# Probability of lies a point in a random triangle

We have a square. we will opt three random point from inside of this square and name it $p_{1},p_{2},p_{3}$ then opt another random point $p_{4}$. what is the probability of that $p_{4}$ lies in triangle $p_{1}p_{2}p_{3}$ ?

I write program and test it 10 times and in each case test it for $10^{7}$ points and the results are:

TEST CASE # 1 = 0.076566
TEST CASE # 2 = 0.076430
TEST CASE # 3 = 0.076308
TEST CASE # 4 = 0.076378
TEST CASE # 5 = 0.076433
TEST CASE # 6 = 0.076340
TEST CASE # 7 = 0.076289
TEST CASE # 8 = 0.076382
TEST CASE # 9 = 0.076402
TEST CASE #10 = 0.076260

• I believe this is sufficient to answer your question. people.missouristate.edu/lesreid/Adv41.html Once you have the expected area, the probability $p_{4}$ lies in your triangle should follow. – JessicaK Jun 1 '14 at 9:23
• @JessicaK. This looks like magics to me ! Thanks for the link. – Claude Leibovici Jun 1 '14 at 9:27
• @JessicaK: Link not available, are there some regional restrictions on that site? Other link to and partial answer at rqna.net/qna/… – Lutz Lehmann Jun 1 '14 at 11:57
• Link's not working for me either. Looks like a related question was asked before, which has that link. – gar Jun 1 '14 at 12:04
• Are you looking for an exact answer, in a formula, or just an approximation? What did the original question ask for, or was it just curiosity? – Henno Brandsma Jun 1 '14 at 13:01

$$\mathbb{P} = \frac{11}{144} \approx 0.0763889$$