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Union and Intersection Part 3-Probability-Assignment Solution, Exercises of Probability and Statistics

Sir Tanika Mukopadhyay taught us Probability at Homi Bhabha National Institute. He gave us assignments so that we can practice what we learned in form of problems. Here is solution to those problems. Its main emphasis is on following points: Probability, Conditional, Multiplying, Occure, Jointly, Output, Input, Union, Mutually, Independent

Typology: Exercises

2011/2012

Uploaded on 08/03/2012

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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
∪∩
∪=
∩∪∩
=
∩+
=
=+
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1

Probability Homework Solution

(a) The results follow directly from the definition of conditional probability.

[ | ] [^ ] [ ]

P A B P A^ B

P B

=^ ∩

If , then [ ] 0 by Corollary 3 and thus [ | ] 0. If , then , and thus [ | ] [^ ]. [ ] If , then , and thus [ | ] [^ ] 1. [ ]

A B P A B P A B

A B A B A P A B P A

P B

A B A B B P A B P B

P B

(b)

If [ | ] [^ ] [ ], then multiplying both sides by [ ], we have: [ ] [ ] [ ] [ ] We then also have that [ | ] [^ ]^ [^ ] [^ ] [ ]. [ ] [ ] We conclude that if [ | ] [ ], then and tend t

P A B P A^ B P A P B

P B

P A B P A P B

P B A P A^ B^ P A P B P B

P A P A

P A B P A B A

o occur jointly.

(i)

[ | ] [^ ] for [ ] 0 [ ] 0 [ ], 0 [ ] 0 [ | ] , [ ] [ ] [ | ] 1 Therefore, 0 [ | ] 1

P A B P A^ B P B

P B

P A B P B P A B

A B B P A B P B P A B

P A B

(ii)

[ | ] [^ ]^ [^ ] 1 [ ] [ ]

B S P S B P S^ B^ P B

P B P B

(iii)

( |^ ) [(^ )^ ]

[ ]

[( ) ( )]

[ ]

[ ] [ ] (^) (Axiom III) [ ] [ | ] [ | ]

P A C B P^ A^ C^ B

P B

P A B C B

P B

P A B P C B

P B

P A B P C B

∪ =^ ∪^ ∩

= ∩^ ∪^ ∩

= ∩^ +^ ∩

2

Let X denote the input and Y the output.