Egzamin dla Aktuariuszy z 28 lutego 1998 r.
Prawdopodobieństwo i Statystyka
Zadanie 1
(
)
2
10
20
2
5
10
2
2
4
2
2
2
4
3
2
1
⋅
⋅
⋅
⋅
⋅
⋅
=
∩
∩
A
A
A
P
( )
2
10
20
2
7
14
2
3
6
2
⋅
=
A
P
(
)
2
10
20
2
7
14
2
2
2
4
2
1
⋅
⋅
⋅
=
∩
A
A
P
(
)
2
10
20
2
5
10
2
2
4
2
3
6
2
3
=
∩
A
A
P
=
7
14
4
3
6
2
10
20
7
14
3
6
4
10
20
2
2
10
20
8
5
10
2
4
3
6
2
10
20
8
7
14
2
4
4
7
14
3
6
10
20
10
20
16
5
10
2
4
2
)
(
.....
4
16
7
14
3
6
8
8
16
5
10
2
4
2
2
A
L
P
→
=
=
⋅
=
Zadanie 2
(
)
(
)
(
)
∑
∑ ∫
∑
∞
=
∞
=
∞
=
−
+
=
+
−
=
=
=
>
=
1
1
1
1
0
0
0
1
0
0
0
0
2
1
2
1
1
)
(
k
k
x
k
k
N
N
n
x
x
x
x
xdx
k
N
P
X
X
X
E
X
X
E
(
)
4
1
2
1
0
+
=
=
X
X
EE
EX
N
N
4
1
2
1
4
1
2
1
4
1
2
1
0
=
−
+
=
−
+
=
EX
ODP
Zadanie 3
Tu jest chyba błąd: wychodzi Be(0,5;0,5)
∫ ∫
Π
=
Π
=
≤
+
1
0
2
arccos
2
2
2
4
t
dr
φ
rd
t
v
u
u
P
bo:
φ
r
v
φ
r
u
sin
cos
=
=
r
=
)
1
,
0
(
,
2
;
0
∈
Π
∈
r
φ
Π
∈
≤
≤
2
;
arccos
cos
cos
2
t
φ
t
φ
t
φ
∫
Π
−
=
−
Π
Π
=
−
Π
Π
=
1
0
arccos
2
1
arccos
2
2
arccos
2
4
t
t
t
r
Π
=
Γ
≅
−
Π
=
−
Π
=
′
−
−
2
1
bo
)
5
,
0
;
5
,
0
(
)
1
(
1
2
1
1
1
2
)
(
1
2
1
1
2
1
Be
t
t
t
t
t
f
Zadanie 4
k – liczba sukcesów
k+4 – liczba porażek
n=2k+4
4
3
2
3
1
4
2
+
+
=
=
k
k
k
k
k
L
p
1
)
1
)(
5
(
)
6
2
)(
5
2
(
9
2
2
3
3
)!
4
2
(
)!
4
(
!
3
2
3
1
)!
5
(
)!
1
(
)!
6
2
(
4
5
1
1
>
+
+
+
+
=
+
+
+
+
+
=
+
+
+
+
k
k
k
k
k
k
k
k
k
k
p
p
k
k
k
k
k
k
2
9
5
6
30
22
4
2
2
>
+
+
+
+
k
k
k
k
(
)
0
5
6
2
45
54
9
60
44
8
2
2
2
>
+
+
−
−
−
+
+
k
k
k
k
k
k
160
licznika
0
)
1
)(
5
(
2
15
10
2
=
∆
>
+
+
+
−
−
k
k
k
k
2
10
160
2
10
160
−
=
−
−
=
B
A
2
k
dla
max
1,32
do
1
1
=
→
≈
>
+
k
k
p
p
8
4
2
2
=
+
⋅
=
n
Zadanie 5
Można ograniczyć się do testu NM
(
)
(
)
05
,
0
2
1
2
1
2
2
2
2
2
1
2
2
1
2
2
2
2
2
=
>
Π
Π
∏
∏
∏
∏
−
−
−
−
−
−
−
+
−
−
t
e
e
e
e
P
µ
y
µ
x
n
µ
y
µ
x
n
i
i
i
i
...
