My Math Forum  

Go Back   My Math Forum > High School Math Forum > Probability and Statistics

Probability and Statistics Basic Probability and Statistics Math Forum


Reply
 
LinkBack Thread Tools Display Modes
May 4th, 2017, 08:41 AM   #1
Newbie
 
Joined: May 2017
From: malaysia

Posts: 1
Thanks: 0

Help with question

In a list of 15 households, 9 own homes and 6 do not own homes. Four households are randomly selected from these 15 households. Find the probability that the number of households in these 4 who own homes is
a) Exactly 3
b) At most 1
c) Exactly 4
jiawei is offline  
 
May 4th, 2017, 10:47 AM   #2
Senior Member
 
romsek's Avatar
 
Joined: Sep 2015
From: Southern California, USA

Posts: 1,412
Thanks: 716

a) $\binom{15}{4}=1365$ ways of selecting 4

$\binom{9}{3}\binom{6}{1}=504$ ways of selecting exactly 3 homeowners

$p=\dfrac{504}{1365}=\dfrac{24}{65}$

b) at most 1 = 1 or 0

There are $\binom{6}{4}=15$ ways to select 0 homeowners and

$\binom{6}{3}\binom{9}{1} = 180$ ways to select 1 homeowner

$p=\dfrac{15+180}{1365} = \dfrac 1 7$

c) you can figure out this one
romsek is offline  
Reply

  My Math Forum > High School Math Forum > Probability and Statistics

Tags
probabilitydistribution, question



Thread Tools
Display Modes


Similar Threads
Thread Thread Starter Forum Replies Last Post
I have a question about sine and cosine transform, and the question also includes a H sCoRPion Differential Equations 0 March 16th, 2015 01:05 PM
Stats question: The Secretary Question word problem pudro Algebra 0 November 10th, 2008 08:12 PM





Copyright © 2017 My Math Forum. All rights reserved.