Doesn't (25/130)^4 assume replacement?
i have been having problems with what should be a trivial probability problem
Suppose there are 130 patients. Each patient was administered one of 6 drugs as follows:
Drug A: 25 patients
Drug B: 31 patients
Drug C: 13 patients
Drug D: 17 patients
Drug E: 28 patients
Drug F: 16 patients
What is the probability that in a random sample of 4 patients, all 4 selected were administered drug A?
My answer is (25/130)^4
The answer in the back is 0.0823 but something tells me it is wrong.
Doesn't (25/130)^4 assume replacement?
Dason (08-05-2012)
Well think of a bag with 2 marbles where one is white and one is black. If you draw 2 marbles without replacement then with probability 1 you'll have 1 white and 1 black. If you draw with replacement then there is probability .25 of getting 2 white, probability .5 of getting 1 white and 1 black, and probability .25 of getting 2 black.
So clearly it makes a difference which one you assume is occurring. But in a random sample of patients I would assume that it is without replacement.
I don't have emotions and sometimes that makes me very sad.
Well you could. But you could just use basic probability rules to figure this one out.
I don't have emotions and sometimes that makes me very sad.
Think about sequentially sampling the patients. What is the probability that the first patient you select was administered drug A? Now - given that the first patient was given drug A what is the probability that the second patient was administered drug A? So on and so forth.
I don't have emotions and sometimes that makes me very sad.
ok I think I may have to use a bunch of bayes forumula?
P(1st given drug A) * P(2nd given drug A | 1st given drug A) * P(3rd given drug A | (1st given drug A and 2nd given drug B))
can I assume independence?
You don't need Bayes formula for this and you don't need to worry about independence. What you have is definitely one way to do it though (if you continue it to the 4th person).
You should probably think about why you aren't actually using Bayes formula for this (what are you actually doing?) and why it doesn't matter if things are independent.
I don't have emotions and sometimes that makes me very sad.
derksheng (08-05-2012)
A: 1st person in sample gets drug A
B: 2nd person in sample gets drug A
and so on
P(A)P(B|A)
=P(A)P(A and B)/P(A)
=P(A and B)
???
ok so I am now completely lost. In my post ^ what would the next logical step be? I don't have P(A and B)
Well what you're actually looking for is P(Person1 had drugA AND Person2 had drugA AND Person3 had drugA AND Person4 had drugA) = P(Person1 had drugA)*P(Person2 had drugA | Person1 had drugA)*P(Person3 had .... | ...)*P(Person4 ... | ...) where you can fill in the ...s
I don't have emotions and sometimes that makes me very sad.
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