Дронов С.В. Задачник по теории вероятностей (первый семестр)
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A n
B k
A B n × k
A B
n
k (k ≤ n)
n(n − 1)...(n − k + 1) =
n!
(n−k)!
.
A
k
n
=
n!
(n − k)!
n k
k = n A
n
n
= n! P
n
n
k n
A
k
n
/P
n
n k
C
k
n
=
n!
(k!(n − k)!
.
• C
0
n
= C
n
n
= 1; C
1
n
= n.
• C
k
n
= C
n−k
n
.
•
P
n
k=0
C
k
n
= 2
n
;
P
n
k=0
(−1)
n
C
k
n
= 0.
• C
k
n
+ C
k−1
n
= C
k
n+1
.
6 × 7 = 42
21×10 = 210
A n
B k
A B n + k
(a + b)
n
(a + b)
n
= (a + b)(a + b)...(a + b) =
n
X
k=0
C
k
a
k
b
n−k
,
C
k
C
0
= C
n
= 1
a
n
b
n
a
j
b
n−j
j n
a (n − j) b
C
j
n
(a + b)
n
=
n
X
j=0
C
j
n
a
j
b
n−j
.
a = b = 1
a = −1, b = 1
k n
(k ≤ n + 1)
k
n
Ω
A ⊆ Ω A
P(A) =
|A|
|Ω|
,
|.|
Ω.
Ω
|Ω| = 4,
Ω = {111, 110, 101, 100, 011, 010, 001, 000},
|Ω| = 8
Ω
k n
k n
k
|Ω| = C
6
45
C
5
6
45 − 6 = 39
39C
5
6
39C
5
6
/C
6
45
≈ 0, 0000287.
n m
k
Ω ⊂ R
n
n
Ω A ⊆ Ω
P(A) =
µ(A)
µ(Ω)
,
µ(.) n
n n
0
00
1
00
(x, y) x y
Ω = {(x, y) : 0 ≤ x, y ≤ 60} ⊂ R
2
|x − y| ≤ 10, (x, y) ∈ Ω x −
10 ≤ y ≤ x + 10,
µ(Ω) = 3600, µ(A) = µ(Ω) − 2 × (50 × 50)/2 = 1100.
P(A) = 11/36.