Function Posted April 1, 2014 Share Posted April 1, 2014 Hello everyone Here's a question from the acceptance exam for Med school: 6,000 people undergo an ABO-bloodtest. 1,846 are not positive for antigen A, nor B. 2,527 are positive for antigen A, while 2,234 are positive for antigen B. How many are positive for both antigens? First thing that came up in my head, was to substract First thing that came up was Boolean algebra, where: [math]6 000 = \neg(A\vee B) + A + B + (A\wedge B)[/math] (O, A, B and AB) [math]1 846 = \neg(A\vee B)[/math] (O) [math]2 527 = A\wedge (A\wedge B)[/math] (A and AB) [math]2 234 = B\wedge (A\wedge B)[/math] (B and AB) But there's a problem: [math]A\wedge (A\wedge B) = (A\wedge A)\wedge B = A\wedge B = 2 527[/math] and [math]B\wedge (A\wedge B) = \cdots = A\wedge B = 2 234[/math] Which obviously isn't right. Can anyone help me on this one? Thanks. F. Link to comment Share on other sites More sharing options...

John Posted April 1, 2014 Share Posted April 1, 2014 (edited) Your third and fourth equations should be [math]2527 = A \vee (A \wedge B)[/math] and [math]2234 = B \vee (A \wedge B)[/math], respectively, though I don't think Boolean algebra really works here, since it deals with truth values not general numbers. I think you should instead look at this in terms of sets and make use of the principle of inclusion-exclusion. We know [math]|A| = 2527[/math], [math]|B| = 2234[/math], and [math] |A \cup B| = 6000 - 1846 = 4154[/math], so we just need to solve for [math]|A \cap B|[/math]. Edited April 1, 2014 by John Link to comment Share on other sites More sharing options...

Function Posted April 1, 2014 Author Share Posted April 1, 2014 Your third and fourth equations should be [math]2527 = A \vee (A \wedge B)[/math] and [math]2234 = B \vee (A \wedge B)[/math], respectively, though I don't think Boolean algebra really works here, since it deals with truth values not general numbers. I think you should instead look at this in terms of sets and make use of the principle of inclusion-exclusion. We know [math]|A| = 2527[/math], [math]|B| = 2234[/math], and [math] |A \cup B| = 6000 - 1846 = 4154[/math], so we just need to solve for [math]|A \cap B|[/math]. Ah yes, thanks The result is thus 607. Is there btw any difference between e.g. [math]\wedge[/math] and [math]\cup[/math]? Link to comment Share on other sites More sharing options...

John Posted April 1, 2014 Share Posted April 1, 2014 If you take a look at the laws of Boolean algebra and set algebra, you'll notice the logical and set operators share quite a few properties. This perhaps becomes clearer if you consider set builder notation, where we say things like [math]A \cup B = \{ x \mid x \in A \vee x \in B \}[/math] or [math]A^{C} = \{ x \mid \neg ( x \in A) \}[/math] (though for the latter, we'd usually say [math]x \not\in A[/math] instead). That is to say, we define our set operations in terms of combinations of sentences connected or modified by logical operators. The difference, then, lies in what objects are used with each set of operators. The logical operators are used for combining or modifying logical statements, while the set operators are used for denoting various collections of elements from sets or their complements. Link to comment Share on other sites More sharing options...

Function Posted April 1, 2014 Author Share Posted April 1, 2014 (edited) If you take a look at the laws of Boolean algebra and set algebra, you'll notice the logical and set operators share quite a few properties. This perhaps becomes clearer if you consider set builder notation, where we say things like [math]A \cup B = \{ x \mid x \in A \vee x \in B \}[/math] or [math]A^{C} = \{ x \mid \neg ( x \in A) \}[/math] (though for the latter, we'd usually say [math]x \not\in A[/math] instead). That is to say, we define our set operations in terms of combinations of sentences connected or modified by logical operators. The difference, then, lies in what objects are used with each set of operators. The logical operators are used for combining or modifying logical statements, while the set operators are used for denoting various collections of elements from sets or their complements. Now, for probabilities; let's define events A and B, then the probability of 'not A' happening, would be [math]P(A^c)[/math] or [math]P(\neg A)[/math], the probability of B and A happening = [math]P(B\cap A)[/math] or [math]P(B\wedge A)[/math]? Edited April 1, 2014 by Function Link to comment Share on other sites More sharing options...

John Posted April 1, 2014 Share Posted April 1, 2014 (edited) The notation for probabilities sort of depends on what exactly you're trying to say, how precise you're trying to be, etc. As an example, take the case of rolling one six-sided die. Then our sample space is [math]\Omega = \{1, 2, 3, 4, 5, 6\}[/math]. Now say we're interested in the probability that the result of a single die roll (denoted [math]X[/math]) will be both less than five and even. As a shorthand, we might write [math]P(X \textnormal{ is less than 5 and } X \textnormal{ is even})[/math] or [math]P(X < 5 \wedge 2 | X)[/math]. If we were writing a paper or something, then we'd probably want to take [math]P[/math] as the function I described in your recent thread, and letting [math]A = \{1, 2, 3, 4\}[/math] and [math]B = \{2, 4, 6\}[/math], we'd write [math]P(A \cap B)[/math]. Of course, it's easy enough to see that [math]A = \{x \in \Omega \mid x < 5\}[/math] and [math]B = \{x \in \Omega \mid 2 | x\}[/math], so writing [math]P(A \wedge B)[/math] probably wouldn't get you lynched by the mathematical community, but it'd look a bit strange. Edited April 1, 2014 by John 1 Link to comment Share on other sites More sharing options...

Function Posted April 1, 2014 Author Share Posted April 1, 2014 (edited) The notation for probabilities sort of depends on what exactly you're trying to say, how precise you're trying to be, etc. As an example, take the case of rolling one six-sided die. Then our sample space is [math]\Omega = \{1, 2, 3, 4, 5, 6\}[/math]. Now say we're interested in the probability that the result of a single die roll (denoted [math]X[/math]) will be both less than five and even. As a shorthand, we might write [math]P(X \textnormal{ is less than 5 and } X \textnormal{ is even})[/math] or [math]P(X < 5 \wedge 2 | X)[/math]. If we were writing a paper or something, then we'd probably want to take [math]P[/math] as the function I described in your recent thread, and letting [math]A = \{1, 2, 3, 4\}[/math] and [math]B = \{2, 4, 6\}[/math], we'd write [math]P(A \cap B)[/math]. Of course, it's easy enough to see that [math]A = \{x \in \Omega \mid x < 5\}[/math] and [math]B = \{x \in \Omega \mid 2 | x\}[/math], so writing [math]P(A \wedge B)[/math] probably wouldn't get you lynched by the mathematical community, but it'd look a bit strange. Ah, well... Very well explained. +1 Thanks. So it's mostly appreciated defining the sample space and sets with conditions for the probability? Edited April 1, 2014 by Function Link to comment Share on other sites More sharing options...

John Posted April 2, 2014 Share Posted April 2, 2014 If I'm reading your question correctly, then yes. The sample space is the set of all possible outcomes for a given experiment. Every individual outcome is a subset of the sample space. For our die roll example, the sample space is {1, 2, 3, 4, 5, 6}. The outcome that the result is odd is the subset {1, 3, 5}, the outcome that the result is 1 or 2 is {1, 2}, the outcome that the result is greater than 5 and odd is the empty set, etc. 1 Link to comment Share on other sites More sharing options...

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