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

  • (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)
P(\(A/E_2\))=P(selecting red ball/2\(^{nd}\)bag is chosen)
P(\(A/E_3\))=P(selecting red ball/3\(^{rd}\)bag is chosen)
=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)
P(\(B/E_2\))=P(selecting red ball/2\(^{nd}\)bag is chosen)
P(\(B/E_3\))=P(selecting red ball/3\(^{rd}\)bag is chosen)


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

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