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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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 even
Answer by venugopalramana(3286) About Me  (Show Source):
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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