−
• Ω
•
(c)
(¯ c)
Ω = c, c¯
{ }
•
{
Ω = 1, 2, 3, 4, 5, 6
}
•
Ω = 0, 1, 2, 3,...
{
}
Ω A,B,...
• • •
{
}
A = cara
B = 2, 4, 6
{
{
C = 0, 1, 2
• • •
∅
∅
}
}
A
•
A
∪B
∪ B = {c, c¯}
•
∪
{
•
∪
{
}
E F = 0, 1, 2, 3, 4, 5
A
A
• • •
∩B
∩B =∅
∩
C D = 4 E F = 1, 2
∩
Ω
• • • •
}
C D = 2, 4, 6
∅
{ }
A
∩B =∅
•
Ac A
∪A
c
=Ω
A
c
∩A =∅
• Ω
A
Ac
Ω
Ω = {1, 2, 3, 4, 5, 6}
A = {2, 4, 6} B = {1, 3, 5} C = {1, 4, 5}
A ∪ B A ∪ C
A∪D
B ∩ A B ∩ C
B∩D
Ac B c C c
Dc
D = {4 }
P(.)
0
≤ P(A) ≤ 1, ∀A ⊂ Ω
P(Ω) P
=1
n j =1
Aj
=
n i=1 P(Aj )
Aj
P(.)
P(A)
= Ω = c, c¯
•
{ }
{c}) = P({c¯}) = 1/2
P(
Ω = 1, 2, 3, 4, 5, 6
•
{
}
{1}) = P({2}) = P({3}) = P({4}) = P({5}) = P({6}) = 1/6
P(
P(A)
•
=
{c}) = 72/135 = 0, 53
P(
P(
{c¯}) = 63/135 = 0, 47
• P(
• •
{1}) = 0, 15, P({2}) = 0, 19, P({3}) = 0, 16, P({4}) = 0, 14, P({5}) = 0, 13, P({6}) = 0, 23
P(A
P(A)
∪ B) P(A
Ω P(A
∪ B) = P(A) + P(B) − P(A ∩ B) P(A
P(A
∩ B) = ∅
∪ B) = P(A) + P(B)
∩ B)
P(B)
Ac P(
P(A
c
)=1
− P(A)
∅) = 0 Ω
P(A)
P(A)
P(A)
= P(A
c
P(B)
P(A
c
∩ B) + P(A ∩ B ). {
} }
Ω = a,b,c,d,e A = a,b,c B = c,d,e B) B) P(A P(A
{ ∩
)
}
{ ∪
= P( a ) + P( b ) + P( c ) = 0, 1 + 0, 1 + 0, 2 = 0, 4
{}
{}
{}
P(B)
{e}) = 0, 2 + 0, 4 + 0, 2 = 0, 8
= P( c ) + P( d ) +
P(
P(A
P(A
Ac
P(Ac )
=1
0, 286 P(D
P(A
∩ B) = 0, 4 + 0, 8 − 0, 2 = 1
P(D
P(A) P(A
c
P(A
P(A
= 100/350 =
= (54 + 38 + 11)/350 = 0, 294
∪ F )
∪ F ) = P(D) + P(F ) − P(D ∩ F ) = 0, 286 + 0, 294 − 0, 109 = 0, 471.
