Thursday, December 9, 2010
Wednesday, December 1, 2010
Unit IV - From Randomness to Probability... Part 1
By now you should have some basic probability rules down...
Like maybe P(A or B) = P(A) + P(B), assuming the two events A and B are independent. And also P(A and B) = P(A)*P(B), assuming the two events A and B are disjoint. And you also should be able to tell the difference between disjoint and independent.
And then the rules had to go and get more complicated (or did they). Lets remove those darn conditions and get a general rule that works for events disjoint, independent or not. The General Addition Rule states that P(A or B) = P(A) + P(B) - P(A and B). This works for both disjoint and, well, joint events. If A and B are disjoint then P(A and B) = 0 so you just get the first addition rule listed above, and if they are not disjoint then subtracting P(A and B) subtracts the overlap, in other words the part of A that was counted in P(A) and in P(B) so therefore counted twice so if not subtracted off, its double counted.
There is also The General Multiplication Rule P(A and B) = P(A)*P(B|A). Which if you didn't know that you probably thinking, "yeah that helps thanks I don't even know how to read that." Ok so you read it like this "The probability of both A and B happening is equal to the probability of A, times the probability of B, given A." I had you up to that given part didn't I. We're talking conditional probability here. P(B|A) means the probability of B, knowing that A has happened. Did you click the link on disjoint and independent above (or right there in front of you now)? The example in there is on drawing a card from a deck of 52. Here's another example with cards. If we draw two cards at random from a well shuffled deck, what is the probability that the first one is red, and the second one is also red. The probability of the first one being red is 26/52 or 1/2, if we don't put the first card back then the probability that the second one is red is not 1/2, its 25/51 because we have one red card out of the deck. So P(A) = 1/2 and P(B|A) = 25/51. So P(A and B) = (1/2)*(25/51).
Here's my favorite (and yes it is possible to have a favorite probability rule, you probably have a favorite quadrilateral and your not even a math teacher, admit it you like rhombus the best don't you). The Complement Rule this basically is a rule because something has got to happen. So I've got kids. If I tell you that the probability that my daughter is sleeping right now (at 10:50 pm), is 90% you should be able to give me the probability that she is not sleeping right? 10%. This is basically the complement rule in action. P(sleeping) = .9, then P(not sleeping) = 1 - .9 = .1. For any event it either happens or it doesn't. So something's got to happen, P(A) = 1 - P(~A) the ~A is read "not A" or the "complement of A." This rule is very helpful in "at least" or "at most" problems. For example if I draw three cards at random from a well shuffled deck of cards what is the probability that I draw at least one red? You could calculate the probability of drawing one red, drawing two reds, drawing three reds and then add them up or figure out the probability that you don't get a red, and subtract from 1. Since drawing no reds is the complement of drawing at least one red.
I'd say that's it in a nut shell, but I'm not a mime, and because there's more... but that's another post.
Like maybe P(A or B) = P(A) + P(B), assuming the two events A and B are independent. And also P(A and B) = P(A)*P(B), assuming the two events A and B are disjoint. And you also should be able to tell the difference between disjoint and independent.
And then the rules had to go and get more complicated (or did they). Lets remove those darn conditions and get a general rule that works for events disjoint, independent or not. The General Addition Rule states that P(A or B) = P(A) + P(B) - P(A and B). This works for both disjoint and, well, joint events. If A and B are disjoint then P(A and B) = 0 so you just get the first addition rule listed above, and if they are not disjoint then subtracting P(A and B) subtracts the overlap, in other words the part of A that was counted in P(A) and in P(B) so therefore counted twice so if not subtracted off, its double counted.
There is also The General Multiplication Rule P(A and B) = P(A)*P(B|A). Which if you didn't know that you probably thinking, "yeah that helps thanks I don't even know how to read that." Ok so you read it like this "The probability of both A and B happening is equal to the probability of A, times the probability of B, given A." I had you up to that given part didn't I. We're talking conditional probability here. P(B|A) means the probability of B, knowing that A has happened. Did you click the link on disjoint and independent above (or right there in front of you now)? The example in there is on drawing a card from a deck of 52. Here's another example with cards. If we draw two cards at random from a well shuffled deck, what is the probability that the first one is red, and the second one is also red. The probability of the first one being red is 26/52 or 1/2, if we don't put the first card back then the probability that the second one is red is not 1/2, its 25/51 because we have one red card out of the deck. So P(A) = 1/2 and P(B|A) = 25/51. So P(A and B) = (1/2)*(25/51).
Here's my favorite (and yes it is possible to have a favorite probability rule, you probably have a favorite quadrilateral and your not even a math teacher, admit it you like rhombus the best don't you). The Complement Rule this basically is a rule because something has got to happen. So I've got kids. If I tell you that the probability that my daughter is sleeping right now (at 10:50 pm), is 90% you should be able to give me the probability that she is not sleeping right? 10%. This is basically the complement rule in action. P(sleeping) = .9, then P(not sleeping) = 1 - .9 = .1. For any event it either happens or it doesn't. So something's got to happen, P(A) = 1 - P(~A) the ~A is read "not A" or the "complement of A." This rule is very helpful in "at least" or "at most" problems. For example if I draw three cards at random from a well shuffled deck of cards what is the probability that I draw at least one red? You could calculate the probability of drawing one red, drawing two reds, drawing three reds and then add them up or figure out the probability that you don't get a red, and subtract from 1. Since drawing no reds is the complement of drawing at least one red.
I'd say that's it in a nut shell, but I'm not a mime, and because there's more... but that's another post.
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