I need help with the second part please!!

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gurly691

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[FONT=&quot] [/FONT]In testing a new drug, researchers found that 10% of all patients using it will have a mild side effect. A random sample of 14 patients using the drug is selected. Find the probability that:

(A) exactly two will have this mild side effect
(B) at least three will have this mild side effect.


N = 14 (14 patients)
X = 2 (exactly 2 will have this mild side effect)
P = 0.10 (prob. of all patients using it will have a mild side effect; success)
Q = 0.90 (1 - p = 1 - 0.10 = 0.9)

(A)

= 14!/(14-2)!2! (0.10)²(0.90)14-2^2
= 14*13*12!/12!2*1(0.10)²(0.90)12^2
= 14*13/2*1(0.01)(0.28)
= 182/2(0.01)(0.28)
= 91(0.01)(0.28) = 0.25 or 25%

Not sure how I do part B....please help!!
 
\(\displaystyle \frac{14!}{(12!)(2!)}\cdot 0.10^{2} \cdot 0.90^{12}\)

You seem to have an extra exponent in there.

p(at least 3) = 1 - p(0, 1, or 2)
 
Is this part right??

OK since p = 0.10 i would do:

0.10(3) + 0.10(4) + 0.10(5) + 0.10(6) + 0.10(7) + 0.10(8) + 0.10(9) + 0.10(10) + 0.10(11) + 0.10(12) + 0.10(13) + 0.10(14) =

0.3 + 0.4 + 0.5 + 0.6 + 0.7 + 0.8 + 0.9 + 1 + 1.1 + 1.2 + 1.3 + 1.4 = 10.2

So 10.2 would be my answer for part B?
 
No. After all the effort to calculate p(2), why would you think p(3) would be a simple multiplication?

No. Red flags should go off in your head. You are calculating a probability. It should be on [0,1]. You honestly believe 10.2 could be the right answer?

You had it at the beginning. Where did you go?
 
This took forever!! Tell me if this is correct please!!

Oh my goodness this took me forever!! But I'm sure I did part B correct now:

P (3) = 0.11284
P (4) = 0.035035
P (5) = 0.0078078
P (6) = 0.00129129
P (7) = 0.000164736
P (8) = 0.000015015
P (9) = 0.00000118118
P (10) = 0.000001585584
P (11) = 0.0000000159432
P (12) = 0.00000000014742
P (13) = 0.00000000000126
P (14) = 0
Add all = 0.15715662385588 or 16%
 
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