n
n
t
t
n
x
y
P
n
N
i
i
H
4
1
2
64
,
1
ln
05
,
0
ln
4
1
2
2
;
0
0
−
=
→
=
>
−
−
≅
∑
∑
4
4 8
4
4 7
6
moc:
95
,
0
4
1
2
64
,
1
4
1
2
2
;
4
1
1
>
−
>
−
−
≅
∑
∑
n
n
n
x
y
P
n
n
N
i
i
H
4
4
4
8
4
4
4
7
6
64
,
1
2
2
1
2
64
,
1
−
<
−
n
n
n
22
51
,
21
03
,
43
2
56
,
6
2
28
,
3
2
2
64
,
1
2
2
1
64
,
1
≥
→
≥
→
>
→
>
→
>
→
−
<
−
n
n
n
n
n
n
n
Zadanie 6
(
)
[ ]
θ
t
e
θ
e
θ
t
e
x
x
θ
e
X
P
t
X
P
t
X
P
t
t
−
−
−
−
=
=
=
≥
=
−
≥
=
≤
−
−
−
∫
1
)
(ln
)
ln
(
1
1
1
)
(
)
(
θ
wykl
e
θ
t
f
θ
t
≅
=
−
∑
=
Γ
≅
−
S
θ
X
i
)
;
5
(
ln
∫
=
=
≤
=
<
=
<
−
θ
c
x
θ
e
x
θ
θ
c
S
P
c
θ
S
P
S
c
θ
P
0
4
5
24
)
(
∫
=
′
=
=
−
...
4
...
3
4
4
x
u
x
u
e
x
x
θ
....
....
05
,
0
1
2
6
24
1
2
3
4
=
+
+
+
+
−
=
−
c
c
c
c
e
c
po wstawieniu wychodzi
2
94
,
3
≈
c
tak samo
(
)
2
18,31
wychodzi
i
1
≤
−
=
>
=
>
θ
c
S
P
θ
c
S
P
θ
θ
P
Zadanie 7
(
) (
)
Y
Y
X
X
E
Y
X
X
E
=
+
+
+
+
+
15
6
5
1
...
...
(
)
(
)
(
)
Y
Y
X
X
E
Y
X
X
Y
X
X
E
Y
X
X
E
3
1
...
2
...
2
...
5
1
5
1
15
6
=
+
+
→
=
+
→
+
+
=
+
+
(
) (
) (
)
µ
Y
Y
µ
Y
Y
X
X
E
Y
X
X
E
Y
X
X
E
ODP
5
3
2
3
1
5
...
...
...
5
1
20
16
15
1
+
=
−
+
=
+
+
−
+
+
+
+
+
=
Zadanie 8
(
)
(
)
( )
(
)
5
,
0
2
1
1
⋅
=
=
=
+
n
n
X
f
P
X
f
P
(
)
(
)
( )
(
)
( )
(
)
5
,
0
2
1
2
1
⋅
=
+
=
=
=
+
n
n
n
X
f
P
X
f
P
X
f
P
( ) (
)
(
)
( )
(
)
( )
(
)
(
)
( )
(
)
5
,
0
2
4
5
,
0
2
1
2
1
⋅
=
+
⋅
=
+
=
=
⋅
+
n
n
n
n
n
X
f
P
X
f
P
X
f
P
X
f
X
f
E
( )
( )
(
)
( )
(
)
2
2
1
=
+
=
=
n
n
n
X
f
P
X
f
P
X
Ef
(
)
(
)
(
)
(
)
(
)
2
2
1
1
1
1
=
+
=
=
+
+
+
n
n
n
X
f
P
X
f
P
X
Ef
( ) (
)
[
]
( )
(
)
( )
(
)
( )
(
)
( )
(
)
[
]
⋅
=
+
=
−
=
+
=
=
+
2
2
1
2
3
1
2
,
cov
1
n
n
n
n
n
n
X
f
P
X
f
P
X
f
P
X
f
P
X
f
X
f
( )
(
)
( )
(
)
( )
(
)
[
]
2
1
2
2
5
,
0
=
+
=
+
=
⋅
n
n
n
X
f
P
X
f
P
X
f
P
[
]
[
]
2
1
2
1
,
5
,
0
5
,
0
1
0
,
Π
Π
=
Π
Π
Π
=
Π
+
Π
Π
=
Π
2
2
1
1
2
5
,
0
5
,
0
1
5
,
0
1
2
2
2
1
=
Π
+
Π
=
Π
+
Π
=
Π
=
Π
3
1
3
2
1
2
(
)(
)
=
⋅
+
⋅
+
−
⋅
+
⋅
=
Π
+
Π
Π
+
Π
−
Π
+
Π
=
∞
→
3
1
2
3
2
2
3
3
2
2
3
1
3
2
3
3
1
2
2
5
,
1
2
3
2
cov
lim
1
2
2
1
2
1
n
9
1
9
25
18
6
3
5
3
5
2
3
2
−
=
−
+
=
−
+
=
Zadanie 9
∫
∞
−
−
+
=
=
c
µ
c
x
µ
c
e
µ
x
EX
µ
c
t
n
n
e
t
X
P
t
P
−
−
−
=
≥
−
=
≤
1
)
(
1
)
(min
µ
c
t
n
e
µ
n
t
f
−
−
=
)
(
min
przesunięty wykładniczy
n
µ
c
E
+
=
→
min
∑
+
=
)
(
µ
c
n
X
E
i
)
(
1
1
1
1
)
(
)
(
A
µ
µ
n
n
n
µ
nc
n
n
n
µ
c
n
A
→
=
−
−
=
+
−
−
−
+
Zadanie 10
(
)
=
+
−
+
=
−
+
+
∑
∑
∑
=
<
=
n
i
j
i
n
i
i
i
j
i
j
i
i
i
n
n
µ
X
c
µ
X
X
c
c
X
c
E
µ
X
c
X
c
E
1
1
2
2
2
2
1
1
2
2
...
(
)
(
)
∑
∑
∑
∑
∑
∑
<
=
<
+
−
+
+
=
+
−
+
+
=
j
i
n
i
j
i
i
i
j
i
i
i
j
i
µ
c
µ
c
µ
µ
γ
c
c
µ
µ
µ
c
µ
µ
µ
γ
c
µ
c
c
1
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
(
)
∑
≠
=
−
+
+
=
∂
∂
i
j
i
j
i
µ
c
µ
µ
γ
c
µ
c
0
2
2
2
2
2
2
2
2
(
)
∑
∑
=
−
+
+
−
0
2
2
)
1
(
2
:
i
po
suma
.
1
2
2
2
2
2
µ
n
c
µ
µ
γ
c
n
µ
i
i
(
)
∑
+
=
+
=
+
+
−
=
2
2
2
2
2
2
2
2
2
2
2
2
2
2
)
1
(
2
2
.
2
γ
n
n
µ
γ
n
µ
µ
n
µ
µ
γ
n
µ
µ
n
c
i
Z 1.
(
)
(
)
2
2
2
2
2
2
2
2
µ
c
µ
µ
γ
c
c
µ
i
i
i
=
+
+
−
∑
(
)
2
2
z
i
i
2
2
2
2
1
rowne
c
czyli
1
2
1
2
γ
n
c
γ
c
µ
γ
c
µ
c
i
i
i
i
+
=
→
−
−
=
−
=
∑
∑
to daje min, można sprawdzić