Томусяк А.А., Трохименко В.С. Теорія ймовірностей
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Ω = { , , , , , , , }
A = { , , , } B = { , , , }
1
n
1
14
1
28
3
7
5
6
7
30
386
729
92
729
729 = 9
3
qm
a
4
105
1
360
4
9
1
4
2
21
A
3
9
A
1
A
2
8
· 4
P (A
1
) =
A
2
8
· 4
A
3
9
=
4
9
.
A
2
P (A
2
) =
7 · 18
9 · 8 · 7
=
1
4
A
3
{1, 2, 6} {1, 3, 5} {1, 8, 9} {2, 3, 4} {2, 7, 9} {3, 6, 9} {4, 5, 9}
{5, 6, 7} 3!
P (A
3
) =
8 · 3!
9 · 8 · 7
=
2
21
.
5
9
C
n
N−M
C
m
N
1
−
23
2
+ 22
2
2 · 24
2
≈
0
,
121
A ·B ·C ABC AB + AC + BC ABC + ABC + ABC
ABC Ω AB
A + B
AB A + B
ω AB
AB A
B A B
A + B
AB A + B
ω A + B
A B A
B A · B
AB
A + B AB AB = A + B
(A+B)(A+B)(A+B) = (A+B)(AA+BA+A B+BB) =
(A + B)(∅ + AB + A B + ∅) = (A + B)(AB + A B) = AAB + BAB +
AA B + BA B = AB + AB + ∅ + ∅ = AB ≈ 0, 87 ≈ 0, 07
A
0
, A
1
, A
2
P (A
0
+ A
1
+ A
2
) = P (A
0
) + P (A
1
) + P (A
2
) =
C
7
21
C
7
28
+
C
6
21
· 7
C
7
28
+
C
5
21
· C
2
7
C
7
28
≈ 0, 36
A
B
C
D
P (A) = 1 −P (A) = 1 −P (B + C + D) =
1−P (B)−P (C)−P ( D)+P (BC)+P (BD)+P (CD)−P (BCD)
P (B) = 0, 72 P (C) = 0, 4 P (D) = 0, 8 P (BC) = 0, 32
P (BD) = 0, 6 P (CD) = 0, 32 P (BCD) = 0, 28 P (A) =
1 − 0, 72 − 0, 4 − 0, 8 + 0, 32 + 0, 6 + 0, 32 − 0, 28 = 0, 04
P (A) + P (B) + P (A B) − 1 1 − P (A + B) + P (A) + P (B)
10
13
,
1
4
,
10
13
,
1
4
,
43
52
≈ 0, 21
kp
1 + (k − 1)p
H
1
H
2
A
P (H
1
) = p P (H
2
) = 1 − p P (A|H
1
) = 1 P (A|H
2
) =
1
k
P (H
1
|A) =
P (H
1
)P (A|H
1
)
P (H
1
)P (A|H
1
) + P (H
2
)P (A|H
2
)
=
=
p
p + (1 − p)
1
k
=
kp
1 + (k − 1)p
.
P (H
1
|A) > P (H
1
) P (H
1
|A)
k
A, B, C P (A + B + C) = P (A) +
P (B) + P (C) − P (A)P (B) − P (B)P (C) + P (A)P (B)P (C)
A
1
A
2
A
3
A
4
A
5
+ A
1
A
2
A
3
A
4
A
5
A
i
i
p(1−α
1
)(1−α
2
)+(1−p)β
1
β
2
A
H
1
H
2
P (A|H
1
) = (1−α
1
)(1−α
2
) P (A|H
2
) = β
1
β
2
p(1−α
1
)(1−α
2
)/P (A)
A
p = 0, 5 ≈
0, 219 A
p =
1
3
≈ 0, 053 ≈ 0, 345
A
p = 3
√
3/4π
R
≈ 0, 194 p = 2/π ≈ 0, 035.
