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Old 10-14-2003, 06:11 PM   #1 (permalink)
Upright
 
stats problem-- any help??

Damn, it's been a long time since stats class, and this problem is killing me:

Can anybody help?


In the US, 85% of people have Rh postive blood
So, 15% have Rh negative blood



If people are randomly tested, one by one, what is the chance that more than eight people will be tested before a Rh negative person is found?
hyperrust is offline  
Old 10-14-2003, 06:36 PM   #2 (permalink)
Riiiiight........
 
So you are given the probability of an event happening. And you are required to find out the probability that you need more than n=8 trials for the event to occur.

so to recap
p = 0.15
n=8

Check out a book on distributions. the distribution you want is a B------- distribution.

and remember that P(A) = 1-P(A")
where A is an event. and A" means the event does not occur.

let n be the number of trials before a Rh negative person is found.
so you want P(n>8)
P(n>8) = 1 - P(n<=7)

hmmm...think i've hinted too much... but there you go... shooo...
dimbulb is offline  
Old 10-14-2003, 06:38 PM   #3 (permalink)
Riiiiight........
 
And technically, this is a probability question, not a statistics question........
dimbulb is offline  
Old 10-14-2003, 06:43 PM   #4 (permalink)
Upright
 
thanks dude/dudette. my girlfriend actually presented me with this PROBABILITY problem (she's taking stats as an elective in grad school)

there's a reason i chose another type of school... thanks for the hint
hyperrust is offline  
Old 10-14-2003, 07:35 PM   #5 (permalink)
Psycho
 
Location: cali
let's just utilize percentages here and say the sample size is 100. so essentially, there are 15 who are Rh negative and 85 who are Rh positive.

you could then think of the problem as arranging them in a line with the probability of the first 9 (since greater than 8) is Rh positive.

so then there are 100! ways of arranging this sample in a line. so then you could also think of it as the possibility of selecting an Rh positive person for the first slot in line as being 85, then for the second slot, there would be 84 possible ways of selecting that, then 83, then 82...77.

so now you have:

(85 * 84 * 83 * 82 * 81 * 80 * 79 * 78 * 77) / 100!

but there are also 85 choose 9 ways of arranging those nine selected slots. so i think, and correct me if i'm wrong, it's something like:

[ (85 * 84 * 83 * 82 * 81 * 80 * 79 * 78 * 77) / 100! ] * ( 85 choose 9 )

DAMN! just had dinner, and dimbulb beat me to it. don't know if my reasoning is right, but his sounds good.
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slant eyes is offline  
Old 10-14-2003, 08:13 PM   #6 (permalink)
Riiiiight........
 
wait wait...i'm sleeping. You don't need the bloody binomial theorem..... hhhaha.....

what is P(n<=7)?
its the same as a failure in the first 7 trials....which is (1-p)^7
so you don't need any fancy smansy combinatorics here.... and there's also the problem of computing 100!......

for many problems of this type, the complement is your goooood friend...
dimbulb is offline  
 

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