document.write( "Question 1118089: https://vle.mathswatch.co.uk/images/questions/question16415.png
\n" ); document.write( "please help \n" ); document.write( "
Algebra.Com's Answer #733321 by math_helper(2461)\"\" \"About 
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Let's look at the four cases:\r
\n" ); document.write( "\n" ); document.write( "Let P(R,R) = probability of selecting red then another red\r
\n" ); document.write( "\n" ); document.write( "What value is P(R,R)?\r
\n" ); document.write( "\n" ); document.write( "On the first pick, there are 7 red and 11 total, so P(R) = 7/11
\n" ); document.write( "On the 2nd pick, there are 6 red and 10 total, so P(R,R) = (7/11)*(6/10) = 42/110
\n" ); document.write( "1. P(R,R) = 42/110
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\n" ); document.write( "\n" ); document.write( "Similarly, using W for white:
\n" ); document.write( "2. P(R,W) = (7/11)*(4/10) = 28/110
\n" ); document.write( "3. P(W,R) = (4/11)*(7/10) = 28/110
\n" ); document.write( "4. P(W,W) = (4/11)*(3/10) = 12/110
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\n" ); document.write( "\n" ); document.write( "Notice these cover all possible outcomes, as expected. We know this because the numerators, when added, sum to 110 and 110/110=1
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\n" ); document.write( "\n" ); document.write( "The probability of one of each color is just the sum of case 2, P(R,W), and case 3, P(W,R):\r
\n" ); document.write( "\n" ); document.write( " 28/110 + 28/110 = \"+highlight%28+56%2F110+%29+\" or about 0.51 \r
\n" ); document.write( "\n" ); document.write( "I hope this helps you.\r
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