1
Probability Homework Solution #3
2.48
(a) The results follow directly from the definition of conditional probability.
[]
[|] []
PA B
PAB PB
∩
=
If , then [ ] 0 by Corollary 3 and thus [ | ] 0.
[]
If , then , and thus [ | ] .
[]
[]
If , then , and thus [ | ] 1.
[]
AB PAB PAB
PA
AB ABA PAB PB
PB
AB ABB PAB PB
∩=∅ ∩ = =
⊂∩= =
⊃∩= ==
(b)
[]
If [ | ] [ ], then multiplying both sides
b
y [ ], we have:
[]
[][][]
[ ] [][]
We then also have that [ | ] [ ].
[] []
We conclude that if [ | ] [ ], then and tend t
PA B
PAB PA PB
PB
PA B PAPB
PA B PAPB
PB A PB
PA PA
PAB PA B A
∩
=>
∩>
∩
=> =
>o occur jointly.
2.49
(i)
[]
[ | ] for [ ] 0
[]
0[ ],0[] 0[|]
,[][] [|]1
Therefore,
0[|]1
PA B
PAB PB
PB
PA B PB PAB
ABB PAB PB PAB
PAB
∩
=>
≤∩ < ⇒≤
∩⊂ ∩ ≤ ⇒ ≤
≤≤
(ii)
[][]
[|] 1
[] []
PS B PB
BS PSB PB PB
∩
⊂⇒ = = =
(iii)
()
[( ) ]
|[]
[( ) ( )]
[]
[][]
(Axiom III)
[]
[|] [|]
PA C B
PA CB PB
PA B C B
PB
PA B PC B
PB
PAB PC B
∪∩
∪=
∩∪∩
=
∩+ ∩
=
=+
docsity.com
