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(Lesson 11: Multiplication Rule; 44)
4.16
Solution to Example 2
Because we are drawing cards with replacement, the draws are independent.
P
A1st and K2nd and K3rd
( )
=
P
A1st
( )
⋅
P
K2nd
( )
⋅
P
K3rd
( )
=
1
13
⋅
1
13
⋅
1
13
=
1
2197
≈
0.000455
( )
Tree
PART B: CONDITIONAL PROBABILITY
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Unformatted text preview: P B A ( ) = the “updated” probability that B occurs, given that A occurs. Technical Note: The idea of “updating” probabilities is a popular idea among Bayesian statisticians, though it is controversial....
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This note was uploaded on 12/29/2011 for the course MATH 119 taught by Professor Kim during the Fall '09 term at SUNY Stony Brook.
 Fall '09
 KIM
 Multiplication

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