Hướng dẫn giải đề thi Xác suất thống kê docx

P(AB) = 0
P(AB) > 0 P(A) > 0 P(B) > 0
x, y ∈ [0; 1] |x − y|  1 0 
P(A|B), P(A)  1
|P(A) − P(A|B)|  1.
P(A|B) =
P(AB)
P(B)



P(A) −
P(AB)
P(B)



 1.
|P(A)P(B) − P(AB)|  P(B).
|P(A)P(B) − P(AB)|  P(A).
|P(A)P(B) − P(AB)|  min(P(A), P(B)).
a =
2
π
F (π/12) =
π/12

−∞
f(x)dx =
−π/2


−∞
f(x)dx +
π/12

−π/2
f(x)dx
=
π/12

−π/2
f(x)dx = a
π/12

−π/2
cos
2
xdx
=
2
π

7π
24
+
1
8

=
7
12

+
1
4π
.
E(X) =
+∞

−∞
xf(x)dx = a
π/2

−π/2
x cos
2
xdx =
2
π
π/2

−π/2
x cos
2
xdx.
π/2

−π/2
x cos
2
xdx = 0 E(X) = 0
f(x, y) = ke

−
1
2
(2x
2
−4x+2+y
2
)
= ke
−(x−1)
2
e
−
y
2
2
.
f(t) =
1
σ
√
2π
e
−
(t−µ)
2
2σ
2
N(µ, σ
2

)
+∞

−∞
f(t)dt = 1
+∞

−∞
e
−
(t−µ)
2
2σ
2
dt = σ
√
2π
+∞

−∞
e
−(x−1)
2
dx =
√
π
+∞

−∞
e

−
y
2
2
dy =
√
2π
+∞

−∞
+∞

−∞
f(x, y)dxdy = 1
1 =
+∞

−∞
+∞

−∞
f(x, y)dxdy = k
+∞

−∞
+∞

−∞
e
−(x−1)

2
e
−
y
2
2
dxdy
= k
+∞

−∞
e
−(x−1)
2
dx
+∞

−∞
e
−
y
2
2
dy
= k
√
π ×
√
2π = k
√

2π.
k =
1
√
2π
X
f
X
(x) =
+∞

−∞
f(x, y)dy = k
+∞

−∞
e
−(x−1)
2
e
−
y
2
2
dy
= ke
−(x−1)
2
+∞


−∞
e
−
y
2
2
dy = k
√
2πe
−(x−1)
2
=
1
√
π
e
−(x−1)
2
.
X ∼ N(1, (1/
√
2)
2
) P(X  1+
√
2) = 1−P(X < 1+
√
2) =
1 − Φ


1 +
√
2 − 1
1/
√
2

= 1 −Φ(2) = 1 −0, 9773 = 0, 0227
X ∼ N(1, (1/
√
2)
2
) E(X) = 1 Var(X) = 1/2
P(1 −
√
5 < X < 1 +
√
5) = P(|X − 1| <
√
5)  1 −
1
2(
√
5)
2
=
9
10
.
• 14 2

+∞

−∞
e
−(x−1)
2
dx =
√
π
+∞

−∞
e
−
y
2
2
dy =
√
2π
+∞

−∞
e
−
(t−µ)
2
2σ
2
dt = σ

√
2π
3
• 1b 10
Var(X
n
) = 9−
13
2n
−(3−
3
2n
)
2
 9 n {X
n
}
{X
n
}