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# A bag contains 5 red and 3 blue balls.The probability that exactly two of the three balls were red,the first ball being red,is

\$\begin{array}{1 1}(A)\;\large\frac{1}{3}\$B)\;\large\frac{4}{7}\\(C)\;\large\frac{15}{28}\\(D)\;\large\frac{5}{28}\end{array}  Can you answer this question? ## 1 Answer 0 votes Options:To find p(2Balls red/first ball is red)We take sample space where the first ball is Red =s(\(R\;B\;R\;,R\;R\;R\;,R\;B\;B\;,R\;R\;B)$
$Frome\; the\; given\; set\; there\; are\; 2\; cases\; in\; which\; there\; are\; 2\; red\; balls\;so\;p(R\;B\;R\;\cup\;R\;R\;B)$
=$\Large\frac{3}{7}\;\times\;\frac{4}{6}\;+\;\frac{4}{7}\;\times\;\frac{3}{6}\;=\;\frac{24}{7}\;=\;\frac{4}{7}$
$'B'\; is\; correct$

edited Jun 4, 2013