This problem can be solved using probability, the equation of the probability of an event A is P(A)= favorable outcomes/possible outcomes. The interception of two probable events is P(A∩B)= P(A)P(B).
There are 12 black marbels, 10 red marbles, and 18 white marbels, all the same size. If two marbles are drawn from the jar without being replaced.
The total of the marbles is 40.
If two marbles are drawn from the jar without being replaced, what would the probability be:
1. of drawing two black marbles?
The probability of drawing one black marble is (12/40). Then, the probability of drawing another black marble after that is (11/39) due we drawing one marble before.
P(Black∩Black) = (12/40)(11/39) = 132/1560, simplifying the fraction:
P(Black∩Black) = 11/130
2. of drawing a white, then a black marble?
The probability of drawing one white marble is (18/40). Then, the probability of drawing then a black marble after that is (12/39) due we drawed one marble before.
P(White∩Black) = (18/40)(12/39) = 216/1560, simplifying the fraction:
P(White∩Black) = 9/65
3. of drawing two white marbles?
The probability of drawing one white marble is (18/40). Then, the probability of drawing another white marble after that is (17/39) due we drawed one marble before.
P(White∩White) = (18/40)(17/39) = 306/1560, simplifying the fraction:
P(White∩White) = 51/260
4. of drawing a black marble, then a red marble?
The probability of drawing one black marble is (12/40). Then, the probability of drawing then a red marble after that is (10/39) due we drawed one marble before.
P(Black∩Red) = (12/40)(10/39) = 120/1560, simplifying the fraction:
P(Black∩Red) = 1/13
, then angle 
is obtuse and triangle with sides a, b, c is obtuse triangle.