Case 1: probability 1 new 2 old p11:(9 * 3)/(12 *1*10) =
Case 2: 2 Probability of new 1 old P 12: (36 * 3 * 6)/(12 *1*10) =100.
Case 3: 3 The new probability p13: (9 * 8 * 7)/(12 *1*10) = 84/220;
Case 4: 3 Old probability p14: (3 * 3 * 2)/(12 *1*10) =1/220.
The second ball taking is based on the first ball taking, so there are four situations in which there are two new balls among the three balls:
Case 1: On the basis of P 1, there are 8 new balls and 4 old ones, so the probability of 2 new balls out of 3 balls is P2 1:
(28*4*6)/ ( 12* 1 1* 10)= 1 12/220;
Case 2: On the basis of P2, there are 7 new balls and 5 old ones, so the probability of taking out 2 new balls out of 3 balls is P22:
(2 1*5*6)/( 12* 1 1* 10)= 105/220;
Case 3: On the basis of P3, there are 6 new and 6 old, so the probability of taking out 3 balls is 2 new P23:
( 15*6*6)/( 12* 1 1* 10)=90/220;
Case 4: On the basis of P4, there are 9 new balls and 3 old balls, so the probability of 2 new balls among 3 new balls is P24:
(36*3*6)/( 12* 1 1* 10)= 108/220.
So:
(1) The probability of two new balls among the three balls taken out for the second time is p: p = p11* p 21+p/2 * p22+p13 * p23+. ...
(2) There are two new balls among the three balls taken out for the second time, and the probability of finding a new ball among the balls taken out for the first time is p':
P ' =(P 1 1 * P 12)/P = 3024/22032 = 0. 1372549 ...