Колодий Н.А., Мазепа Е.А. Сборник задач и упражнений по теории вероятностей
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+∞
R
−∞
f
ξ
(x)dx = 1
f
ξ
(x) =
d
dx
F
ξ
(x) x
f
ξ
(x)
f : R 7→ R
+
+∞
R
−∞
f(x)dx = 1
p
k
(1−p)
2
k = 1, 2, . . .
0 < p < 1
1
n(n+1)
n = 1, 2, . . . p(1 − p)
n−1
n = 1, . . .
0 < p < 1
2
k
k!
k = 0, 1, . . .
p
k
= p
k
(1 − p)
2
0 < p
k
< 1
P
k
p
k
= 1
X
k
p
k
=
∞
X
k=1
p
k
(1 − p)
2
= (1 − p)
2
∞
X
k=1
p
k
=
p(1 − p)
2
1 − p
= p(1 − p) < 1.
p
k
=
1
k(k+1)
0 < p
k
< 1
X
k
p
k
=
∞
X
k=1
1
k(k + 1)
= lim
n→∞
n
X
k=1
1
k(k + 1)
=
= lim
n→∞
1 −
1
2
+
1
2
−
1
3
+ ··· +
1
n
−
1
n + 1
=
= lim
n→∞
1 −
1
n + 1
= 1.
F (x) = e
−e
−x
F (x) =
2
π
arctg x
F (x) =
0
x < −
π
2
,
cos x
−
π
2
6 x < 0,
1
x ≥ 0;
F (x) =
0
x < −
π
2
,
cos x
−
π
2
6 x <
π
3
,
1
x ≥
π
3
.
P
(ξ 6
π
4
)
P
(ξ ≥ −
π
3
)
P
(−
π
2
< ξ <
π
3
)
F
1
(x) F
2
(x)
c
1
c
2
F (x) = c
1
F
1
(x) + c
2
F
2
(x)
ξ
0
ξ
0
c
P
(ξ < ln 2)
P
(|ξ| < 1)
P
(ξ > −1)
f(x) = cx
2
e
−x
x ≥ 0; f(x) =
2c
e
x
+ e
−x
.
f(x) = −|x| + 1 |x| 6 1 0 |x| > 1
f(x) = x(1 − x) 0 < x 6 1 0
f(x) =
2x
(1+x
2
)
2
x > 0 0 x 6 0
ξ
λ =
1
3
P
{ξ > 3}
P
{ξ > t + 3 | ξ > t}
n
A = { } B = {
}
a
i
i
i = 1, n, a
i
= 1, 6 ξ
(a
1
, a
2
, . . . , a
n
)
ξ {0, 1, 2, . . . , n}
A
(a
1
, a
2
, . . . , a
n
)
ξ = 0, 2, 4, . . . , 2k k ∈ Z, 0 6 2k 6 n
P
(A) =
X
k∈Z, 062k6n
P
n
{ξ = 2k} =
X
k∈Z, 062k6n
C
2k
n
1
2
n−2k
1
2
2k
=
=
1
2
n
X
k∈Z, 062k6n
C
2k
n
=
1
2
n
2
n−1
=
1
2
.
P
(B) = 1 −
P
(A) =
1
2
.
n
a
1
+a
2
a
1
a
2
N
m N
a
1
ξ
f
ξ
(x) =
1
π
1
1+x
2
P
{ξ ≥ 1}
P
{|ξ| 6 1}
ξ a < c < d <
b
P
{ξ ∈ (c, d) } > 0
F (x) ξ
P
{ξ ∈ [c, b] | ξ ∈ (a, d)}
x
2
+y
2
= 1 A
ξ
ξ
ξ
M
ξ =
+∞
Z
−∞
xdF
ξ
(x)
F
ξ
ξ
c
M
(c) = c
ξ 6 η
M
ξ 6
M
η
M
|ξ| < ∞
c
M
(cξ) = c
M
ξ
M
|ξ| < ∞
M
|η| < ∞
M
(ξ + η) =
M
ξ +
M
η
|
M
ξ| 6
M
|ξ|
ξ > 0
M
ξ = 0
ξ = 0
ξ
X ξ g : R 7→ R
X X
P
x∈X
|g(x)|
P
{ξ = x} < ∞
M
g(ξ) =
X
x∈X
g(x)
P
{ξ = x}. (7.1)
ξ
f
ξ
g : R 7→ R
M
g(ξ) =
+∞
Z
−∞
g(x)f
ξ
(x)dx (7.2)
M
ξ
k
k ξ
M
(ξ −
M
ξ)
k
k
ξ
k |ξ|
k ξ
ξ
D
ξ
D
ξ =
M
(ξ −
M
ξ)
2
.
