Дронов С.В. Задачник по теории вероятностей (первый семестр)
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λ
m
e
−λ
/m!
m, λ
m, λ
ϕ(x) =
1
√
2π
e
−x
2
/2
x ϕ(x) x ϕ(x) x ϕ(x) x ϕ(x)
P
a ≤
k − np
√
npq
≤ b
!
≈ Φ(b) − Φ(a),
Φ(x) =
1
√
2π
x
Z
−∞
exp{−x
2
/2} dx −
Φ(x)
x
Φ(x)
x
Φ(x)
Φ(−∞) = 0, Φ(x) + Φ(−x) = 1, Φ
n = 2, k = 1, 2, p = 1/6, q = 5/6 P
n
(1) +
P
n
(2) = 11/36.
3
∆v
3
∆v
n =
2000
∆v
; p = ∆v × 10
−4
, k ≥ 1.
Q =
n
X
k=1
P
n
(k) = 1 − P
n
(0) = (1 − ∆v × 10
−4
)
2000/∆v
.
∆v = 1 Q = 1 −0, 9999
2000
∆v
P
n
(0) ≈
0, 2
0
0!
e
−0,2
≈ 0, 8187,
Q ≈ 0, 1813.
∆v n
k ξ
ξ > k.
P(ξ > k) ≤ 0, 01.
ξ
P(ξ > k) = P
ξ−np
√
npq
>
k−1000×1/2
√
1000×1/2×1/2
≈
≈ 1 − Φ
k−500
10
√
5
≤ 0, 01.
500 − k
10
√
5
≤ −1, 64 ⇒ k ≥ 523.
P
10000
(5200) ≈
1
√
2500
ϕ
5200 − 10000 × 1/2
√
2500
!
=
ϕ(4)
50
≈ 4 × 10
−6
,
P(k ≥ 5200) = P
k − np
√
npq
≥ 4
!
≈ 1 − Φ(4) ≈ 3.18 × 10
−5
,
k n
n ≥ k.
p
p
p
p
p(1−p) ≤ 1/4
< Ω , = , P >
ξ : Ω → R =
∀x ∈ R {ω ∈ Ω : ξ(ω) < x} ∈ =.
ξ
F (x) = P(ξ < x).
F (x)
• x < y ⇒ F (x) ≤ F (y);
• lim
x→−∞
F (x) = 0; lim
x→∞
F (x) = 1;
• (∀x
0
) lim
x→x
0
−0
F (x) = F (x
0
).
ξ
ξ x
1
x
2
··· x
n
p
1
p
2
··· p
n
p
j
= P(ξ = x
j
), j = 1, 2, ..., n;
P
p
j
= 1.
p x
F (x) =
x
Z
−∞
p(t)dt,
F ξ p(t)
p
∀t p(t) ≥ 0;
∞
R
−∞
p(t)dt = 1.
F p
x F
0
(x) = p(x).
ξ
P(ξ = k) = C
k
3
0, 4
k
0, 6
3−k
, k = 0, 1, 2, 3,
ξ
ξ
x ≤ 0 ⇒ F (x) = 0;
0 < x ≤ 1 ⇒ F (x) = P(ξ = 0) = 0, 216;
1 < x ≤ 2 ⇒ F (x) = P(ξ = 0) + P(ξ = 1) = 0, 648;
2 < x ≤ 3 ⇒ F (x) = P(ξ = 0) + P(ξ = 1) + P(ξ = 2) = 0, 936;
x > 3 ⇒ F (x) = 1.
ξ
p(x) =
x
2
, 0 ≤ x ≤ A,
A ξ
2
A
1 =
∞
Z
−∞
p(t)dt =
A
Z
0
x
2
dx =
A
3
3
⇒ A =
3
√
3.
ξ
2
x > 0.
P(ξ
2
< x) = P(|ξ| <
√
x) =
√
x
Z
−
√
x
p(t)dt =
√
x
Z
0
p(t)dt.
0 ≤
√
x ≤
3
√
3 ⇒ P(ξ
2
< x) =
√
x
R
0
t
2
dt =
1
3
x
3/2
;
√
x >
3
√
3 ⇒ P(ξ
2
< x) =
3
√
3
R
0
t
2
dt = 1.
G(x) = P(ξ
2
< x) =
x ≤ 0
1
3
x
3/2
, 0 < x <
3
√
9,
x ≥ 3
g(x) =
x ≤ 0 x >
3
√
9,
√
x
2
ξ
x ≤ 0 ⇒ p(t) = 0 t < x F (x) = 0;
0 < x ≤
3
√
3 ⇒ F (x) =
x
R
0
t
2
dt = x
3
/3;
x >
3
√
3 ⇒ F (x) =
3
√
3
R
0
p(t)dt = 1.
ξ
F (x) =
x < 2,
(x−2)
2
, 2 ≤ x < 3.
x ≥ 3.
1 ≤ ξ ≤ 2, 5
F (x) = A + B x.
A B
ξ
p(t) =
A
e
t
+ e
−t
.
A
ξ
ξ
ξ π/4 π/2 3π/4
sin ξ
α, β ≥ 0 F
1
, F
2
F (x) = αF
1
(x) + βF
2
(x)
ξ f(x)
η = 3ξ.
ξ
p(t) =
1
σ
√
2π
exp
(
−
t
2
2σ
2
)
;
η = 1/ξ
ξ
p(t) =
(
t 0 < t ≤
√
2
η = g(ξ)
g(x) =
0, 5 < x ≤ 1
x
2