document.write( "Question 1083927: From a box containing 25 item 5 of which are defective, 4 are chosern at random ,let X be the number of defective items found.Obtain the probability distribution of X if
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document.write( "I. the items are chosen with replacement.
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document.write( "II. the items are chosen with out replacement. \n" );
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Algebra.Com's Answer #698010 by Boreal(15235)![]() ![]() You can put this solution on YOUR website! With replacement, the probability of \n" ); document.write( "0 defective is (4/5)^5=0.328 \n" ); document.write( "1 defective is 5(4/5)^4(1/5) \n" ); document.write( "2 defective is 10(4/5)^3(1/5)^2 \n" ); document.write( "3 defective is 10(4/5)^2(1/5)^3 \n" ); document.write( "4 defective is 5(4/5)(1/5)^4 \n" ); document.write( "5 defective is (1/5)^5 \n" ); document.write( "---------------------------- \n" ); document.write( "without replacement \n" ); document.write( "0 defective is (20/25)(19)24)(18/23)(17/22)(16/21) or 20C5/25C5=20!/15!5! divided by 25!/20!5!=0.2918 \n" ); document.write( "1 defective is 20C4*5C1/25C5 \n" ); document.write( "2 defective is 20C3*5C2/25C5 \n" ); document.write( "3 defective is 20C2*5C3/25C5 \n" ); document.write( "4 defective is 20C1*5C4/25C5 \n" ); document.write( "5 defective is 1/25C5=5/25*4/24*3/23*2/22*1/21=5C5/25C5=1/25!/5!20!=5!20!/25!=the first part.\r \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " |