logo

Ask Questions, Get Answers

 
X
 Search
Want to ask us a question? Click here
Browse Questions
Ad
Home  >>  CBSE XII  >>  Math  >>  Probability
0 votes

Three bags contain a number of red and white balls as follows:

Bag 1:3 red balls,Bag 2:2 red balls and 1 white ball. Bag 3:3 white balls. The probability that bag i will be chosen and a ball is selected from it is $\Large \frac{i}{6}$, i=1,2,3.What is the probability that \[(i)\;a \;red\;ball\;will\;be\;selected?\;(ii)\;a\;white\;ball\;is\;selected?\]
Can you answer this question?
 
 

1 Answer

0 votes
Toolbox:
  • (1) \(E_1\)=1\(^{st}\)bag is chosen
  • \(E_2\)=2\(^{nd}\)bag is chosen
  • \(E_3\)=3\(^{rd}\)bag is chosen
  • A-Red ball selected
  • B-White ball selected
  • Probability i\(^{th}\) bag chosen=\(\large\frac{2}{6}\) i=\(1\;2\;3\)
  • p(\(E_1\))=\(\large\frac{1}{6}\) p(\(E_2\))=\(\frac{2}{6}\) p(\(E_3\))=\(\frac{3}{6}\)
  • P(A)=P(\(E_1\))P(\(A/E_1\))+P(\(E_2\))P(\(A/E_2\))+P(\(E_3\))P(\(A/E_3\))
  • P(B)=P(\(E_1\))P(\(B/E_1\))+P(\(E_2\))P(\(B/E_2\))+P(\(E_3\))P(\(B/E_3\))
P(\(A/E_1\))=P(selecting red ball/1\(^{st}\)bag is chosen)
=\(\large\frac{3}{3}\)=1
P(\(A/E_2\))=P(selecting red ball/2\(^{nd}\)bag is chosen)
=\(\large\frac{2}{3}\)
P(\(A/E_3\))=P(selecting red ball/3\(^{rd}\)bag is chosen)
=\(\frac{0}{3}\)
=P(A)=\(\large\frac{1}{6}\;\times1+\;\frac{2}{6}\;\times\;\frac{2}{3}\; +\;\frac{3}{6}\;\times0\)
P(\(B/E_1\))=P(selecting red ball/1\(^{st}\)bag is chosen)
=\(\large\frac{0}{6}\)
P(\(B/E_2\))=P(selecting red ball/2\(^{nd}\)bag is chosen)
=\(\large\frac{1}{3}\)
P(\(B/E_3\))=P(selecting red ball/3\(^{rd}\)bag is chosen)
=\(\large\frac{3}{3}\)=1
p(B)==\(\large\frac{1}{6}\;\times0+\;\frac{2}{6}\;\times\;\frac{2}{3}\;+\;\frac{3}{6}\;\times1\)
=\(\large\frac{9+2}{18}\)=\(\large\frac{11}{18}\)

 

answered Mar 1, 2013 by poojasapani_1
edited Jun 4, 2013 by poojasapani_1
 

Related questions

Ask Question
student study plans
x
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App
...