]

now count the number of ways you can move the 2 and 0

Answer 24

would that be\(\binom 5 2\) ways to move the 2 and 0? no they are distinct...
I really didn't think of using multinomials

Answer 25

no...the order of the 2 and 0 matters

Answer 26

yeah I see that, one sec, blanking out lol

Answer 27

the number of permutations of 2,1,1,1,0 are... ?
that's what you are asking, right?

Answer 28

yes

Answer 29

why can I not think of a formula for this... sorry let me wallow in my stupidity for a sec

Answer 30

how many choices do you have for the 2...then how many do you have for the 0

Answer 31

5 and 5 respectively, and 5 choose 3 for the remaining 1's ?

Answer 32

so we multiply those factors?

Answer 33

just move the 2 and 0...the 1's then only have 1 way to be placed

Answer 34

no 3! for the remaining 1's... I think

Answer 35

oh, so just 25?

Answer 36

;)...no...little smaller than that

Answer 37

oh darn 20

Answer 38

yes...so take 20 times the probability I wrote above

Answer 39

5 places to put the 0, 4 to put the 2

Answer 40

gotchya, never would have occurred to me to use multinomials. For the sake of completeness, do you see an outright error in my argument?

Answer 41

yes...there are a few things wrong...the obvious one is that they you wrote it the 2nd choice cant be a repeat of the 1st (you have 4/5)...but it is possible that if the first two choices you get the same thing

Answer 42

I hope you can translate that it didn't come out very well ;)

Answer 43

but given she only missed one, doesn't the ordering of that not matter?
if she gets stamp1 twice in a row, that's
1*1/5*4/5*3/5*2/5 ways, same thing, no?
or does the condition change that? or am I making no sense at all?

Answer 44

oh wait it's not the same lol

Answer 45

for the first one you have 5 out of 5 possibilities...for the second you still have 5 out of 5 possibilities for the 3...now it gets hard since we would have to do cases...did you already have a repeat or not....

Answer 46

I didn't mean ways, I meant chance above

Answer 47

Its a nerd convention lol

Answer 48

ok, I think I get the picture, I'll let y'all know theofficil solution when it is released

Answer 49

I have a feeling the solution they give will not use multinomials, but we shall see :)
Thanks, Doc

Answer 50

I get...after simplifying ...\[\frac{240}{601}\approx.39993\]

Answer 51

that seems like a high probability to me intuitively, that it's a 40% chance she's missing only one given she doesn't have them all, but those kinds of things are hard to estimate sometimes

Answer 52

when I first did it, I didn't use the multimomial. I just thought it would be easier to explain the solution using them

Answer 53

Thank you, your solution makes perfect sense. If you feel like typing out your other solution I'll be happy to see it, but if you don't feel like going to the trouble, no worries. Or if it uses anything past basic pmf introduction, which is where I am, I won't follow anyway.

Answer 54

I don't think it would make much sense. ;)

Answer 55

I trust that :D

Answer 56

if you have matlab...or one of its clones you could try running this

-------------------------
function p=stamp(n)

t=0;
w=0;
for k=1:n
x=ceil(5*rand(1,5));
y=sum(ismember(1:5,x));
if y<5
w=w+1;
if y==4
t=t+1;
end
end

end

p=t/w;
------------------

it will give the conditional probability

Answer 57

I like to check my answers by doing simulations

Answer 58

Nice, unfortunately my copy of matlab suddenly crashed permanently and no longer compiles anything. Thanks for the CS approach though :)

Answer 59

Here's an online Matlab that is open to everyone - http://www.compileonline.com/execute_matlab_online.php

There is also Octave, that is a free and open source version of Matlab - https://www.gnu.org/software/octave/
(doesn't have all the libraries of Matlab, but is good for home-grown code)

Answer 60

Nice work @zarkon!

Answer 61

Here is the solution as provided by MIT, for all those who want to see another solution. Variety is, after all, the spice of life.
Given that Ann does not start out with a complete collection, the number of unique sequences of Ann's stamps is \(5^5−5!\), each of which is still equally likely in the conditional universe. Thus, we need only count the number of sequences with exactly 1 missing stamp and then divide by \(5^5−5!\).

There are 5 possibilities for the missing stamp, 4 possibilities for the duplicate stamp, \(\binom52\) ways to place the 2 duplicate stamps in the sequence of 5 stamps, and \(3!\) ways to place the remaining 3 stamps in the remaining 3 positions in the sequence. Therefore, the probability that Ann is missing one stamp is
\[{5⋅4⋅\binom52⋅3!\over5^5−5!}={5!\binom52\over5^5−5!}=\frac{240}{601}≈0.39933.
\]

Answer 62

@Zarkon I like your explanation better