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  #1  
Old 07-30-2011
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Stats What's the probability of getting at least one positive result?

New test is 95% negative on patients who do not have the disease. If test is used in 8 healthy volunteers, what is the probability of getting at least one positive result?

a) 1- 0.05 X 8
b) 0.05 X 8
c) 0.05 ^ 8
d) 0.95 ^ 8
e) 1 - 0.95 ^ 8

^ means degree (like 2^3=8)
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Old 07-31-2011
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should be E
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nothing is impossible!
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Old 07-31-2011
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The answer is E) 1 - 0.95 ^ 8

Explanation:

Short version

P(+) = p = 0.05
P(-) = q = 0.95
n = 8

The probability of getting AT LEAST one positive = P(X>=1)
This is the same as:

1 - P(X=0) = 1 - 0.95^8

P(X=0) = always q^n

------------
Long version

This follows a Binomial distribution, where:

P(X=0) = nCr * p^r * (q)^(n-r)
P(X=0) = (n combinations r) * p^r * (q)^(n-r)
P(X=0) = 8C0 * 0.05^0 * (0.95)^(8-0)
P(X=0) = 1 * 0.95^8 = 0.95^8

Last edited by bebix; 07-31-2011 at 12:54 PM. Reason: sorry...should say "always q^n" and NOT p^n
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Old 07-31-2011
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Quote:
Originally Posted by bebix View Post
The answer is E) 1 - 0.95 ^ 8

Explanation:

Short version

P(+) = p = 0.05
P(-) = q = 0.95
n = 8

The probability of getting AT LEAST one positive = P(X>=1)
This is the same as:

1 - P(X=0) = 1 - 0.95^8

P(X=0) = always p^n

------------
Long version

This follows a Binomial distribution, where:

P(X=0) = nCr * p^r * (q)^(n-r)
P(X=0) = (n combinations r) * p^r * (q)^(n-r)
P(X=0) = 8C0 * 0.05^0 * (0.95)^(8-0)
P(X=0) = 1 * 0.95^8 = 0.95^8
wondering wats the diffrence b/w C. and E. ???? isnt E.1-.95 and C.0.05 same ???
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Old 07-31-2011
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Quote:
Originally Posted by pass7 View Post
wondering wats the diffrence b/w C. and E. ???? isnt E.1-.95 and C.0.05 same ???
@pass7

The probability of getting ALL positive results
c) 0.05 ^ 8 => P(X=8)

The probability of getting AT LEAST one positive result
e) 1 - 0.95 ^ 8 => P(X>=1)

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Old 07-31-2011
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E.) 1-all negative= at least one positive

please correct me if i am wrong.
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Old 07-31-2011
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Correct Answer

Quote:
Originally Posted by bebix View Post
The answer is E) 1 - 0.95 ^ 8

Explanation:

Short version

P(+) = p = 0.05
P(-) = q = 0.95
n = 8

The probability of getting AT LEAST one positive = P(X>=1)
This is the same as:

1 - P(X=0) = 1 - 0.95^8

P(X=0) = always p^n

------------
Long version

This follows a Binomial distribution, where:

P(X=0) = nCr * p^r * (q)^(n-r)
P(X=0) = (n combinations r) * p^r * (q)^(n-r)
P(X=0) = 8C0 * 0.05^0 * (0.95)^(8-0)
P(X=0) = 1 * 0.95^8 = 0.95^8


E is correct;
Heard the concept in lecture, but got wrong 'could not apply'.
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Old 07-31-2011
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Quote:
Originally Posted by jahn77 View Post
E is correct;
Heard the concept in lecture, but got wrong 'could not apply'.
Well, now you can!!!
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Old 07-31-2011
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Quote:
Originally Posted by bebix View Post
The answer is E) 1 - 0.95 ^ 8

Explanation:

Short version

P(+) = p = 0.05
P(-) = q = 0.95
n = 8

The probability of getting AT LEAST one positive = P(X>=1)
This is the same as:

1 - P(X=0) = 1 - 0.95^8

P(X=0) = always p^n

------------
Long version

This follows a Binomial distribution, where:

P(X=0) = nCr * p^r * (q)^(n-r)
P(X=0) = (n combinations r) * p^r * (q)^(n-r)
P(X=0) = 8C0 * 0.05^0 * (0.95)^(8-0)
P(X=0) = 1 * 0.95^8 = 0.95^8
1 - P(X=0) = 1 - 0.95^8

P(X=0) = always p^n

Could you please explain this bit in words? I am not able to follow this concept.
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Old 07-31-2011
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Quote:
Originally Posted by donofitaly View Post
1 - P(X=0) = 1 - 0.95^8

P(X=0) = always q^n

Could you please explain this bit in words? I am not able to follow this concept.
P(+) = p = 0.05
P(-) = q = 0.95
n = 8

P(X=0) = Probability of getting ALL negative results
P(X=0) = always q^n (this is a fact!!!...now, if you really really want to understand why this is always true, just read the explanation below about binomial distribution and probabilities!)

------------------------------
This follows a Binomial distribution, where:

P(X=0) = nCr * p^r * (q)^(n-r)
P(X=0) = (n combinations r) * p^r * (q)^(n-r)
P(X=0) = 8C0 * 0.05^0 * (0.95)^(8-0)
P(X=0) = 1 * 0.95^8 = 0.95^8 ==> q^n

------------------------------

Then, the probability of getting AT LEAST one positive is equal to:

P(X>=1)...and this is ALWAYS the same as 1 - P(X=0).

Finally, at least one will be equal to => 1 - q^n => 1 - (0.95)^8

For more about binomial distribution
http://en.wikipedia.org/wiki/Binomial_distribution
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