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(Lesson 11: Multiplication Rule; 4-4)
4.16
Solution to Example 2
Because we are drawing cards with replacement, the draws are independent.
P
A-1st and K-2nd and K-3rd
(
)
=
P
A-1st
(
)
⋅
P
K-2nd
(
)
⋅
P
K-3rd
(
)
=
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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- Fall '09
- KIM
- Multiplication, Conditional Probability, Probability, Probability theory, Probability space
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