P(B) =
0, 2
P(A
∩ B) = 0, 1
)
∩ B)
∩ Bc)
c
∪ B)
P(A) P(A
P(D)
∪ B)
c P(A
P(A
= 0, 3
P(F ) P(D
∪ B) = P(A) + P(B) −
= 200/350 = 0, 571
P(S )
∩
= 50/350 = 0, 143 F ) = 38/350 = 0, 109
{}
− P(A) = 1 − 0, 4 = 0, 6
A∩B ∩ B) = P({c}) = 0, 2 A∪B
P(C )
{}
∩ B) = 0, 1
p
= 0, 2
P(B)
=p
P(A
∪ B) = 0, 5
P(A
|B)
P(A P(B)
=0
P(A
|B)
P(A
∩ B) , |B) = P(AP(B)
P(B)
>0
|B)
Ω = (1, 1), ..., (1, 6), (2, 1), ..., (2, 6), (3, 1), ..., (3, 6), (4, 1), ..., (4, 6), (5, 1),..., (5, 6), (6, 1),..., (6, 6)
{
}
A = (3, 4) B= P(A) = 1/36 P(B) = 18/36 = 1/2
{
}
= (1, 1),..., (1, 6), (3, 1), ..., (3, 6), (5, 1), ..., (5, 6) P(A
{
∩ B) = 1/36
∩ B) = 1/36 = 1 = 0, 056. |B) = P(AP(B) 1/2 18 = {(1, 1), (2, 1),..., (6, 1), (1, 3), (2, 3),..., (6, 3), (1, 5), (2, 5), ..., (6, 5)} = {(2, 1), ..., (2, 6), (4, 1),..., (4, 6), (6, 1),..., (6, 6)} P(D) = 18/36 = 1/2 P(C ∩ D) = 9/36 = 1/4 P(C ∩ D) 1/4 1 P(C |D) = = = = 0, 5. P(D) 1/2 2 P(A
C = D= P(C )
= 18/36 = 1/2
}
Ω
P(A
∩ B) = P(A|B)P(B),
P(B)
>0
P(A
∩ B) = P(B|A)P(A),
P(A)
>0
P(V
|A1) = 3/4 = 0, 75 P(B |A2) = 2/5 = 0, 4 P(B)
= P(B A1)+ P(B A2) = P(B A1)P(A1)+ P(B A2)P(A2) = 1/4 1/2+ 2/5 1/2 = 1/8+ 1/5 = 0, 325
∩
∩
P(M P(F
∩ N ) = P(N |M )P(M ) = 0, 2 × 0, 4 = 0, 08 ∩ N ) = P(N |F )P(F ) = 0, 5 × 0, 6 = 0, 30
P(A
|B) = P(A)
|
|
×
×
P(A
P(F |A) = P(F ) |A) = 762/3155 = 0, 242 P(F |S ) = 275/1503 = 0, 183
P(F
P(F
|S ) = P(F )
= 1037/4658 = 0, 223 P(F ) = 1037/4658 = 0, 223
P(F
|A) = P(F ) P(F |S ) = P(F )
0, 4
P(B)
= 0, 8
P(A) =
0, 5
A
B
P(A|B) =
0, 3
P(B)
= 0, 8
P(A) =
0, 3
B
Ac
= 0, 2
C 1 C 2
P(F )
P(A|B) =
P(A)
C i
∩ B) = P(A)P(B)
P(B) =
0, 2
A
B
C k
Ω
∩ C = ∅ ∀i = j j
k i=1
C i = Ω Ω C 2 C 1
C 1 C 2 P(C k ) P(A
C k
C 4
C 5
C 3
C 6
Ω
P(C 1 ) P(C 2 ) P(A
|C )
|C 1)
P(A
k
P(A) = P(A
∩ C 1 ) + P(A ∩ C 2 ) + ... + P(A ∩ C k ) = P(A|C 1 )P(C 1 ) + P(A|C 2 )P(C 2 ) + ... + P(A|C k )P(C k )
|C 2)
P(A
B Bc B) + P(A P(A) = P(A
∩
∩
Bc )
|B) = 0, 2
c
P(A
|B ) = 0, 3
P(B)
Ω = P(A B)P(B) + P(A B c )P(B c ) = 0, 2
|
|
C 1 C 2
= 0, 8
P(A) P(A)
× 0, 8 + 0, 3 × 0, 2 = 0, 16 + 0, 06 = 0, 22
C k
Ω
P (C 1 )
|
P (C 2 ) P (C k ) P (A C 2 ) P (A C k )
|
P (A C 1 )
|
P(C j
|A) = P(A|C )P(C ) . =1 P(A|C )P(C ) P(C |A) P(A|C ) j
k i
j
i
i
j
P(H )
= 0, 015
P(H P )
|
P(H P )
|
= =
P(H
∩ P ) =
j
P(P
|H ) = 0, 995
|H c ) = 0, 01
P(P
H )P(H ) H ) + P(P H c ) P(P H )P(H ) 0, 995 0, 015 = c c P(P H )P(H ) + P(P H )P(H ) 0, 995 0, 015 + 0, 01 P(P )
|
P(P
P(P
|
|
∩
|
∩
×
×
× 0, 985
=
0, 0149 = 0, 601 0, 0248
P(X =
xi ) = p(xi ) = pi , i = 1, 2,...
