Want to ask us a question? Click here

Browse Questions

Ad |

0 votes

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 \[(i)\;a \;red\;ball\;will\;be\;selected?\;(ii)\;a\;white\;ball\;is\;selected?\]

0 votes

- (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)==\(\large\frac{1}{6}\;\times0+\;\frac{2}{6}\;\times\;\frac{2}{3}\;+\;\frac{3}{6}\;\times1\)

=\(\large\frac{9+2}{18}\)=\(\large\frac{11}{18}\)

Ask Question

Tag:MathPhyChemBioOther

Take Test

x

JEE MAIN, CBSE, NEET Mobile and Tablet App

The ultimate mobile app to help you crack your examinations

...