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A bag contains 5 red marbles and 3 black marbles.Three marbles are drawn one by one with out replacement.What is the probability that at least one of the three marbles drawn be black,if the first marble is red?

1 Answer

Toolbox:
  • let\((E_1)\)be the event 1st marble is red
  • \((E_2)\)at least one is black
  • First marble is red can occur in following ways
  • \((E_1)\)={\(R\;B\;B)(R\;B\;R)(R\;R\;B)(R\;R\;R)\)}
  • At least one black
  • \((E_2)\)={\(B\;B\;B)(B\;R\;B)(B\;B\;R)(B\;B\;R)(R\;B\;B)(R\;B\;R)(R\;R\;B)\)}
  • \((E_1\;\cap\;E_2)\)={\(R\;B\;R)(R\;B\;B)(R\;R\;B)\)}
  •  
There are 5R and 3B marbles
P\((E_1)\)={P\((R\;B\;B)+P(R\;B\;R)+P(R\;R\;B)+P(R\;R\;R)\)}
=\(\Large\frac{5}{8}\;\times\;\frac{3}{7}\;\times\;\frac{2}{6}\;+\;\frac{5}{8}\;\times\;\frac{3}{7}\;\times\;\frac{4}{6}\;+\;\frac{5}{8}\;\times\;\frac{4}{7}\;\times\;\frac{3}{6}\;+\;\frac{5}{8}\;\times\;\frac{4}{7}\;\times\;\frac{3}{6}\)
We can also do
P(first ball is red)=\(\large\frac{5}{8}\)
P\((E_1)\)
P\((E_2)\)=P(at least 1 black)
=1-P(no black)
=1-p\((R\;R\:R)\)
=1-\(\large\frac{5}{8}\times\)\(\large\frac{4}{7}\times\)\(\large\frac{3}{6}\)
=\(\large\frac{23}{28}\)
=\(\Large\frac{5}{8}\;\times\;\frac{3}{7}\;\times\;\frac{4}{6}\;+\;\frac{5}{8}\;\times\;\frac{3}{7}\;\times\;\frac{2}{6}\;+\;\frac{5}{8}\;\times\;\frac{4}{7}\;\times\;\frac{3}{6}\)
=\(\large\frac{5}{8}\times\)\(\large\frac{5}{7}\)
=\(\Large\frac{25}{56}\)
Requier probability=\(\large\frac{25}{56}\times\)\(\large\frac{8}{5}\)
=\(\Large\frac{5}{7}\)

 

answered Feb 24, 2013 by poojasapani_1
edited Jun 5, 2013 by poojasapani_1
 

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