120/300
20/100
=
/300
/300
,
P(F ) > 0 E F
P(E | F ) :=
P(E ∩ F )
P(F )
.
P(E ∩ F ) = P(E | F )P(F )
{(2, 6), (3, 5), (4, 4), (5, 3), (6, 2)}
5/36 A
B P(A ∩B)
1/36 P(B) = 1/6
P(A | B) =
1/36
1/6
= 1/6
A B
P(B) = 3/5 P(A ∩ B)
P(A ∩ B) =
3
2

2
0
/
5
2
P(A | B) =
3/10
3/5
= 1/2
B A
bb, bg, gb, gg 1/4 P(A ∩ B) = P(bb) = 1/4
P(B) = P(bb, bg, gb) = 3/4
1/4
3/4
= 1/3
D T
P(D | T ) =
P(D ∩T )
P(T )
=
(0.99)(0.003)
(0.99)(0.003) + (0.01)(0.997)
≈ 23%.
3/10
0.3
P(E | F ) P(F | E)
P(E ∩ F ) = P(E | F )P(F ),
P (F | E) =
P (F ∩E)
P (E)
=
P(E | F )P(F )
P (E)
D C
P(D) = 0.36, P(C) = 0.30, P(C | D) = 0.22 P(D | C)
P(D | C) = P(D ∩ C)/P(C) P(D ∩ C) = P(C | D)P(D) =
(0.22)(0.36) = 0.0792 P(D | C) = 0.0792/0.3 = 0.264 = 26.4%
A W M
P(A | W ) = 0.30, P(A | M) = 0.25, P(W ) = 0.60
P(W | A)
P(W | A) =
P(W ∩A)
P(A)
.
P(W ∩A) = P(A | W )P(W ) = (0.30)(0.60) = 0.18.
P(A)
P(A) = P(W ∩A) + P(M ∩A).
P(M ∩A) = P(A | M)P(M) = (0.25)(0.40) = 0.10.
P(A) = P(W ∩A) + P(M ∩A) = 0.18 + 0.10 = 0.28.
P(W | A) =
P(W ∩A)
P(A)
=
0.18
0.28
.
P(E) > 0
P(F | E) =
P(E | F )P(F )
P(E | F )P(F ) + P(E | F
c
)P(F
c
)
.
P(E) = P((E ∩ F ) ∪(E ∩ F
c
)) = P(E ∩ F ) + P(E ∩ F
c
).
P(F | E) =
P(E ∩ F )
P(E)
=
P(E | F )P(F )
P(E ∩ F ) + P(E ∩ F
c
)
=
P(E | F )P(F )
P(E | F )P(F ) + P(E | F
c
)P(F
c
)
.
1/3
1/3
1/2 1/2
N
n N
n
P (N | n) N n
P (N | n) = 0 P (N | N) = 1
P (N | n) n 0 N
N = 200 P (200 | 50)
y (n) := P (200 | n) 200
n y(0) = 0 y(200) = 1
n n + 1 n − 1
1/2
y(n) =
1
2
y(n + 1) +
1
2
y(n − 1).
y(n) + y(n − 1)
y(n + 1) − y(n) = y(n) − y(n − 1).
y(n)
x y(n) y(0) = 0
y(200) = 1 y(n) = n/200 y(50) = 1/4
y(n)
y (n) = y (n) − y (0) = (y (n) − y (n − 1)) + ... + (y (1) − y (0))
= n (y (1) − y (0)) = ny (1) .
y (0) = 0 y (1)
y (200) = 1 y (1) = 1/200 y (n) = n/200 y (50) = 1/4
z (n) n
z (n)
z
z(n) =
1
2
z(n + 1) +
1
2
z(n − 1).
0
0 z (0) 6= 0
z (n) − z (0) = (z (n) − z (n − 1)) + ... + (z (1) − z (0))
= n (z (1) −z (0)) ,
z (n) = n (z (1) − z (0)) + z (0) .
a := z (1) − z (0) b := z (0) n
z (n) = an + b.
a b n
0 6 z (n) 6 1 a = 0
z (1) = z (0) ,
z (n) = z (0)
n z (200) = 1
z (n) = 1 n.
40% 40%
U = U =
U ⊆ M
c
U ∩M
c
= U
P(U | M
c
) =
P (U ∩M
c
)
P (M
c
)
=
P (U)
1 − P (M)
=
4/10
6/10
=
2
3
.