pi 0
x1
x2
x3
p1
p2
p3
≤p ≤1 i
pi P(X
= 1) = P(X = 2) = ... = P(X = 6) = 1/6
p = 1 i
i
pi
0 2 . 0
5 1 . 0 s e d a d i l i b a b o r P
0 1 . 0
5 0 . 0
0 0 . 0
2
3
4
5
6
7
8
9
10
11
12
Soma dos valores das duas faces dos dados
x F (x) = P(X
F (2) = P(X
≤ x) =
p(x ) i
xi ≤x
≤ 2) = P(X = 1) + P(X = 2) = 0, 245 + 0, 288 = 0, 533.
F (x) = P(X
≤ x) = 0, 533,
F (x) =
2 0 0, 245 0, 533 0, 789 0, 934 1
≤x<3 1 2 3 4
x < 1; x < 2; x < 3; x < 4; x < 5; x 5.
≤ ≤ ≤ ≤
≥
0 . 1
8 . 0
) x ( F
6 . 0
4 . 0
2 . 0
0 . 0
0
1
2
3
4
5
6
Número de doses de vacina
µ
E(X )
n
µ=E
x p(x ) (X ) = i
i
i=1
σ2
V (x)
n
2
σ = V (x) = E(X
− µ)
2
(x − µ) p(x ) = 2
i
i
i=1
n
2
2
2
σ = V (x) = E(X ) + ( E(X )) ,
E
x p(x ) (X ) = 2
2 i
i
i=1
11
E(X)
=
xi p(xi ) = 2
i=1
+9 11
E(X
2
)
=
× 4/36 + 10 × 3/36 + 11 × 2/36 + 12 × 1/36 = 7 x2i p(xi ) = 22
i=1
+92 V (X)
=
× 1/36 + 3 × 2/36 + 4 × 3/36 + 5 × 4/36 + 6 × 5/36 + 7 × 6/36 + 8 × 5/36 2
2
2
2
2
2
× 4/36 + 10 × 3/36 + 11 × 2/36 + 12 × 1/36 = 54, 833 E(X ) − (E(X)) = 51, 167 − 7 = 54, 833 − 49 = 5, 833 2
2
2
2
× 1/36 + 3 × 2/36 + 4 × 3/36 + 5 × 4/36 + 6 × 5/36 + 7 × 6/36 + 8 × 5/36
2
2
5
E(X)
=
xi p(xi ) = 1
i=1
× 0, 245 + 2 × 0, 288 + 3 × 0, 256 + 4 × 0, 145 + 5 × 0, 066 = 2, 499
5
E(X
2
)
=
x2i p(xi ) = 12
2
i=1
V (X)
P(X P(X
=
= x) =
= 4)
P(X
E(X
2
2x+1 25
)
2
2
2
× 0, 245 + 2 × 0, 288 + 3 × 0, 256 + 4 × 0, 145 + 5 × 0, 066 = 7, 671 2
− (E (X))
= 7, 671
2
− 2, 499
= 1, 426
, x = 0, 1, 2, 3, 4
≤ 1)
P(2
≤ X < 4)
P(X
> −1)
X
X
F (x) =
P(X
≤ 50)
P(X
≤ 40)
P(40
0
x < −10;
0, 25 0, 75
− 10 ≤ x < 30; 30 ≤ x < 50;
1
x ≥ 50.
≤ X < 60)
P(X
< 0)
P(0
≤ X < 10)
P(−10
< X < 10)
X
F (x) =
X
P(X
0
x < 10;
0, 2
10 ≤ x < 12;
0, 5
12 ≤ x < 13;
0, 9
13 ≤ x < 25;
1
x ≥ 25.
