SOLUTION: Dear Math Expert , can you help me with this problem :) Three distinct integers are chosen at random from the first 20 positive integers compute the probability that a- their su

Algebra ->  Probability-and-statistics -> SOLUTION: Dear Math Expert , can you help me with this problem :) Three distinct integers are chosen at random from the first 20 positive integers compute the probability that a- their su      Log On


   

Question 1070161: Dear Math Expert , can you help me with this problem :)
Three distinct integers are chosen at random from the first 20 positive integers compute the probability that
a- their sum is even
b- their product is even
Found 2 solutions by Fombitz, stanbon:
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
1 3 5 7 9 11 13 15 17 19 - 10 odds
2 4 6 8 10 12 14 16 18 20 - 10 evens
Look at the possible choices for the 8 outcomes,
EEE-Even
EEO-Odd
EOE-Odd
EOO-Even
OEE-Odd
OEO-Even
OOE-Even
OOO-Odd
So 4/8 are even sums, 4/8 are odd sums
.
.
.
Now products,
EEE-Even
EEO-Even
EOE-Even
EOO-Even
OEE-Even
OEO-Even
OOE-Even
OOO-Odd
So 7/8 are even product, 1/8 are odd products

Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
Three distinct integers are chosen at random from the first 20 positive integers compute the probability that
a- their sum is even:::
all three are even:: 10C3 = 120
2 are odd and one is even:: 10C2*10 = 450
P(sum is even) = (120+450)/20C3 = 370/1140 = 37/114
-----------------------------------------------
b- their product is even
all three are odd:: 10C3 = 120
----
P(product is even) = (20C3-120)/20C3 = 1020/1140 = 51/57 = 17/19
------------
Cheers,
Stan H.
------------------