Egzamin dla Aktuariuszy z 24 marca 2001 r.
Prawdopodobieństwo i Statystyka
Zadanie 1
)
5
,
0
(
2
1
,
1
)
2
(
2
2
2
wykl
χ
σ
S
≅
Γ
≅
≅
(
)
63212
,
0
1
2
2
1
2
2
2
2
≈
−
=
≤
=
≤
−
e
σ
S
P
σ
S
P
Zadanie 2
1
X
- numer pierwszej kuli
=
=
wpp
0
2
1
2
2
X
X
X
Y
=
∨
=
=
wpp
0
3
2
3
1
3
3
X
X
X
X
X
Y
=
∨
=
∨
=
=
wpp
0
4
3
4
2
4
1
4
4
X
X
X
X
X
X
X
Y
=
=
wpp
0
1
2
2
2
X
X
X
Z
=
=
=
wpp
0
lub
2
3
1
3
3
3
X
X
X
X
X
Z
=
∨
=
∨
=
=
wpp
0
3
4
2
4
1
4
4
4
X
X
X
X
X
X
X
Z
i
i
i
X
Z
Y
=
+
(
)
(
)
∑
∑
=
=
=
=
=
=
=
=
10
1
10
1
1
1
2
2
20
11
100
10
2
11
10
1
10
k
k
k
k
X
P
k
X
Z
E
EZ
(
)
(
)
∑∑
∑∑
= =
= =
=
⋅
=
+
=
=
=
=
=
=
10
1
10
1
2
1
10
1
10
1
2
1
3
3
10
11
2
10
2
11
100
1
100
1
10
10
,
,
k
j
k
j
j
k
j
X
k
X
P
j
X
k
X
Z
E
EZ
(
)
∑∑∑
=
=
=
=
=
20
33
(..)
,
,
3
2
1
4
4
P
m
X
j
X
k
X
Z
E
EZ
20
11
5
,
5
2
2
2
−
=
−
=
EZ
EX
EY
20
33
5
,
5
,
10
11
5
,
5
4
3
−
=
−
=
EY
EY
(
)
7
,
18
9
,
18
20
33
10
11
20
11
4
5
,
5
4
3
2
1
≈
≈
−
−
−
⋅
=
+
+
+
=
Y
Y
Y
X
E
ODP
Zadanie 3
(
)
( )
( )
∫
∫
∞
∞
−
−
∂
∂
=
+
→
+
=
=
>
x
x
x
x
x
t
f
t
Y
E
e
x
x
e
t
f
t
Y
E
x
X
Y
E
)
(
)
2
(
2
)
(
( )
( )
x
x
Y
E
e
xe
e
e
x
Y
E
x
x
x
x
+
=
→
−
−
=
−
→
−
−
−
−
1
2
(
)
( )
(
)
(
)
3
2
1
2
=
+
=
+
=
=
=
X
X
E
X
Y
XE
E
X
XY
EE
EXY
2
1
1
)
1
(
=
+
=
+
=
X
E
EY
COV(X,Y)=3-2=1
Nie można obliczyć vary
Odpowiedź (E) jest prawidłowa
Zadanie 4
V
nzl
X
10
1
,
0
2
≅
σ
N
X
)
9
(
10
2
2
χ
σ
V
≅
)
9
(
3
10
3
10
)
9
(
10
9
1
10
2
2
t
V
X
V
σ
σ
X
t
σ
V
σ
X
≅
=
→
≅
6021
,
0
8331
,
1
3
05
,
0
3
0
≈
→
=
→
=
>
c
c
c
V
X
P
Zadanie 5
(
) (
)
αβ
α
β
β
β
α
X
X
X
P
X
X
X
P
−
+
−
=
=
=
=
=
≠
≠
=
2
)
1
(
3
,
2
,
1
1
,
1
,
1
1
2
3
1
2
3
(
)
=
+
+
=
≠
≠
)
2
,
3
,
2
(
)
3
,
2
,
3
(
)
3
,
2
,
1
(
1
,
1
1
2
P
P
P
X
X
P
αβ
α
β
β
α
αβ
α
β
αβ
α
αβ
αβ
α
α
β
βα
αβ
α
β
β
α
β
α
β
βα
−
+
−
=
−
+
−
+
+
−
+
−
=
−
+
−
+
−
+
−
=
2
)
1
(
2
2
2
2
)
1
(
)
1
(
)
1
(
2
2
2
2
)
1
(
2
2
2
)
1
(
β
β
α
αβ
α
β
αβ
α
β
β
αβ
ODP
=
−
−
+
−
+
−
=
Zadanie 6
∑
Γ
≅
)
,
( θ
n
X
i
(
)
=
−
≤
≤
−
=
≤
≤
=
≤
−
≤
θ
n
θ
θ
X
θ
n
θ
θ
P
θ
θ
θ
P
θ
θ
θ
θ
P
1
99
100
1
101
100
99
100
ˆ
1
101
100
01
,
1
ˆ
99
,
0
40000
1
99
100
96
,
1
95
,
0
1
99
100
1
101
100
≈
→
−
=
→
=
−
≤
≤
−
=
n
n
n
X
n
P
Zadanie 7
Atom punkt
θ
x
=
θ
e
θ
X
P
−
−
=
=
1
)
(
dalej wykładniczy
tn
e
t
P
t
P
−
−
=
>
−
=
≤
1
)
(min
1
)
(min
tn
ne
t
f
−
=
)
(
min
n
θ
e
θ
P
θ
P
−
−
=
>
−
=
=
1
)
(min
1
)
(min
2
2
ˆ
2
ˆ
)
(
θ
θ
E
θ
θ
E
θ
R
+
−
=
(
)
∫
∞
−
−
−
−
+
+
=
−
+
=
θ
θ
n
θ
n
n
θ
xn
θ
e
n
e
n
θ
e
θ
ne
x
θ
E
2
2
2
2
2
2
2
1
ˆ
(
)
∫
∞
−
−
−
+
=
−
+
=
θ
θ
n
n
θ
xn
θ
e
n
e
θ
xne
θ
E
1
1
ˆ
θ
n
θ
n
θ
n
θ
n
e
n
θ
θ
e
n
θ
θ
e
n
e
n
θ
θ
R
−
−
−
−
=
+
−
−
+
+
=
2
2
2
2
2
2
2
2
2
2
)
(
Zadanie 8
(
)
(
)
3
2
1
2
3
2
1
2
X
od
nzl
bo
var
5
var
X
X
X
X
X
X
X
v
+
→
+
=
+
=
(
)
(
)
(
)
=
−
=
=
+
−
=
∧
=
=
=
+
=
−
−
−
−
10
5
5
5
5
2
1
1
2
2
1
2
!
