If the distribution is continuous, the distribution density can be integrated (double integration) in the area range of p≥2root(q), 1≤p≤6, 1≤q≤6. But now it is not a continuous distribution, and the integral becomes a sum. Assuming that the dice are equal, the probability of generating various numbers is the same, and the combination of P and Q has 6x6=36 possibilities, and the probability of each occurrence is 1/36 (equivalent to distribution density), then when p≥2root(q), 1≤p≤6, 1 ≤ Q ≤ is calculated.
Q= 1: P is desirable, 2,3,4,5,6, then1/36+1/36+1/36+1/36+1.
Q=2: P desirable, 3, 4, 5, 6, 4/36.
Q=3: P is desirable, 4, 5, 6, and 3/36 is used.
Q=4: P is desirable, 4, 5, 6, and 3/36 is used.
Q=5: P desirable, 5, 6, 2/36.
Q=6: P desirable, 5, 6, 2/36.
(The above is equivalent to the first multiple integral)
Add it again (equivalent to the second double integral) as follows:
5/36 + 4/36 + 3/36 + 3/36 + 2/36 +2/36 = 19/36。 That is, the probability of real number solution is 19/36.
(2) Find the probability that the equation in (1) has two identical real number solutions.
At this time, it is required that b 2-4ac > 0. That is, p > 2 roots (q)
Compared with the above-mentioned solution with real numbers, the equation will only occur when q= 1 4, so two cases of p=2 when q = 1 and p=4 when q=4 are removed, that is, the probability is 17/36.