Answer:
Step-by-step explanation:
There are first 20 positive integers 1,2...20. Three distinct integers are chosen at random.
Total no of ways of drawing 3 integers = [tex]20C3 = 1140[/tex]
To Compute the probability that: (a) their sum is even
a) Sum can be even if either all 3 are even or 1 is even and 2 are odd.
There are in total 10 odd and 10 even.
Ways of sum even = [tex]10C3 + 10C2 (10C1)\\= 120+45(10)\\= 570[/tex]
Prob = [tex]\frac{570}{1140} =0.50[/tex]
b) their product is even
Here any one should be even or both
SO no of ways are
=[tex]10C2 +10C2(10C1)\\= 45 +450\\=495[/tex]
Prob = [tex]\frac{495}{1140} =0.4342[/tex]
