document.write( "Question 59056: three distinct integers are chosen at random from the first 20 positive interger compute the pobabability that
\n" ); document.write( "a- their sum is even
\n" ); document.write( "b- their product is even \n" ); document.write( "

Algebra.Com's Answer #40502 by venugopalramana(3286) $\"\"$ $\"About$

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three distinct integers are chosen at random from the first 20 positive interger compute the pobabability that
\n" ); document.write( "a- their sum is even
\n" ); document.write( "b- their product is even
\n" ); document.write( "IN THE FIRST 20 +VE INTEGERS WE HAVE 10 EVEN AND 10 ODD INTEGERS
\n" ); document.write( "WE HAVE COMBINATION OF 3 FROM 20 IN 20C3 WAYS =20*19*18/(3*2*1)=1140 WAYS
\n" ); document.write( "A....SUM IS EVEN
\n" ); document.write( "IT CAN HAPPEN IN 2 WAYS
\n" ); document.write( "1..ALL 3 ARE EVEN ...POSSIBILITIES ARE 10C3=10*9*7/(3*2*1)=105 WAYS
\n" ); document.write( "2..2 ODD AND 1 EVEN ..POSSIBILITIES = 10C2*10C1=(10*9/2)(10)=450 WAYS
\n" ); document.write( "TOTAL POSSIBILITIES = 105+450=555 WAYS
\n" ); document.write( "PROBABILITY= 555/1140 = 111/228
\n" ); document.write( "B...PRODUCT IS EVEN
\n" ); document.write( "IT CAN HAPPEN IN 3 WAYS
\n" ); document.write( "1..ALL 3 EVEN...POSSIBILITIES = 10C3=105 WAYS
\n" ); document.write( "2..2 EVEN & 1 ODD...POSSIBILITIES = 10C2*10C1=450 WAYS
\n" ); document.write( "3..1 EVEN & 2 ODD....POSSIBILITIES = 10C1*10C2 = 450 WAYS
\n" ); document.write( "TOTAL POSSIBILITIES = 105+450+450 = 1005
\n" ); document.write( "PROBABILITY = 1005/1140 = 67/76 \n" ); document.write( "

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