real huge numbers

getback0 · 2011-03-23 04:32:34

I am having trouble matching my answer to the key provided -

Question:

Suppose each of 100 professors in a large mathematics department picks at random one of 200 courses. What is the probability that two professors pick the same course?

My Solution:

Sample space outcomes = 200 to the power 100

Assuming professor-1 and professor-2 of the 100 professors were to pick the same course of the 200 available, this could happen in

n(E1) = C(200, 1) * P(199, 98) .... (1)

Since there are C(100, 2) ways of picking professor-1 and professor-2 , we have -

n(E) = C(100, 2) * n(E1) .... (2)

Hence, probability is

P(E) = n(E)/n(S)

P(E) = [ C(200, 1) * P(199, 98) * C(100, 2) ] / (200 ^ 100 )

Microsoft Excel calculates this to be -

P(E) = 3.26715E-12 --> in other words, highly unlikely!

Key provided

Could you please help me determine the problem in my calculation???

bobbym · 2011-03-23 04:53:56

Hi getback0;

This is a birthday problem and is figured like this:

That is the probability that no two professors have picked the same course. So the probability that 2 or more have is the complement.

There is a really cool formula for this that I do not use:

getback0 · 2011-03-23 05:04:57

Hi bobbym! Hope this finds you well - THANKS AS always for getting back!

However, I disagree with the solution given the verbiage - if the probability in question was for two or more professors to pick the same course, the question is misleading since it says "two professors picking the same course"

Said another way, what would be your take if the question was "only 2 professors pick the same course"?

bobbym · 2011-03-23 05:12:57

Hi getback0;

I agree that the question is poorly worded. Since they gave the answer for 2 or more and they did not say only 2, I solved it in that manner.

Hope this finds you well.

I am old and grouchy as usual.

THANKS AS always for getting back!

At the beginning of each remaining day the men in the white coats wheel me in front of the monitor. If I do not answer questions they beat me and refuse to feed me. Good place! Only 500 a day and all the prescription drugs I can swallow.

Exactly 2 is a very difficult problem( that is why they always ask for 2 or more ) which has been apparently only solved by one guy. I am looking at his stuff right now, please hold on.

getback0 · 2011-03-23 06:59:09

not sure if it sounds like binomial to me (not disagreeing - just not sure) - for the following reasons -

experiment following binomial distribution should have -

1. n identical trials
2. each should have only 2 outcomes - success and failure
3. probability of success on each trial should stay the same
4. all trials should be independent of each other

IT COULD WELL BE JUST ME - as with these problems, it just seems one needs to have the 'eye' to fit a problem to an established distribution!

getback0 · 2011-03-23 07:04:14

sorry - i didnt realize your response changed - was responding to your older post -

hey AND I HAVE NEVER FOUND YOU grouchy! To me, you are an extremely kind-hearted person who takes all the effort to help out random people (who may not always be nice in return).

THANKS SIR!

bobbym · 2011-03-23 07:14:27

Hi getback0;

I think you are overestimating me, I just want to eat.

Anyway according to Finch, Not Peter Finch, he says to compute just 2 in a birthday type problem:

With n = 100 and d = 200:

http://books.google.com/books?id=aFDWuZ … ch&f=false

Math Is Fun Forum

#1 2011-03-23 04:32:34

real huge numbers

#2 2011-03-23 04:53:56

Re: real huge numbers

#3 2011-03-23 05:04:57

Re: real huge numbers

#4 2011-03-23 05:12:57

Re: real huge numbers

#5 2011-03-23 06:59:09

Re: real huge numbers

#6 2011-03-23 07:04:14

Re: real huge numbers

#7 2011-03-23 07:14:27

Re: real huge numbers

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