Prime Suspects
Probability puzzles require you to weigh all the possibilities and pick the most likely outcome.
Two hundred people in an auditorium are asked to think of a single digit number from 1 to 9 inclusive and write it down. All those who wrote down a prime number are now asked to leave. Ninety people remain behind in the hall. How many of these are expected to have written down an odd number?
HintRemember that 1 is not a prime number.
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Answer
Those that remain behind must have written {1,4,6,8,9} and from this only {1,9} are odd. The probability of an odd number is thus 2/5.
Expected number of odds is 2/5 * 90 = 36.
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Comments
smiley_lover
Mar 17, 2006
 they didn't all write a number grr... or is my math wrong 
HibsMax
Mar 17, 2006
 DOH, helps if you don't count 1 as a prime number! I came up with 1/4 * 90 instead of 2/5 * 90. 
cms271828
Mar 17, 2006
 Nice and easy.
You shouldn't really count 1 as a prime number because its not.
Heres why...
The integers form a UFD, in a UFD primes and irreducibles are the same. Since units aren't irreducibles, and 1 is a unit, then 1 can't be prime. 
norcekri
Mar 18, 2006
 This assumes that each of the 9 numbers is equally likely to be chosen. Experimental evidence shows this to be false. On what did you base your assertion that it's true? 
cms271828
Mar 18, 2006
 Assuming equal likeliness for digits 19 is the obvious thing to do, in practice you might find number 7 is chosen more because some ppl like the number 7.
But for sake of simplicity we can assume equal likeliness.
When tossing a coin, its not equally likely to get a head or tail, but it practically is, so we just call it 1/2. 
lelrod
Mar 18, 2006
 The fact that only 90 people remain indicates that there is not equal probability of choosing any number. If there wereequal probability, 111 would remain. Knowing that, why would the answer be based on equal probability? 
Jimbo
Mar 19, 2006
 What experimental evidence Norcekri? How many trials? Is it culturally biased? Do you agree with Lelrod's assertion that there must be 111 if all selections are equally likely? 
Vigo95
Mar 22, 2006
 you guys are great and smart ! since the language questions are my forte , my answer to this one was ... who cares ! 
nenad
Mar 30, 2006
 I also think that solution is not clear  it is based on 'equal probability' for all numbers, and yet out of 200 people only 90 remained, which is NOT expected 5/9=111 (ie primes 2,3,5,7 out, 5 of 9 remains in). It is much closer to 4/9=88, indicating that one more 'prime' was used, and only candidate is #1. That would result in 4,6,8,9 remaining, and 1/4*90= only 22 would have odd(#9) 
(user deleted)
Apr 06, 2006
 Hey,
The problem is quite simple. Lemme explain...
The numbers which the remaining 90 people must have written are 1,4,6,8,9
now... the to odds here are 1 and 9... Probably of selecting them is 1/5 + 1/5 = 2/5
So the number of guys who might have written 1 or 9 is 2/5*90 = 36... Cool ! 
Dedrik
Aug 09, 2006
 Well of course it didn't out to be exactly 111 that remained, that would be unlikely compared to something else happening. But I shouldn't think that would give you a better clue as to the distribution of the numbers left (1,4,6,8,9) so why not assume eaqually likely, seems better than guessing a random distribution. 
masquerademe235
Aug 28, 2007
 Great teaser!! 
SpanshDST
Feb 09, 2008
 Awsome teaser 
javaguru
Dec 11, 2008
 Easy one.
To cms: Maybe a disproportionate number of people left because they liked 7 and it's prime. Nevertheless, they're gone now and no longer influencing the outcome. 
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