Description

Patrick Gardocki
2-19
a)
A′ = 3,5,6,8
b)
A ∩ B = 1,2
c)
(A ∩ B) ∪ C = 1,2,3,4,5
d)
(B ∪ C)′ = 7,8
e)
(A ∩ B)′ ∪ C = 1,3,4,5,6,7,8
2-81
a)
528
P(A) = 628 = 0.2448461
52ˆ8/62ˆ8
## [1] 0.2448461
b)
P(B) = 106288 = 4.580011e − 7
10ˆ8/62ˆ8
## [1] 4.580011e-07
c)
P(B ≥ 1) = 1 − P(A) = 0.7551539
1 – 52ˆ8/62ˆ8
## [1] 0.7551539
d)
×102×526
P(B = 2) = 628 = 0.2535389
(choose(8,6)*10ˆ2*52ˆ6)/(62ˆ8)
## [1] 0.2535389 2-90
a)
P
(choose(5,1)*36ˆ5)/36ˆ6
## [1] 0.1388889
b)
P
(choose(5,1)*36ˆ5)/36ˆ6
## [1] 0.1388889
c) P
(5/36) * (5/36)
## [1] 0.01929012
d)
P(A ∪ B) = P(A) + P(B) − P(A ∩ B) =
(5/36) + (5/36) – (25/36ˆ2)
## [1] 0.2584877 2-141
a)
P(A|B′) = PP(A(B∩B′)′) = 1−PP(A(B)) = 1−PP(A(B))
P
∴ P(A|B′) =
((52ˆ8/62ˆ8)/(1-(10ˆ8/62ˆ8)))
## [1] 0.2448462
b)
P(A′ ∩ B) = P(A′|B)P(B) = P(B) = 621088 = 4.580011e − 7
B is a subset of A’ so P(A′|B) = 1
10ˆ8/62ˆ8
## [1] 4.580011e-07
c)
Let T be passwords with exactly 2 integers and I be passwords with at least 1 integer.
P Since P(T) is a subset of P
From 2-81 part d, P(T) = 0.2535 From 2-81 part c, P(I) = 0.7551

0.2535/0.7551
## [1] 0.3357171
2-179
a)
P(F|S) = 0.6;P(F|V ) = 0.04
P(F) = P(F|S)P(S) + P(F|V )P(V ) = 0.6 × 0.2 + 0.04 × 0.8 = 0.152
0.6*0.2 + 0.8*0.04
## [1] 0.152
b)
P(S|F) = P (FP|S(F)P)(S) = 0.06.×1520.2 = 0.78947
(0.6*0.2)/0.152
## [1] 0.7894737
c)
P(V |F′) = P (FP′|(VF)′P)(V ) = 01.−960×.1520.8 = 0.90566
(0.96*0.8)/(1-0.152)
## [1] 0.9056604
Q2
131 43 122 41 41 = 54,912
choose(13,1) * choose(4,3)*choose(12,2)*choose(4,1)ˆ2
## [1] 54912