Explanation:
when picking (ahem, buying) a soccer ball, there are 75 total possibilities.
the chance to pick a bad one is then
4/75 = 0.053333333...
now we are buying 3 balls out of the 75.
how many possible options do we have ?
picks cannot be repeated, and the sequence does not matter, so we need combinations :
C(75,3) = 75! / (3! × (75-3)!) = 75×74×73/(3×2) = 25×37×73 =
= 67,525 possible groups of 3 balls.
from all these possibilities, how many have 2 good and 1 bad ball ?
as many as options we have to pick 2 balls out of 71 and 1 out of 4 :
C(71,2) × C(4,1) = 71! / (2 × (71-2)!) × 4! / (1 × (4-1)!) =
= 71×70/2 × 4 = 71×70×2 = 9,940 possibilities
so, the total probability to buy 2 good and 1 bad ball is
9940/67525 = 1988/13505 = 0.147204739...