√
D
ξ
ξ σ
ξ
0 6
D
ξ =
M
ξ
2
− (
M
ξ)
2
6
M
ξ
2
.
c
D
(cξ) = c
2
D
ξ
D
(ξ) = 0
ξ =
c
D
(ξ + c) =
D
ξ
ν
k
=
M
ξ
k
µ
k
=
M
(ξ −
M
ξ)
k
µ
2
= ν
2
−ν
2
1
, µ
3
= ν
3
−3ν
1
ν
2
+2ν
3
1
, µ
4
= ν
4
−4ν
1
ν
3
+6ν
2
1
ν
2
−3ν
4
1
.
Mo ξ Me ξ
γ
1
γ
2
Me ξ
P
(ξ 6 Me ξ) >
1
2
P
(ξ > Me ξ) >
1
2
.
P
(ξ 6 Me ξ) =
P
(ξ > Me ξ) =
1
2
,
Me ξ
γ
1
=
M
(ξ−
M
ξ)
3
σ
3
γ
2
=
M
(ξ−
M
ξ)
4
σ
4
− 3 γ
2
= 0
M
ξ
D
ξ
ξ
ξ
ξ
λ > 0
p
k
=
P
(ξ = k) =
λ
k
k!
e
−λ
k = 0, 1, 2, . . . ξ
M
ξ =
∞
X
k=0
k
λ
k
k!
e
−λ
= e
−λ
∞
X
k=1
λ
k
(k − 1)!
= λe
−λ
∞
X
k=1
λ
k−1
(k − 1)!
=
= λe
−λ
e
λ
= λ.
M
ξ
2
=
∞
X
k=0
k
2
λ
k
k!
e
−λ
= e
−λ
∞
X
k=1
k
λ
k
(k − 1)!
=
= e
−λ
∞
X
k=2
λ
k
(k − 2)!
+ e
−λ
∞
X
k=1
λ
k
(k − 1)!
= λ
2
e
−λ
∞
X
k=2
λ
k−2
(k − 2)!
+ λ =
= λ
2
e
−λ
e
λ
+ λ = λ
2
+ λ.
D
ξ =
M
ξ
2
− (
M
ξ)
2
= λ
2
+ λ − λ
2
= λ.
ξ
n ∈ N 0 < p < 1 p
k
=
P
(ξ = k) =
C
k
n
p
k
q
n−k
k = 0, n q = 1 −p ξ
M
ξ =
n
X
k=0
kC
k
n
p
k
q
n−k
= p
∂
∂p
n
X
k=0
C
k
n
p
k
q
n−k
!
=
= p
∂
∂p
(p + q)
n
= pn(p + q)
n−1
q=1−p
= pn.
M
ξ
2
=
n
X
k=0
k
2
C
k
n
p
k
q
n−k
= p
∂
∂p
n
X
k=0
kC
k
n
p
k
q
n−k
!
=
= p
∂
∂p
pn(p + q)
n−1
= pn(p + q)
n−1
+ p
2
n(n − 1)(p + q)
n−2
q=1−p
=
= pn + p
2
n(n − 1).
D
ξ =
M
ξ
2
− (
M
ξ)
2
= pn + p
2
n(n − 1) − (np)
2
=
= pn − p
2
n = pn(1 − p) = pnq.
3/5 ξ n
ξ
M
ξ
D
ξ
k p
k
q
k
= 1−p
k
ξ n
M
ξ
D
ξ
ξ
ξ
M
ξ
D
ξ
N ξ
M
ξ
ξ
λ > 0
M
1
1+ξ
p ξ
ξ
M
ξ
D
ξ
ξ f
ξ
(x) = ce
−2|x|
c
M
ξ
n
c
c > 0
+∞
Z
−∞
f
ξ
(x) dx = c
+∞
Z
−∞
e
−2|x|
dx = 2c
+∞
Z
0
e
−2x
dx = −ce
−2x
+∞
0
= 1.
c = 1
M
ξ
n
=
+∞
Z
−∞
x
n
f
ξ
(x) dx =
+∞
Z
−∞
x
n
e
−2|x|
dx.
n = 2k + 1 k ∈ N
M
ξ
2k+1
= 0
n = 2k
M
ξ
n
= 2
+∞
Z
0
x
2k
e
−2x
dx =
+∞
Z
0
t
2
2k
e
−t
dt =
Γ(2k + 1)
2
2k
=
(2k)!
2
2k
.
Γ(α)
ξ f
ξ
(x) = cxe
−2x
, x > 0 c
M
ξ
n
ξ f
ξ
(x) = ce
−x
2
x > 0 c
M
ξ
ξ
f
ξ
(x) = a cos x
x ∈ [−
π
2
,
π
2
] a
P
(|ξ| <
π
4
)
M
ξ
ξ
M
ξ = 4
D
ξ = 3
ξ
x
k
=
1
2
k
p
k
=
1
2
k
k = 1, 2, 3, . . .
M
g(ξ)
g(ξ) = ξ
k
ξ
e
x
(1+e
x
)
2
, −∞ < x < ∞.
ξ
f
ξ
(x) =
1
2
e
−|x|