document.write( "Question 1105631: A box contains with 15 red balls, 10 white balls, 20 blue balls and 5 yellow balls.
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document.write( "Two balls were selected one after one with replacement.
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document.write( "Calculate the probability for following incidents.
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document.write( "1) both the balls to be white
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document.write( "2) first ball to be red and second to be white
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document.write( "3) none of the balls to be yellow
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document.write( "4) either to be red or white or red and white
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document.write( "5) second ball NOT to be a blue ball
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document.write( "6) first ball to be a yellow
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document.write( "7) at least one ball to be a blue ball
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document.write( "8) more ever to be one red ball
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document.write( "9) first ball to be a white ball and second ball NOT to be a white ball
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document.write( "10) only one ball to be a red ball
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Algebra.Com's Answer #720501 by rothauserc(4718) You can put this solution on YOUR website! There are 15 + 10 + 20 + 5 = 50 balls in the box \n" ); document.write( ": \n" ); document.write( "The balls are selected with replacement \n" ); document.write( ": \n" ); document.write( "1) (10/50) * (10/50) = 1/25 \n" ); document.write( ": \n" ); document.write( "2) (15/50) * (10/50) = 3/50 \n" ); document.write( ": \n" ); document.write( "3) (45/50) * (45/50) = 81/100 \n" ); document.write( ": \n" ); document.write( "4) (15/50) + (10/50) + (3/50) = 28/50 = 14/25 \n" ); document.write( ": \n" ); document.write( "5) 1 - (20/50) = 3/5 \n" ); document.write( ": \n" ); document.write( "6) (5/50) = 1/10 \n" ); document.write( ": \n" ); document.write( "7) (20/50) + (20/50) = 4/5 \n" ); document.write( ": \n" ); document.write( "8) ? \n" ); document.write( ": \n" ); document.write( "9) (10/50) * (1 - (10/50)) = (1/5) * (4/5) = 4/25 \n" ); document.write( ": \n" ); document.write( "10) (15/50) * (1 - (15/50)) = (3/10) * (7/10) = 21/100 \n" ); document.write( " |