6/60 = a/10 = b/10 = c/20 = d/20, and it is easy to get a=b= 1 and c=d=2.
C(m, n) is used to indicate the number of n combinations extracted from m elements.
Extract all the possible numbers of two students = c (2,6) =15 (this step can be listed manually, and it is convenient to use the combination number here).
P (at least 1 person listened to three games) =P (exactly 1 person listened to three games) +P (two people listened to three games) = 8/15+1/5 = 9//kloc-0.
(2) It is easy to find the average value A = (1× 0+/kloc-0 /×1+2× 2+2× 3)/6 =1/6.
Variance s? =[(0- 1 1/6)? +( 1- 1 1/6)? +2×(2- 1 1/6)? +2×(3- 1 1/6)? ]/6=34/54≈0.63
(3) According to the meaning of the question * * *, there are 6 seats, and the number of seats in the table changes to 0, 2, 4 and 6, and the proportion remains unchanged.
Estimable:
Average A'=2A= 1 1/3.
Variance s? =4s？ =68/27≈2.52