what is the detailed answer for this question?

Question

Students have been organized in two groups for Chinese language courses – 10 students in the first group and 9 in the second. 2 students in the first group and 7 in the second have prior experience with Chinese. However, before the first class, 3 students from the first group have been moved to the second. Only 2 students of the second group have come to the first class. What is the probability that all of them have prior experience with Chinese?

Answer 1

If all here means that both of the people from the second group speak Chinese, we add 3 from the first group, so there will be 12 in the second group, with from 7 to 9 who speak Chinese. We will determine each of these probabilities, and with each, the probability that both selected speak Chinese

Of the 3 moved to the second group, where 2 of 10 speak Chinese,

P(none who moves speaks Chinese) is C(8,3)/C(10.3) = 8*7*6/10*9*8 = 7/15

P(1 who moves speaks Chinese) is C(8,2)C(2,1)/C(10,3) = 8*7/2*2/((10*9*8)/(3*2*1) = 7/15

P(2 who move speak Chinese) is C(8,1)C(2,2)/C(9,3) = 8/((10*9*8)/6) = 1/15

Thus, 7/15 of the time, 7 of 12 in group 2 speak Chinese, 7/15 of the time, 8 speak Chinese, and 1/15 of the time, 9 speak Chinese. (by the way, a check for these probabilities is the expected value: 7/15 * 1 + 1/15 * 2 = 9/15 = .6. 3 selected * 2 Chinese of 10 = 3 * 2/10 = .6)

Then, if 7 of 12 speak Chinese, P(2 Chinese speakers are moved) is 7*6/12*11 = 7/22

If 8 of 12 speak Chinese, P(2 Chinese speakers are moved) is 8*7/12*11 = 14/33

If 9 of 12 speak Chinese, P(2 Chinese speakers are moved) is 9*8/12*11 = 6/11

Thus, P(2 Chinese speakers are moved) =

7/15 * 7/22 + 7/15 * 14/33 + 1/15 * 6/11 =

49/330 + 98/495 + 2/55 =

(49*3 + 98 * 2 + 2 * 18)/990 =

379/990

Then, the probability of