p = 3
√
3/2π
A
p = 0, 05 P
20
(> 3) =
1 − (P
20
(0) + P
20
(1) + P
20
(2)) = 1 − (0, 1897 + 0, 3771 + 0, 3580) =
1 −0, 9240 = 0, 0760
≈ 0, 021 ≈ 0, 245 ≈ 0, 180 ≈ 0, 307
N = 30 M = 20 n = 5
³
p =
2
3
´
λ = 8 n = 9
³
n = 20, p =
1
8
´
M(X) = 3, 35 D(X) = 1, 0275
σ(X) = 1, 01
X
P
27
64
27
64
9
64
1
64
M(X) =
3
4
D(X) =
9
16
σ(X) =
3
4
X
n = 3 p =
1
4
X
P 10
−1
9 · 10
−2
9
2
· 10
−3
9
3
· 10
−4
9
4
· 10
−5
9
5
· 10
−6
X
P 9
6
· 10
−7
9
7
· 10
−8
9
8
· 10
−9
9
9
· 10
−10
X
P
91
216
61
216
37
216
19
216
7
216
1
216
X
p
k
k = 1, 6
X
k 1 − p
1
µ
5
6
¶
3
p
1
1 − p
1
=
µ
5
6
¶
3
p
1
=
91
216
1 −(p
1
+ p
2
)
µ
4
6
¶
3
p
2
1−(p
1
+p
2
) =
µ
4
6
¶
3
p
2
=
61
216
1−(p
1
+p
2
+p
3
) =
µ
3
6
¶
3
p
3
=
37
216
1 − (p
1
+ p
2
+ p
3
+ p
4
) =
µ
2
6
¶
3
p
4
=
19
216
1 − (p
1
+ p
2
+ p
3
+ p
4
+ p
5
) =
µ
1
6
¶
3
p
5
=
7
216
1 − (p
1
+ p
2
+ p
3
+ p
4
+ p
5
+ p
6
)
p
6
=
1
216
M(X) =
q
p
D(X) =
q
p
2
X p
0 < p < 1 0, 1, 2, . . .
p
n
= P (X = n) = pq
n
q = 1 − p M(X) =
∞
P
n=0
np
n
=
∞
P
n=1
npq
n
= pq(1+2q+3q
2
+···+nq
n−1
+···) =
pq
(1 − q)
2
=
q
p
D(X) = M(X
2
) − (M(X))
2
=
∞
P
n=1
n
2
pq
n
−
q
2
p
2
= pq
∞
X
n=1
n
2
q
n−1
−
q
2
p
2
=
pq(1 + q)
(1 − q)
3
−
q
2
p
2
=
q + q
2
p
2
−
q
2
p
2
=
q
p
2
(−1; 1)
∞
P
n=0
x
n
=
1
1 − x
∞
P
n=1
nx
n−1
=
µ
1
1 − x
¶
0
=
1
(1 − x)
2
∞
P
n=1
n
2
x
n−1
=
µ
x
(1 − x)
2
¶
0
=
1 + x
(1 − x)
3
M(X) = D(X) = λ
∞
P
n=0
x
n
n!
= e
x
P (X = 1) = 0, 2
P (X = 2) = 0, 3 P (X = 3) = 0, 5 M(Y ) = −5, 6
D(Y ) = 2, 44 P (X = 1) = x P (X = 2) = y
P (X = 3) = 1 −x −y M (X) = x + 2y + 3(1 −x −y) = 2, 3
M(X
2
) = x + 4y + 9(1 − x − y) = 5, 9 p =
1
4
3
4
P (X > 20) ≈ 0, 4
80p(1 − p) = 15 P (X > 20)
e
−1
P (Y < x) = P (−5X + 2 < x) = P
µ
X >
x − 2
−5
¶
=
=
½
e
λ
5
(x−2)
, x 6 2,
1, x > 2;
M(Y ) = −
5
λ
+ 2 D(Y ) =
25
λ
2
a = 0
b = 1 c = 8 d = 1
x = 0,
1
4
,
√
2
4
M(X) = 0
D(X) =
1
4
a
2
σ(X) =
1
2
a M(X) = D(X) = k + 1
σ(X) =
√
k + 1 F (x, y) =
=
0, x 6 0 y 6 0,
1
2
(sin x + sin y − sin(x + y)), 0 < x 6
π
2
, 0 < y 6
π
2
,
1
2
(1 + sin x − cos x), 0 < x 6
π
2
, y >
π
2
,
1
2
(1 + sin y − cos y), x >
π
2
, 0 < y 6
π
2
,
1, x >
π
2
, y >
π
2
,
M(X) = M (Y ) =
π
4
D(X) = D(Y ) =
π
2
16
+
π
2
− 2
Y = πX
2
X
f(y) =
1
2
√
πy
, π 6 x 6 4π,
0,
M(Y ) =
7π
3
D(Y ) =
34
45
π
2
Y = πX
2
F (y) = P (Y < y) = P (πX
2
< y) = P
³
X
2
<
y
π
´
=
=
0, y 6 π,
P
µ
X <
r
y
π
¶
, π < y 6 4π,
1, y > 4π;
=