Browse Questions

# 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 $(i)\;a \;red\;ball\;will\;be\;selected?\;(ii)\;a\;white\;ball\;is\;selected?$

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)==$\large\frac{1}{6}\;\times0+\;\frac{2}{6}\;\times\;\frac{2}{3}\;+\;\frac{3}{6}\;\times1$
=$\large\frac{9+2}{18}$=$\large\frac{11}{18}$

edited Jun 4, 2013