13 52
A = B =
P (B | A) P(A) P(A ∩B)
52
13
4
2
·
48
11
P(A) =
4
2
·
48
11
52
13
.
52
13
·
39
13
A B
4
2
·
48
11
·
2
1
·
37
12
P(A ∩ B) =
4
2
·
48
11
·
2
1
·
37
12
52
13
·
39
13
.
P (B | A) =
P (A ∩ B)
P(A)
=
4
2
·
48
11
·
2
1
·
37
12
.
52
13
·
39
13
4
2
·
48
11
.
52
13
=
2 ·
37
12
39
13
P( < $25K) =
212
500
+
36
500
=
248
500
= 0.496.
P ( > $25K | > $25K) =
54/500
(198 + 54)/500
=
54
252
= 0.214
P ( > $25K | < $25K) =
36/500
248/500
= 0.145.
E, F ∈ F P(E), P(F ) > 0
P(E ∩ F ) = P(E)P(F | E)
P(E) = P(E | F )P(F ) + P(E | F
c
)P(F
c
)
P(E
c
| F ) = 1 −P(E | F )
P(F | E) =
P(E∩F )
P(E)
E
E ∩ F E ∩ F
c
(i)
P (E) = P (E ∩ F ) + P (E ∩ F
c
)
= P (E | F ) P (F ) + P (E | F
c
) P (F
c
) .
F = E P(E | E
c
) = 0
4
5
1
7
B C
A = P(B) = P(C) =
1
2
P (A ∩ C) = P (C) P (A | C) =
1
2
·
1
7
=
1
14
.
P(E ∩ F ) = P(E)P(F | E)
E
1
, E
2
, . . . , E
n
∈ F
P (E
1
∩ E
2
∩ ... ∩ E
n
)
= P (E
1
) P (E
2
| E
1
) P (E
3
| E
1
∩ E
2
) ···P (E
n
| E
1
∩ E
2
∩ ··· ∩ E
n−1
) .
5 8
3
B
1
, B
2
, B
3
, ... B
i
i
P(B
1
∩ B
2
∩ B
3
) = P(B
1
)P(B
2
| B
1
)P(B
3
| B
1
∩ B
2
) =
5
13
6
14
7
15
.
1
R
i
= i
P ( ) = P(B
1
∩R
2
∩R
3
) + P(R
1
∩B
2
∩R
3
) + P(R
1
∩R
2
∩B
3
) = 3
5 · 8 · 9
13 · 14 · 15
.
S F
1
, . . . , F
n
F F
c
F
1
, . . . , F
n
⊆ S S =
S
n
i=1
F
i
E ∈ F
P (E) =
n
X
i=1
P (E | F
i
) P (F
i
) .
0.4
0.2
30%
A
1
=
A =
P (A
1
) = P (A
1
| A) P (A) + P (A
1
| A
c
) (1 − P (A))
= 0.4 · 0.3 + 0.2(1 − 0.3) = 0.26
P (A | A
1
) =
P (A ∩ A
1
)
P (A
1
)
=
P (A) P (A
1
| A)
0.26
=
0.3 · 0.4
0.26
=
6
14
.
F
1
, . . . , F
n
⊆ S S =
S
n
i=1
F
i
E ⊆ S j = 1, . . . , n
P (F
j
| E) =
P (E | F
j
) P (F
j
)
P
n
i=1
P (E | F
i
) P (F
i
)
35% 25%
40%
P (D) = P (I) P (D | I) + P (II) P (D | II) + P (III) P (D | III)
= 0.35 · 0.02 + 0.25 · 0.01 + 0.4 · 0.03 =
215
10, 000
.
P (III | D) =
P (III) P (D | III)
P (D)
=
0.4 · 0.03
215/10, 000
=
120
215
.
m
p
K C
P (K | C) =
P (C | K) P (K)
P (C | K) P (K) + P (C | K
c
) P (K
c
)
=
1 · p
1 · p +
1
m
(1 − p)
=
mp
1 + (m − 1)p
.
A = { }
B = { } C = { }
P (A | C)?
P (B | C)
A C B C
9 6
Y
X
A
40%
17% 20%
1/2 1/3
1/4 1/3
∗
p
1 − p 0 < p < 1 N
n
∗
F P
F
(E) := P (E | F )
P
F
∗
P
F
E F
E (E
c
| F ) = 1 −E (E | F ) .