≤ 12)
P(X
< 12)
P(12
≤ X ≤ 20)
P(X
> 18)
X
CO 2 CO 2
CO 2
f (x)
CO 2
X
f (x)
≥ 0, ∀x ∈ (−∞, ∞) ∞ −∞
f (x) P(a
b a
≤ X ≤ b) = f (x)dx =
f (x)dx = 1
f (x)
P(X
a
b
= k) = 0
k [a, b], [a, b), (a, b] e (a, b)
a
P(X
f (x) = x/8, 3 < x < 5 P(X > 3, 5) P(4 < X < 5)
< 4)
< 4) =
P(X
> 3, 5) =
< 4, 5) =
P(X
< 3, 5)
dx =
5 x 3,5 8
1 8
dx =
2
4
1 8
=
3
x2 2
5 x 4 8
dx =
1 8
x2
4,5 x 3 8
dx =
1 8
x2
< X > 5) =
P(X
x2
− − − −
4 x 3 8
P(X
P(4
b
2
2
1 8
16 2
5
3,5 5 4
4,5 3
1 8
=
=
9 2
1 8
=
> 4, 5) P(X < 3, 5 X > 4, 5) = (1 = 0, 2031 + 0, 2969 = 0, 5
=
1 8
16 2
< 4, 5)
P(X
< 3, 5)
P(X
> 4, 5)
= 0, 4375
12,25 2
25 2
25 2
7 16
P(X
=
20,25 2
9 2
12,75 16
=
9 16
=
= 0, 7969
= 0, 5625 11,25 16
= 0, 7031
P(X
∪
− P(X > 3, 5)) + (1 − P(X < 4, 5)) = (1 − 0, 7969) + (1 − 0, 7031)
X x
F (x) = P(X f (x)
≤ x) =
F (x) f (x) =
d F (x) dx
0 F (x) = 0,12x P(X
P(1
< 2, 8)
< X < 2)
P(X
< 2, 8) = F (2, 8) = 0, 2
−∞
f (u)du,
−∞
× 2, 8 = 0, 56 P(X > 1, 5) = 1 − F (1, 5) = 1 − 0, 2 × 1, 5 = 1 − 0, 3 = 0, 7
x < 0; 0 x < 5; x 5;
≤
≥
P(X
> 1, 5)
P(X
<
−2)
P(X
> 6)
P(X
<
−2) = F (−2) = 0 P(X > 6) = 1 − F (6) = 1 − 1 = 0 P(1 < X < 2) = P(X < 2) − P(X < 1) = F (2) − F (1) = 0, 2 × 2 − 0, 2 × 1 = 0, 4 − 0, 2 = 0, 2
f (x) ∞
E(X )
=µ=
xf (x)dx
−∞
f (x) ∞
2
σ = V (x) =
− µ)2f (x)dx
(x
−∞
σ 2 = E(X 2 )
− µ2
∞
2
E(X
)=
x2 f (x)dx
−∞
f (x) = x/8, 3 < x < 5 5
E(X)
=
5
xf (x)dx =
3
E(X
2
3
5
)
=
V (X)
=
5
x2 f (x)dx =
3
x 1 x dx = 8 8
3
E(X
2
)
2
− (E(X))
x3 3
1 8
3
x 4
2
− 4, 083
=
4
x 1 x2 dx = 8 8
= 17
5
125 3
5
=
3
1 8
− 273
625 4
=
98 = 4, 083 24
=
− 814
544 = 17 32
= 0, 329
f (x) = 1, 5x2 , −1 < x < 1 P(X
X
> 0)
P(X
> 0, 5)
x
P(X
P(−0, 5
< X < 0, 5)
P(X
< −2)
> x) = 0, 05
X
F (x) =
X
P(X
0
x < −2;
0, 25x + 0, 5
− 2 ≤ x < 2; x ≥ 2;
1
< 1, 8)
P(X
> −1, 5)
P(X
X
Y f (x) =
8 9
y− 0
4 9
0, 5 ≤ y < 2;
< −2)
P(−1
< X < 1)
P(Y
< 0, 8)
P(Y
> 1, 5|Y ≥ 1)
Y
X
f (x) = X σ2
1 √ e σ 2π
−(x−µ)2 2σ 2
µ
,
X
µ
f(x)
x
µ
f (x)
µ
→0
x
→ ±∞
f (x)
x=µ µ
σ2
b
P(a
< X < b) =
a
(µ, σ 2 )
σ2
−∞ < x < ∞, σ2 > 0, µ ∈ R
