Hướng dẫn giải đề thi Xác suất thống kê docx

H­íng dÉn gi¶i ®Ò thi x¸c suÊt

C©u 1.

a) C¸c em tù gi¶i.

b) Tr­êng hîp P(AB) = 0 lµ hiÓn nhiªn, ta chØ cÇn xÐt tr­êng hîp

P(AB) > 0. Khi ®ã P(A) > 0 vµ P(B) > 0. Ta chó ý r»ng, víi hai sè

thùc x, y ∈ [0; 1] th× ta lu«n cã bÊt ®¼ng thøc |x − y| 6 1, do ®ã, v× 0 6

P(A|B), P(A) 6 1 nªn ta suy ra

|P(A) − P(A|B)| 6 1.

V× P(A|B) = P(AB)

P(B)

nªn

P(A) −

P(AB)

P(B)

 6 1.

Hay

|P(A)P(B) − P(AB)| 6 P(B). (1)

T­¬ng tù ta cã bÊt ®¼ng thøc

|P(A)P(B) − P(AB)| 6 P(A). (2)

Tõ (1) vµ (2), ta suy ra

|P(A)P(B) − P(AB)| 6 min(P(A), P(B)).

C©u 2.

a) TÝnh ®­îc a =

2

π

.

Ta cã

F(π/12) =

π/

Z

12

−∞

f(x)dx =

−Z

π/2

−∞

f(x)dx +

π/

Z

12

−π/2

f(x)dx

=

π/

Z

12

−π/2

f(x)dx = a

π/

Z

12

−π/2

cos2 xdx

=

2

π

7π

24

+

1

8



=

7

12

+

1

4π

.

b) Ta cã

E(X) = Z

+∞

−∞

xf(x)dx = a

Z

π/2

−π/2

x cos2 xdx =

2

π

Z

π/2

−π/2

x cos2 xdx.

V× Z

π/2

−π/2

x cos2 xdx = 0 nªn E(X) = 0.

C©u 3.

a) Ta cã

f(x, y) = ke− 1

2

(2x

2−4x+2+y

2

) = ke−(x−1)2

e

−

y

2

2 .

V× f(t) = 1

σ

√

2π

e

−

(t−µ)

2

2σ2

lµ hµm mËt ®é cña ph©n phèi chuÈn N (µ, σ2

) nªn

theo tÝnh chÊt cña hµm mËt ®é ta cã Z

+∞

−∞

f(t)dt = 1 hay Z

+∞

−∞

e

−

(t−µ)

2

2σ2 dt = σ

√

2π.