document.write( "Question 1062257: A box contains six red pens and three blue pens. (a) A pen is selected at random, the colour is noted and the pen is not returned to the box. This procedure is performed a second, then a third time.(a) Find the probability of obtaining (i) three red pens, (ii) two red pens and one blue pen, in any order, (iii) more than one blue pen. \n" ); document.write( "

Algebra.Com's Answer #677116 by Fombitz(32388) $\"\"$ $\"About$

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Three red pens,
\n" ); document.write( " $\"P=%286%2F9%29%285%2F8%29%284%2F7%29=5%2F21\"$
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\n" ); document.write( "Look at the possible outcomes,
\n" ); document.write( "RRB
\n" ); document.write( "RBR
\n" ); document.write( "BRR
\n" ); document.write( "There are three of them so,
\n" ); document.write( " $\"P=3%286%2F9%29%285%2F8%29%283%2F7%29=15%2F28\"$
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\n" ); document.write( "We'll do this one backwards.
\n" ); document.write( "Find the probability of no blue pens or only one blue pen,
\n" ); document.write( "No blue pens would be all red from above, $\"P=5%2F21\"$
\n" ); document.write( "One blue pen would be the second answer, $\"P=15%2F28\"$
\n" ); document.write( "So then no blue pens or one blue pen would be,
\n" ); document.write( " $\"P=5%2F21%2B15%2F28=20%2F84%2B45%2F84=65%2F84\"$
\n" ); document.write( "So then the complement of that probability is your answer,
\n" ); document.write( " $\"P=1-65%2F84=84%2F84-65%2F84=19%2F84\"$ \n" ); document.write( "

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