document.write( "Question 1118084: https://vle.mathswatch.co.uk/images/questions/question2752.png
\n" );
document.write( "please help me \n" );
document.write( "
Algebra.Com's Answer #733309 by math_helper(2461)![]() ![]() You can put this solution on YOUR website! Since there is only one blue one: \n" ); document.write( "P(1 blue and 1 red) = 1 - P(both red) = \n" ); document.write( "\n" ); document.write( "= \n" ); document.write( "= \n" ); document.write( "= \n" ); document.write( "\n" ); document.write( "So, to sanity check this: \n" ); document.write( "Say n=1: P(1 blue and 1 red) = 2/(1+1) = 1 (ok, we must get one of each) \n" ); document.write( "If n=2: P(1 blue and 1 red) = 2/(2+1) = 2/3 (ok, b/c 1/3 of the time you'd expect two reds) \n" ); document.write( "If n=100: P(1 blue and 1 red) = 2/(100+1) = 2/101 (seems ok, very unlikely to happen given all those reds, much more likely to get two reds) \r \n" ); document.write( "\n" ); document.write( "To do part(b), set the equation above equal to 0.125 (= 1/8) then solve that equation for n. \r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " |