Egzamin dla Aktuariuszy z 21 czerwca 1997 r.
Prawdopodobieństwo i Statystyka
Zadanie 1
49 – miejsce na asy
4! – ustawienie asów
48! – ustawienie pozostałych
50
51
52
!
4
52
51
50
49
49
!
4
!
52
!
48
!
4
49
⋅
⋅
=
⋅
⋅
⋅
⋅
=
⋅
⋅
Zadanie 2
9375
,
0
max
max
=
≥
∧
≤
c
θ
θ
P
c>1
)
;
0
( θ
t
∈
8
)
(max
=
≤
θ
t
t
P
4142
,
1
9375
,
0
1
1
)
(
8
≈
→
=
−
=
−
c
c
c
θ
F
θ
F
Zadanie 3
(
) (
)
(
)
=
−
=
=
+
=
=
+
=
=
+
=
−
−
−
−
50
50
20
30
50
2
1
1
2
1
2
1
1
!
50
50
!
20
)!
50
(
30
50
50
50
e
e
k
e
k
N
N
P
k
N
N
N
P
N
N
k
N
P
k
k
12
5
3
5
2
50
var
5
3
5
2
50
3
2
5
3
50
50
!
50
3
2
!
)!
50
(
30
50
50
50
50
50
50
=
⋅
=
→
=
=
−
=
−
−
−
k
k
k
k
k
k
e
k
k
e
Zadanie 4
(
)
2
1
)
,
cov(
)
,
cov(
7
2
1
2
2
=
→
−
+
=
+
X
X
X
Y
σ
Y
X
m
x
σ
Y
X
m
x
( )
(
)
( )
(
)
4
4
1
8
7
2
1
var
8
var
var
9
var
2
2
=
→
+
=
+
+
=
+
=
=
X
X
σ
σ
X
X
Y
E
X
Y
E
Y
z tego:
2
4
2
1
)
,
cov(
=
=
Y
X
Zadanie 5
(
)
(
)
=
>
=
Π
Π
∑
∑
∑
−
+
−
−
−
−
−
∏
∏
t
e
e
P
e
e
P
i
i
i
i
i
x
n
x
x
x
n
x
n
8
8
2
8
8
1
0
2
2
2
2
2
2
1
2
2
1
01
,
0
2
ln
4
ln
8
4
)
4
;
0
(
8
2
=
+
>
=
>
−
=
>
=
≅
−
∑
∑
∑
n
t
X
P
t
n
x
P
t
e
P
n
N
i
i
n
x
i
8
7
6
n
n
t
n
n
t
n
n
t
n
X
P
i
2
33
,
2
2
ln
4
33
,
2
2
2
ln
4
01
,
0
2
2
ln
4
2
⋅
=
+
=
+
→
=
+
>
∑
moc
9
,
0
2
33
,
2
)
4
;
(
1
>
⋅
>
≅
=
∑
n
X
P
n
n
N
i
µ
8
7
6
28
,
1
2
2
33
,
2
9
,
0
2
2
33
,
2
−
≤
−
⋅
→
>
−
⋅
>
n
n
n
n
n
n
Y
P
2
28
,
1
33
,
2
n
≤
+
22
,
7
≥
n
53
12
,
52
=
→
≥
n
n
Zadanie 6
[
]
[
]
∫ ∫
∫
∫
∫
∞
∞
−
−
−
+
−
=
+
=
+
=
+
+
=
−
−
−
=
+
1
0
1
0
1
0
1
0
1
0
2
2
)
1
2
(
1
)
(
x
x
y
x
y
x
y
x
x
y
x
x
dx
x
x
x
e
e
ye
e
e
xe
dydx
e
y
x
Zadanie 7
(
) (
)
4
1
3
)
1
(
4
3
3
)
1
(
3
)
1
(
3
)
1
(
)
0
(
)
(
0
0
2
1
0
4
3
2
1
0
2
2
θ
θ
θ
θ
θ
θ
θ
θ
θ
θ
X
P
θ
f
θ
X
f
X
θ
f
−
=
−
−
=
−
−
=
=
=
=
=
∫
∫
∫
=
=
−
=
−
=
−
=
−
=
−
⋅
=
1
0
1
0
1
0
5
4
4
3
2
6
,
0
20
12
20
4
5
12
5
1
4
1
12
5
4
12
12
3
)
1
(
4
θ
θ
θ
θ
θ
θ
θ
ODP
Zadanie 8
9
8
1
9
2
8
x
λ
x
λ
e
x
λ
e
λ
L
i
i
−
−
∑
=
=
∑
=
−
+
+
−
=
8
1
9
9
ln
ln
2
ln
8
ln
i
i
x
λ
x
λ
x
λ
λ
L
∑
∑
=
=
⋅
=
−
=
−
+
−
=
∂
∂
8
1
9
1
9
0
10
2
8
i
i
i
i
λ
x
λ
x
λ
x
λ
λ
∑
∑
=
=
=
→
=
−
9
1
9
1
10
ˆ
0
10
i
i
i
i
x
λ
x
λ
Zadanie 9
(
)
+
+
−
=
∑
=
50
26
2
50
26
2
25
...
25
24
1
i
i
x
x
x
ODP
(
)
(
)
=
+
+
−
+
+
⋅
=
+
+
−
+
+
=
+
+
⋅
=
+
+
98
50
...
...
333
,
3
24
25
...
...
500
...
4
,
10
25
...
2
50
1
2
50
2
1
2
25
1
2
25
2
1
50
1
25
1
x
x
x
x
x
x
x
x
x
x
x
x
→
(
)
008
,
2314
25
4
,
10
25
333
,
3
24
50
500
98
240
4
,
10
25
500
...
50
26
2
2
2
50
26
=
⋅
−
⋅
−
+
=
=
⋅
−
=
+
+
∑
=
i
i
x
x
x
417
,
0
25
240
25
008
,
2314
24
1
2
=
−
=
ODP
Zadanie 10
Łańcuch Markowa:
3
2
2
1
3
2
2
1
3
2
)
1
,
1
(
=
+
=
P
3
2
)
1
,
2
(
=
p
3
2
)
1
,
3
(
=
p
6
1
3
1
2
1
)
2
,
1
(
=
=
p
3
1
)
2
,
2
(
=
p
p(3,2)=0
6
1
3
1
2
1
)
3
,
1
(
=
=
p
p(2,3)=0
3
1
)
3
,
3
(
=
p
szukamy rozkładu stacjonarnego i
1
p
[
]
[
]
3
2
1
3
2
1
,
,
3
1
0
3
2
0
3
1
3
2
6
1
6
1
3
2
,
,
p
p
p
p
p
p
=
=
=
→
=
+
=
+
=
+
+
1
3
1
2
3
3
1
2
2
1
1
3
2
1
4
1
4
1
3
1
6
1
3
1
6
1
3
2
3
2
3
2
p
p
p
p
p
p
p
p
p
p
p
p
p
p
1
3
2
1
=
+
+
p
p
p
1
4
1
4
1
1
1
1
=
+
+
p
p
p
3
2
1
2
3
1
1
=
→
=
p
p