From an urn containing 7 white balls, 5 black and 2 red balls, 5 balls are drawn at random

From an urn containing 7 white balls, 5 black and 2 red balls, 5 balls are drawn at random. Find the probability of a random event among the drawn black and red balls equally.

Except for the option when all the balls are white, then the following options are possible: out of 5 balls there will be 2 red, 2 black and one white or 1 red, 1 black and three white.
The total number of balls is 7 + 5 + 2 = 14.
The total number of ways to remove 5 balls out of 14:
n = C (14.5) = 14! / (5! (14 – 5)!) = 14! / (5! 9!) = 2002;
The number of ways to remove 2 red, 2 black and one white:
m1 = C (5.2) C (2.2) C (7.1) = (5! / ((2! 3!) 1 7 = 70;
The number of ways to remove 1 red, 1 black and 3 white:
m2 = C (5.1) C (2.1) C (7.3) = 5 2 (7! / (3! 4!) = 35;
The total number of ways to get an equal number of black and red balls:
m = m1 + m2 = 70 + 35 = 105;
The probability of taking out an equal number of black and red balls:
P = m / n = 105/2002 = 0.052.
Answer: 0.052.