document.write( "Question 1026005: An urn contains 3 red and 4 green balls. 4 balls are drawn, one after another subject to the following rule. If the ball drawn is red we keep is out of the urn; if its green we put it back. Find the probability that the 4th ball drawn is red. \n" ); document.write( "

Algebra.Com's Answer #641296 by Fombitz(32388) $\"\"$ $\"About$

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Look at the possible 16 outcomes and count the ones that have a red ball for 4th draw,
\n" ); document.write( "RRRR <--- Couldn't happen since only there is not a fourth red ball
\n" ); document.write( "RRRG
\n" ); document.write( "RRGR <--
\n" ); document.write( "RRGG
\n" ); document.write( "RGRR <--
\n" ); document.write( "RGRG
\n" ); document.write( "RGGR <--
\n" ); document.write( "RGGG
\n" ); document.write( "GRRR <--
\n" ); document.write( "GRRG
\n" ); document.write( "GRGR <--
\n" ); document.write( "GRGG
\n" ); document.write( "GGRR <--
\n" ); document.write( "GGRG
\n" ); document.write( "GGGR <--
\n" ); document.write( "GGGG
\n" ); document.write( "There are 7 possible outcomes with a fourth draw of red, so then the probability would be,
\n" ); document.write( " $\"P=7%2F16\"$
\n" ); document.write( " \n" ); document.write( "

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