Answer:
0.6 or 60%.
Explanation:
First find the number of favorable outcomes (being trained on any station except the most desirable) and the total number of possible outcomes (being trained on any 4 stations).
Assuming all stations are equally likely to be selected for training, if there are 10 stations in total, the probability of not being trained on the most desirable station is as follows:
1) Favorable outcomes: Selecting 4 stations out of the 9 remaining stations (excluding the most desirable one): C(9, 4).
2) Total outcomes: Selecting any 4 stations out of the 10 available stations: C(10, 4).
P(not trained on most desirable) = Favorable outcomes / Total outcomes
= C(9, 4) / C(10, 4)
C(9, 4) = 9! / (4! * (9-4)!) = 9! / (4! * 5!) = (9 * 8 * 7 * 6) / (4 * 3 * 2 * 1) = 126
C(10, 4) = 10! / (4! * (10-4)!) = 10! / (4! * 6!) = (10 * 9 * 8 * 7) / (4 * 3 * 2 * 1) = 210
P(not trained on most desirable) = 126 / 210 = 6 / 10 = 0.6
Therefore, the probability of not being trained on the most desirable station is 0.6 or 60%.