MyWikiBiz, Author Your Legacy — Friday November 29, 2024
Jump to navigationJump to search
244 bytes added
, 02:44, 20 April 2009
Line 5,175: |
Line 5,175: |
| The condition that "men are just as apt to be black as things in general" can be expressed in terms of conditional probabilities as <math>\operatorname{P}(\mathrm{b}|\mathrm{m}) = \operatorname{P}(\mathrm{b}),</math> which means that the probability of the event <math>\mathrm{b}\!</math> given the event <math>\mathrm{m}\!</math> is equal to the unconditional probability of the event <math>\mathrm{b}.\!</math> | | The condition that "men are just as apt to be black as things in general" can be expressed in terms of conditional probabilities as <math>\operatorname{P}(\mathrm{b}|\mathrm{m}) = \operatorname{P}(\mathrm{b}),</math> which means that the probability of the event <math>\mathrm{b}\!</math> given the event <math>\mathrm{m}\!</math> is equal to the unconditional probability of the event <math>\mathrm{b}.\!</math> |
| | | |
− | Thus, for example, it is sufficient to observe in the Othello setting that P(''b''|''m'') = 1/4 while P(''b'') = 1/7 in order to cognize the dependency, and thereby to tell that the ostensible arrow is anaclinically biased.
| + | In the Othello example, it is enough to observe that <math>\operatorname{P}(\mathrm{b}|\mathrm{m}) = \tfrac{1}{4}</math> while <math>\operatorname{P}(\mathrm{b}) = \tfrac{1}{7}</math> in order to recognize the bias or dependency of the sampling map. |
| | | |
− | This reduction of a conditional probability to an absolute probability, in the form P(''A''|''Z'') = P(''A''), is a familiar disguise, and yet in practice one of the ways that we most commonly come to recognize the condition of independence P(''AZ'') = P(''A'')P(''Z''), via the definition of a conditional probability according to the rule P(''A''|''Z'') = P(''AZ'')/P(''Z''). To recall the familiar consequences, the definition of conditional probability plus the independence condition yields P(''A''|''Z'') = P(''AZ'')/P(''Z'') = P(''A'')P(''Z'')/P(''Z''), to wit, P(''A''|''Z'') = P(''A''). | + | This reduction of a conditional probability to an absolute probability in the form <math>\operatorname{P}(A|Z) = \operatorname{P}(A)</math> is a familiar disguise, and yet in practice one of the ways that we most commonly come to recognize the condition of independence <math>\operatorname{P}(AZ) = \operatorname{P}(A)P(Z),</math> via the definition of a conditional probability according to the rule <math>\operatorname{P}(A|Z) = \frac{\operatorname{P}(AZ)}{\operatorname{P}(Z)}.</math> To recall the familiar consequences, the definition of conditional probability plus the independence condition yields <math>\operatorname{P}(A|Z) = \frac{\operatorname{P}(AZ)}{P(Z)} = \frac{\operatorname{P}(A)\operatorname{P}(Z)}{\operatorname{P}(Z)},</math> to wit, <math>\operatorname{P}(A|Z) = \operatorname{P}(A).</math> |
| | | |
| As Hamlet discovered, there's a lot to be learned from turning a crank. | | As Hamlet discovered, there's a lot to be learned from turning a crank. |