A. A. Sveshnikov - Problems in Probability Theory, Mathematical Statistics and Theory of Random Functions

ANSWERS AND SOLUTIONS I RANDOM EVENTS 1. RELATIONS AMONG RANDOM EVENTS By definition AA = A, AA = A. The event A is a particular case of B.B = A6, C = A5. (a) A certain event U, (b) an impossible event V. (a) At least one book is taken, (b) at least one volume from each of the three complete works is taken, (c) one book from the first work or three books from the second, or one from the first and three from the second, (d) two volumes from the first and second works are taken, (e) at least one volume from the third work and one volume from the first work and three from the second, or one from the second and three from the first. The selected number ends with 5. means that all items are good, means that one or none of them is defective.

Using the properties of events (BB = B, BB = B, B = U, BU = B, B = V, BV = B), we get A = BC. (a) A means reaching the interior of the region SA, means hitting the exterior of SA. Then AB = U; that is, A = V, B = U. (b) AB means reaching the region SAB common to SA and SB; means falling outside SA. Then AB = V; that is, A = U, B = V (c) AB means reaching the common region SAB; AB means hitting SAB; SAB = SAB only if SA = SB; that is, A = B. X = Use the equalities = B, = A. The equivalence is shown by passing to the complementary events.

The equalities are proved by passage from n to n + 1. No, since . Use the equality C means a tie. C = (A 1A2)(B1B2B1B3B2B3). 2. 4/9. p = 0.25 since the first card may belong to any suit. 1/65 0.00013. 23/240. 23/240.

The succession of draws under such conditions is immaterial and therefore p = 2/9. One may consider that for control the items are taken from the total lot; p = (nk)/(n + mk). One may consider one-digit numbers, (a) 0.2, (b) 0.4, (c) 0.04. (a) N = a + 10b. This condition is satisfied only if a is even and a + b is divisible by 9, p = 1/18, (b) N = a + 10b + 100c. This number should be divisible by 4 and by 9; that is, a + b + c is divisible by 9, a + 2b is divisible by 4(m = 22), p = 11/360. pk = (k = 1, 2, 3, 4, 5), Pl = 0.0556, p2 = 0.0025, p3 = 0.85104, p4 = 0.2 10 5 , p5 = 0.2 10 7. The favorable combinations: (a) (7, 7, 7); (b) (9, 9, 3), (9, 6, 6); (c) (2, 8, 11), (2, 9, 10), (3, 7, 11), (3, 8, 10), (4, 6, 11), (4, 7, 10), (4, 8, 9), (6, 7, 8) and, therefore, m = 4 + 24 + 43 8 = 564; p = 0.079. It is necessary to get nm nickels from 2n buyers. It is necessary to get nm nickels from 2n buyers.

The number of possible cases is , where N is the number of cases when it is impossible to sell 2n tickets, is the number of cases in which the first nickel came from the (2