S = {(i, j) | i, j = 1, 2, 3, 4, 5, 6}
A = {(1, 2) , (2, 1)}
B = {(1, 6) , (2, 5) , (3, 4) , (4, 3) , (5, 2) , (6, 1)}
C = {(1, 1) , (1, 2) , (1, 3) , (1, 4) , (1, 5) , (1, 6) , (2, 1) , (3, 1) , (4, 1) , (5, 1), (6, 1)}
P (A | C) =
P (A ∩ C)
P(C)
=
2/36
11/36
=
2
11
.
P (B | C) =
P (B ∩ C)
P(C)
=
2/36
11/36
=
2
11
.
P(A) = 2/36 6= P (A | C)
P (B) = 6/36 6= P(B | C)
E
F P (F | E)
P(E)
11
36
{(6, 3), (6, 4), (6, 5), (6, 6), (3, 6), (4, 6), (5, 6)} P (E ∩ F ) =
7
36
P (F | E) =
P (E ∩ F )
P(E)
=
7/36
11/36
=
7
11
.
B
i
i
P(B
1
) = P(B
2
) =
1
2
P(G | B
1
) =
1
2
P(G | B
2
) =
2
5
P(B
1
| G) =
P(G | B
1
)P(B
1
)
P(G | B
1
)P(B
1
) + P(G | B
2
)P(B
2
)
=
1/4
1/4 + 1/5
=
5
9
.
E F
P(E) = 0.6 P(F | E) = 0.15 P(E ∩ F) =
(0.6) (0.15) = 0.09 P(E
c
∩F ) = 0 P(F ) = P(E ∩F ) + P(E
c
∩F )
P(F ) = 0.09
E F
P(E) = 0.6 P(F | E) = 0.8
P(E ∩ F ) = (0.6) (0.8) = 0.48 P(E
c
∩ F ) = 0 P(F ) =
P(E ∩ F ) + P(E
c
∩ F P(F ) = 0.48
M C
P (M | C) =
P (C | M) P(M)
P (C | M) P(M) + P (C | M
c
) P(M
c
)
=
(0.05) (0.49)
(0.05) (0.49) + (0.0025) (0.51)
≈ 0.9505.
H 5000 X
X Y Y
P (H) = P (H | X) P(X) + P (H | Y ) P(Y )
= (0.99) (0.6) + (0.95) (0.4)
= 0.974.
P (Y | H) =
P (H | Y ) P(Y )
P (H)
=
(0.95) (0.4)
0.974
≈ 0.39.
P (X | H
c
) =
P (H
c
| X) P(X)
P (H
c
)
=
P (H
c
| X) P(X)
1 − P (H)
=
(1 − 0.99) (0.6)
1 − 0.974
=
(0.01) (0.6)
0.026
≈ 0.23
D A
A B C
P (A | D) =
P (D | A) P(A)
P (D | A) P(A) + P (D | B) P(B) + P (D | C) P(C)
=
(0.05) (0.25)
(0.05) (0.25) + (0.04) (0.35) + (0.02) (0.4)
= 0.362.
C K
E
G
P (K | C) =
P(C | K)P(K)
P(C)
=
P(C | K)P(K)
P(C | K)P(K) + P(C | E)P(E) + P(C | G)P(G)
=
1 ·
1
2
1 ·
1
2
+
1
3
·
1
4
+
1
4
·
1
4
=
24
31
≈ .774,
77.4%
+ D
P (D | +) =
P (+ | D) P (D)
P (+ | D) P (D) + P (+ | D
c
)P (D
c
)
=
(0.95) (0.01)
(0.95) (0.01) + (0.005) (0.99)
= 0.657.
P
F
(E) = P (E | F ) 0 1
P
F
P
F
(S) = P (S | F ) =
P (S ∩ F )
P (F )
=
P (F )
P (F )
= 1.
{E
i
}
∞
i=1
, E
i
∈ F
P
F
∞
[
i=1
E
i
!
= P
∞
[
i=1
E
i
| F
!
=
P ((
S
∞
i=1
E
i
) ∩ F )
P (F )
=
P (
S
∞
i=1
(E
i
∩ F ))
P (F )
=
P
∞
i=1
P (E
i
∩ F )
P (F )
=
∞
X
i=1
P (E
i
∩ F )
P (F )
=
∞
X
i=1
P
F
(E
i
) .
(E ∪ F ) ∩ G = (E ∩ G) ∪ (F ∩ G)
{E
i
∩ F }
∞
i=1