# SOLUTION: three distinct integers are chosen at random from the first 20 positive interger compute the pobabability that a- their sum is even b- their product is even

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 Click here to see ALL problems on Probability-and-statistics Question 59056: three distinct integers are chosen at random from the first 20 positive interger compute the pobabability that a- their sum is even b- their product is evenAnswer by venugopalramana(3286)   (Show Source): You can put this solution on YOUR website!three distinct integers are chosen at random from the first 20 positive interger compute the pobabability that a- their sum is even b- their product is even IN THE FIRST 20 +VE INTEGERS WE HAVE 10 EVEN AND 10 ODD INTEGERS WE HAVE COMBINATION OF 3 FROM 20 IN 20C3 WAYS =20*19*18/(3*2*1)=1140 WAYS A....SUM IS EVEN IT CAN HAPPEN IN 2 WAYS 1..ALL 3 ARE EVEN ...POSSIBILITIES ARE 10C3=10*9*7/(3*2*1)=105 WAYS 2..2 ODD AND 1 EVEN ..POSSIBILITIES = 10C2*10C1=(10*9/2)(10)=450 WAYS TOTAL POSSIBILITIES = 105+450=555 WAYS PROBABILITY= 555/1140 = 111/228 B...PRODUCT IS EVEN IT CAN HAPPEN IN 3 WAYS 1..ALL 3 EVEN...POSSIBILITIES = 10C3=105 WAYS 2..2 EVEN & 1 ODD...POSSIBILITIES = 10C2*10C1=450 WAYS 3..1 EVEN & 2 ODD....POSSIBILITIES = 10C1*10C2 = 450 WAYS TOTAL POSSIBILITIES = 105+450+450 = 1005 PROBABILITY = 1005/1140 = 67/76