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 #641308 by Edwin McCravy(20062)\"\" \"About 
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document.write( "The other tutor paid no attention to the rule:\r\n" );
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\n" ); document.write( "If the ball drawn is red we keep is out of the urn;
\n" ); document.write( "if it's green we put it back.
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document.write( "He wrongly assumed that all had the same probability.\r\n" );
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document.write( "There are 7 ways to draw out 4 with the 4th one R:\r\n" );
document.write( "RRGR,RGRR,RGGR,GRRR,GRGR,GGRR, and GGGR\r\n" );
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document.write( "1. RRGR\r\n" );
document.write( "We begin with 3 R's and 4 G's.\r\n" );
document.write( "The probability of drawing the 1st one R is 3/7.\r\n" );
document.write( "If we do, there remain 2 R's and 4 G's.\r\n" );
document.write( "The probability of drawing the 2nd one R is 2/6.\r\n" );
document.write( "If we do, there remain 1 R and 4 G's.\r\n" );
document.write( "The probability of drawing the 3rd one G is 4/5.\r\n" );
document.write( "If we do we put it back, and there remain 1 R & 4 G's.\r\n" );
document.write( "The probability of drawing the 4th one R is 1/5.\r\n" );
document.write( "P(RRGR) = (3/7)(2/6)(4/5)(1/5) = 4/175\r\n" );
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document.write( "2. RGRR\r\n" );
document.write( "We begin with 3 R's and 4 G's.\r\n" );
document.write( "The probability of drawing the 1st one R is 3/7.\r\n" );
document.write( "If we do, there remain 2 R's and 4 G's.\r\n" );
document.write( "The probability of drawing the 2nd one G is 4/6.\r\n" );
document.write( "If we do we put it back, and there remain 2 R's & 4 G's.\r\n" );
document.write( "The probability of drawing the 3rd one R is 2/6.\r\n" );
document.write( "If we do, there remain 1 R and 4 G's.\r\n" );
document.write( "The probability of drawing the 4th one R is 1/5.\r\n" );
document.write( "P(RRGR) = (3/7)(4/6)(2/6)(1/5) = 2/105\r\n" );
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document.write( "3. RGGR\r\n" );
document.write( "We begin with 3 R's and 4 G's.\r\n" );
document.write( "The probability of drawing the 1st one R is 3/7.\r\n" );
document.write( "If we do, there remain 2 R's and 4 G's.\r\n" );
document.write( "The probability of drawing the 2nd one G is 4/6.\r\n" );
document.write( "If we do we put it back, and there remain 2 R's & 4 G's.\r\n" );
document.write( "The probability of drawing the 3rd one G is 4/6.\r\n" );
document.write( "If we do we put it back, and there remain 2 R's & 4 G's.\r\n" );
document.write( "The probability of drawing the 4th one R is 2/6.\r\n" );
document.write( "P(RGGR) = (3/7)(4/6)(4/6)(2/6) = 4/63\r\n" );
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document.write( "4. GRRR \r\n" );
document.write( "We begin with 3 R's and 4 G's.\r\n" );
document.write( "The probability of drawing the 1st one G is 4/7.\r\n" );
document.write( "If we do we put it back, and there remain 3 R's & 4 G's.\r\n" );
document.write( "The probability of drawing the 2nd one R is 3/7.\r\n" );
document.write( "If we do, there remain 2 R's and 4 G's.\r\n" );
document.write( "The probability of drawing the 3rd one R is 2/6.\r\n" );
document.write( "If we do, there remain 1 R and 4 G's.\r\n" );
document.write( "The probability of drawing the 4th one R is 1/5.\r\n" );
document.write( "P(RGGR) = (4/7)(3/7)(2/6)(1/5) = 4/245\r\n" );
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document.write( "5. GRGR\r\n" );
document.write( "We begin with 3 R's and 4 G's.\r\n" );
document.write( "The probability of drawing the 1st one G is 4/7.\r\n" );
document.write( "If we do we put it back, and there remain 3 R's & 4 G's.\r\n" );
document.write( "The probability of drawing the 2nd one R is 3/7.\r\n" );
document.write( "If we do, there remain 2 R's and 4 G's.\r\n" );
document.write( "The probability of drawing the 3rd one G is 4/6.\r\n" );
document.write( "If we do we put it back, and there remain 2 R's & 4 G's.\r\n" );
document.write( "The probability of drawing the 4th one R is 2/6.\r\n" );
document.write( "P(GRGR) = (4/7)(3/7)(4/6)(2/6) = 8/147.\r\n" );
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document.write( "6. GGRR\r\n" );
document.write( "We begin with 3 R's and 4 G's.\r\n" );
document.write( "The probability of drawing the 1st one G is 4/7.\r\n" );
document.write( "If we do we put it back, and there remain 3 R's & 4 G's.\r\n" );
document.write( "The probability of drawing the 2nd one G is 4/7.\r\n" );
document.write( "If we do we put it back, and there remain 3 R's & 4 G's.\r\n" );
document.write( "The probability of drawing the 3rd one R is 3/7.\r\n" );
document.write( "If we do, there remain 2 R's and 4 G's.\r\n" );
document.write( "The probability of drawing the 4th one R is 2/6.\r\n" );
document.write( "P(GRGR) = (4/7)(4/7)(3/7)(2/6) = 16/343.\r\n" );
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document.write( "7. GGGR\r\n" );
document.write( "We begin with 3 R's and 4 G's.\r\n" );
document.write( "The probability of drawing the 1st one G is 4/7.\r\n" );
document.write( "If we do we put it back, and there remain 3 R's & 4 G's.\r\n" );
document.write( "The probability of drawing the 2nd one G is 4/7.\r\n" );
document.write( "If we do we put it back, and there remain 3 R's & 4 G's.\r\n" );
document.write( "The probability of drawing the 3rd one G is 4/7.\r\n" );
document.write( "If we do we put it back, and there remain 3 R's & 4 G's.\r\n" );
document.write( "The probability of drawing the 4th one R is 3/7.\r\n" );
document.write( "P(GRGR) = (4/7)(4/7)(4/7)(3/7) = 192/2401.\r\n" );
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document.write( "4/175+2/105+4/63+4/245+8/147+16/343+192/2401 = 163558/540225\r\n" );
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document.write( "A little over 30% of the time. \r\n" );
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document.write( "Edwin

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