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#1
07-30-2011
 USMLE Forums Guru Steps History: Step 1 Only Posts: 326 Threads: 20 Thanked 220 Times in 124 Posts Reputation: 232
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)
 The above post was thanked by: bebix (07-31-2011), dr_lizard (07-31-2011), pass7 (07-31-2011), usluipek (07-31-2011)

#2
07-31-2011
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should be E
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#3
07-31-2011
 USMLE Forums Master Steps History: 1+CK+CS+3 Posts: 1,357 Threads: 194 Thanked 3,259 Times in 881 Posts Reputation: 3269

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
 The above post was thanked by: jahn77 (07-31-2011), pass7 (07-31-2011)
#4
07-31-2011
 USMLE Forums Guru Steps History: 1+CK+CS+3 Posts: 406 Threads: 30 Thanked 140 Times in 108 Posts Reputation: 150

Quote:
 Originally Posted by bebix 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 ???
#5
07-31-2011
 USMLE Forums Master Steps History: 1+CK+CS+3 Posts: 1,357 Threads: 194 Thanked 3,259 Times in 881 Posts Reputation: 3269

Quote:
 Originally Posted by pass7 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)

 The above post was thanked by: pass7 (07-31-2011)
#6
07-31-2011
 USMLE Forums Guru Steps History: 1+CK+CS Posts: 460 Threads: 278 Thanked 161 Times in 79 Posts Reputation: 171

E.) 1-all negative= at least one positive

please correct me if i am wrong.
#7
07-31-2011
 USMLE Forums Guru Steps History: Step 1 Only Posts: 326 Threads: 20 Thanked 220 Times in 124 Posts Reputation: 232

Quote:
 Originally Posted by bebix 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'.
#8
07-31-2011
 USMLE Forums Master Steps History: 1+CK+CS+3 Posts: 1,357 Threads: 194 Thanked 3,259 Times in 881 Posts Reputation: 3269

Quote:
 Originally Posted by jahn77 E is correct; Heard the concept in lecture, but got wrong 'could not apply'.
Well, now you can!!!
 The above post was thanked by: jahn77 (07-31-2011)
#9
07-31-2011
 USMLE Forums Guru Steps History: 1 + CS Posts: 405 Threads: 55 Thanked 523 Times in 155 Posts Reputation: 533

Quote:
 Originally Posted by bebix 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.
#10
07-31-2011
 USMLE Forums Master Steps History: 1+CK+CS+3 Posts: 1,357 Threads: 194 Thanked 3,259 Times in 881 Posts Reputation: 3269

Quote:
 Originally Posted by donofitaly 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

http://en.wikipedia.org/wiki/Binomial_distribution
 The above post was thanked by: donofitaly (08-01-2011)

 Tags Biostatistics-Epidemiology, Step-1-Questions

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