∼ N (µ, σ2)
f (x)
Y
1 √ e σ 2π
−(x−µ)2 2σ 2
dx
X
∼ N (µ, σ2) X
Z = Z
−µ
σ
∼ N (0, 1)
X
P(a
µ
< X < b)
a−−µµ< X X −−µ µ< b −b −µ)µ < < σ σ σ a − µ b−µ
=
P (a
=
P
=
P
σ
σ
σ
P(Z
< z)
f(x)
0
P(Z
z
> 0) = P(Z < 0) = 0, 5 1, 5) = P(Z > 1, 5) P(Z < P(Z < 1, 5) = 1 P(Z < 1, 5)
−
X
∼ N (2, 9)
P(2
< X < 5)
−
−
P(X
> 4)
x
∈ [a, b]
∼ N (6, 1) Z ∼ N (0, 1) X
P(2 P(X
> 4) = P
P(5 P(X
< X < 5) = P
−2 X√ 9
< X < 7) = P
< 3) = P
P(0
X −6 1
P(5
P(0
< z < a) = 0, 3212
−2 2√
−2 < 5√ −2 < X√
9
9
−2 > 4√ 9
< X < 7)
5−6 1
<
6 < 3− 1
9
P(X
< 3)
= P(0 < Z < 1) = 0, 8413
= P(Z > 2/3) = P(Z > 0, 67) = 1 X −6
6 < 7− 1
− 0, 5 = 0, 3413
− 0, 7486 = 0, 2514
− × P(0 < Z < 1) = 2 × (0, 8413 − 0, 5) = 0, 6826 = P(Z < −3) = 1 − P(Z < 3) = 1 − 0, 9987 = 0, 0013 1
= P ( 1 < Z < 1) = 2
< z < a) = 0, 3212
f(x)
f(x)
0
1
x
P(2
0
x
< X < 5)
f(x)
P(X
> 4)
P(X
< 3)
f(x)
−1
0
1
x
P(5
X
P(X
0.67
−3
< X < 7)
∼ N (10/3, 2/3)
b
P(X
< b) = 0, 025
c
P(X
> c) = 0, 050
d
P(X
e
P(3
< X < e) = 0, 500
√
b−10/3 1/3
< b) = 0, 025
→P
> d) = 0, 350
X −10/3 1/3
<
√
=P Z<
b−10/3 1/3
√
= 0, 025
0
x
b−10/3 1/3
√
P(X
c−10/3 1/3
√
P(X
d−10/3 1/3
√
P(3
= 0, 39
10 3
1 3
X −10/3 1/3
→ c = 0, 39
< X < e) = 0, 50
c−10/3 1/3
√
1/3 + 10/3 = 2, 20
=P Z>
c−10/3 1/3
= 0, 05
c−10/3 1/3
= 0, 35
√
1/3 + 10/3 = 4, 28
→ √ × → √ P
>
→ c = 1, 64 ×
> d) = 0, 35
−1, 96 → b = −1, 96 ×
→ P X√− / /
> c) = 0, 05 = 1, 64
=
>
d−10/3 1/3
√
=P Z>
√
1/3 + 10/3 = 3, 56
3−10/3
P
1/3
<
X −10/3 1/3
√
e−10/3 1/3
− √ = P −1 < Z < e√ / = 0, 50 / P(X < −1) = 0, 1587 0, 5 e− / √ / = 0, 41 → c = 0, 41 × 1/3 + 10/3 = 3, 57 0, 41
<
10 3 1 3
10 3
0, 6587
1 3
¯ µ X σ/ n
−√ →
n→∞
N (0, 1)
x
µ
x ¯
σ
√
µ
σ/ n
n > 30
¯ X X X µ = 5, 4
σ2
¯ > 5) = P P(X
= 4, 44
¯ X
− 5, 4
4, 44/40
>
5
− 5, 4
4, 44/40
= P(Z >
−1, 20) = 1 − P(X < 1, 20) = 0, 8849.
X ∼ N (4, 1) P(X
≤ 4)
P(4
< X < 5)
P(2
< X < 5)
P(5
≤ X ≤ 7)
P(X
X ∼ N (−5, 10) P(−5
< X ≤ −2)
P(X
≤ 0)
P(X
> −6)
P(−7
≤ X ≤ −6)
X ∼ N (5/4, 1/9) P(X
≤ 7/5)
P(0
≤ X ≤ 6/5)
P(X
≤ 3/5)
P(X
≥ 3)
N (2, 2)
≤ 1)
P(2
≤ X ≤ 2)
