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

## 1 Answer

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
edited Jun 4, 2013

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