testing header
Math Goodies is a free math help portal for students, teachers, and parents.
Free Math
Interactive Math Goodies Software

Buy Math Goodies Software
testing left nav
Math Forums @ Math Goodies
Math Forums @ Math Goodies
Home | Profile | Active Topics | Members | Search | FAQ
Save Password
Forgot your Password?

 All Forums
 Homework Help Forums
 Probability and Statistics
 I need help with the second part please!!
 New Topic  Reply to Topic
 Printer Friendly
Author Previous Topic Topic Next Topic  

Junior Member

4 Posts

Posted - 10/26/2011 :  18:24:44  Show Profile  Reply with Quote
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)

= 14!/(14-2)!2! (0.10)(0.90)14-2^2
= 14*13*12!/12!2*1(0.10)(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%

But I'm not sure how to do part B.....please help!!
Go to Top of Page


3059 Posts

Posted - 10/27/2011 :  08:59:01  Show Profile  Reply with Quote
"At least three" is the complement of "No more than two".
Go to Top of Page
  Previous Topic Topic Next Topic  
 New Topic  Reply to Topic
 Printer Friendly
Jump To:
Math Forums @ Math Goodies © 2000-2004 Snitz Communications Go To Top Of Page
This page was generated in 0.03 seconds. Snitz Forums 2000
testing footer
About Us | Contact Us | Advertise with Us | Facebook | Blog | Recommend This Page

Copyright © 1998-2014 Mrs. Glosser's Math Goodies. All Rights Reserved.

A Hotchalk/Glam Partner Site - Last Modified 18 Dec 2014