5
10
)!
5
(
5
!
5
5
5
5
e
e
j
e
j
X
X
P
j
X
j
X
P
X
X
j
X
P
j
j
)
5
,
0
;
5
(
2
1
5
10
!
5
)!
5
(
!
5
5
10
5
10
5
dwum
j
e
j
j
e
≅
=
−
=
−
−
(
)
4
5
2
1
2
1
5
5
var
2
1
2
=
⋅
⋅
=
=
+
X
X
X
ODP=1,25+5=6,25
Zadanie 9
X
E
EX
X
E
−
>=
<
(
)
(
)
∫
∫
∑ ∫
∑
∑
∞
=
+
∞
=
∞
=
−
−
+
−
−
−
−
−
−
=
−
=
=
+
+
=
2
1
3
2
1
1
1
1
)
1
(
1
...
2
k
k
k
k
k
λ
k
λ
k
λ
k
λ
x
λ
x
λ
x
λ
e
ke
e
e
k
e
λ
k
e
λ
e
λ
X
E
...
2
2
+
+
=
−
−
λ
λ
e
e
u
...
2
3
2
+
+
=
−
−
−
λ
λ
λ
e
e
ue
(
)
(
)
2
2
1
1
...
1
λ
λ
λ
λ
λ
λ
λ
e
e
u
e
e
e
e
e
u
−
−
−
−
−
−
−
−
=
→
−
=
+
+
=
−
(
)( )
1
1
1
1
1
1
2
=
−
→
−
=
−
=
−
−
=
−
−
−
−
−
X
E
e
X
E
e
e
e
e
e
e
X
E
λ
λ
λ
λ
λ
λ
λ
)
ln(
)
1
ln(
1
c
c
λ
c
c
e
λ
−
+
=
+
=
λ
EX
1
=
)
(
ln
)
1
ln(
1
A
c
c
c
ODP
=
−
−
+
=
Zadanie 10
9
4
)
2
(
9
4
)
0
(
9
1
)
2
(
:
2
=
=
=
−
p
p
p
S
81
16
)
4
(
81
32
)
2
(
81
24
)
0
(
81
8
)
2
(
81
1
)
4
(
:
4
=
=
=
=
−
=
−
p
p
p
p
p
S
3
3
3
3
3
3
3
6
9
64
)
6
(
9
192
..
64
128
)
4
(
9
240
..
16
128
96
)
2
(
9
160
..
32
96
32
)
0
(
9
60
..
24
32
4
)
2
(
9
12
..
4
8
)
4
(
9
1
)
6
(
:
=
=
+
=
=
+
+
=
=
+
+
=
=
+
+
=
−
=
+
=
−
=
−
p
p
p
p
p
p
p
S
(
) (
)
(
)
=
≤
≤
=
=
=
≤
≤
=
=
≤
≤
=
=
5
,...,
5
,
2
...
5
,...,
5
,
2
5
,...,
5
,
2
6
1
10
8
1
10
9
1
10
S
S
S
P
S
S
S
P
S
S
S
P
r
[
]
k
S
k
k
k
k
odpada
k
−
=
−
−
∈
−
=
=
=
=
−
=
2
4
;
4
;
4
;
2
,
0
,
2
,
4
;
6
4
(
)
(
) (
)
∑
−
=
=
=
−
=
=
≤
=
=
5
6
6
4
6
10
2
5
,
2
k
k
S
P
k
S
P
S
S
P
(
) (
) (
) (
) (
) (
) (
) (
)
=
−
=
=
+
=
=
+
=
=
+
=
−
=
=
2
4
0
2
2
0
4
2
6
4
6
4
6
4
6
4
S
P
S
P
S
P
S
P
S
P
S
P
S
P
S
P
2265
,
0
9
60
16
9
160
32
9
240
24
9
192
81
8
5
5
5
3
≈
⋅
+
⋅
+
⋅
+
=