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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?

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}$
=$\Large\frac{5}{7}$

edited Jun 5, 2013