document.write( "Question 473930: Help!\r
\n" ); document.write( "\n" ); document.write( "A shipment of 11 televisions contains 4 regular and 7 deluxe models. The manufacturer failed to mark the model designation on the cartons. If 3 cartons are selected at random, what is the probability that exactly 2 of them are the deluxe model? (to three decimal places.)
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Algebra.Com's Answer #325062 by Maths68(1474)\"\" \"About 
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Solution:-\r
\n" ); document.write( "\n" ); document.write( "Regular 1
\n" ); document.write( "Deluxe 2
\n" ); document.write( "Total 3 \r
\n" ); document.write( "\n" ); document.write( "Now
\n" ); document.write( "Using nCr=n! /(n-r)! r!\r
\n" ); document.write( "\n" ); document.write( "3 things can be chosen out of 11 = 11C3 ways.
\n" ); document.write( " = 11!/(11-3)!3!
\n" ); document.write( " = 11!/ 8!3!
\n" ); document.write( " =(11*10*9*8!) / (8! * 3*2*1)
\n" ); document.write( " = (11*10*9) /(3*2*1)
\n" ); document.write( " = 990 /6
\n" ); document.write( " = 165 ways
\n" ); document.write( "1 Regular can be chosen out of 4 = 4C1
\n" ); document.write( " = 4! /(4-1)!1!
\n" ); document.write( " =4*3!/3!1!
\n" ); document.write( " = 4 ways
\n" ); document.write( "2 Delux can be chosen out of 7 = 7C2
\n" ); document.write( " =7!/(7-2)!2!
\n" ); document.write( " =7!/5!2!
\n" ); document.write( " =7*6*5!/5!2!
\n" ); document.write( " =7*6/2
\n" ); document.write( " =42/2
\n" ); document.write( " =21 ways\r
\n" ); document.write( "\n" ); document.write( "P(exactly 2 are deluxe model) = (7C2 * 4C1) / 11C3
\n" ); document.write( " = (21*4) /165
\n" ); document.write( " =84/165
\n" ); document.write( " =0.509
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