Introduction to Statistics: Difference between revisions

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At this point it's worth remembering that all events are ''not'' disjoint events. For events that are not disjoint, we end up with the following probability definition.

How can we see this from example? Well, let's consider drawing from a deck of cards. I'll define two ''events'': "drawing a Queen", and "drawing a Spade". ~~We can tell off the bat~~ that these are not disjoint events, because you can draw a queen that is also a spade. There are four queens in the deck, so if we perform the experiment of drawing a card, putting it back in the deck and shuffling (what statisticians refer to as ''sampling with replacement'', we will end up with a probability of <math>\frac{1}{13}</math> for a queen draw. By the same argument, we obtain a probability for drawing a spade as <math>\frac{1}{4}</math>. The expression <math>P(A\cup B)</math> here can be translated as "the chance of drawing a queen or a spade". If we assume naively (as I used to) that for this case <math>P(A\cup B) = P(A) + P(B)</math>, we can simply add our probabilities together for "the chance of drawing a queen or a spade" as <math>\frac{1}{13}+\frac{1}{4}</math>. If we were to gather some data experimentally, we would find that our results would differ from the prediction -- the probability observed would be slightly less than <math>\frac{1}{13}+\frac{1}{4}</math>. Why? Because we're counting the queen of spades twice in our expression, once as a spade, and again as a queen. We need to count it only once, as it can only be drawn with probability of <math>\frac{1}{52}</math>. [[Still confused?]]

How can we see this from example? Well, let's consider drawing from a deck of cards. I'll define two ''events'': "drawing a Queen", and "drawing a Spade". It is immediately clear that these are not disjoint events, because you can draw a queen that is also a spade. There are four queens in the deck, so if we perform the experiment of drawing a card, putting it back in the deck and shuffling (what statisticians refer to as ''sampling with replacement'', we will end up with a probability of <math>\frac{1}{13}</math> for a queen draw. By the same argument, we obtain a probability for drawing a spade as <math>\frac{1}{4}</math>. The expression <math>P(A\cup B)</math> here can be translated as "the chance of drawing a queen or a spade". If we assume naively (as I used to) that for this case <math>P(A\cup B) = P(A) + P(B)</math>, we can simply add our probabilities together for "the chance of drawing a queen or a spade" as <math>\frac{1}{13}+\frac{1}{4}</math>. If we were to gather some data experimentally, we would find that our results would differ from the prediction -- the probability observed would be slightly less than <math>\frac{1}{13}+\frac{1}{4}</math>. Why? Because we're counting the queen of spades twice in our expression, once as a spade, and again as a queen. We need to count it only once, as it can only be drawn with probability of <math>\frac{1}{52}</math>. [[Still confused?]]

'''Proof:''' If <math>A</math> and <math>B</math> are not disjoint, we have to avoid the double counting problem by exactly specifying their union.

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 At this point it's worth remembering that all events are ''not'' disjoint events.  For events that are not disjoint, we end up with the following probability definition.<br>
 <math> P(A\cup B) = P(A) + P(B) - P(A\cap B)</math><br>
-How can we see this from example?  Well, let's consider drawing from a deck of cards.  I'll define two ''events'': "drawing a Queen", and "drawing a Spade".  We can tell off the bat that these are not disjoint events, because you can draw a queen that is also a spade.  There are four queens in the deck, so if we perform the experiment of drawing a card, putting it back in the deck and shuffling (what statisticians refer to as ''sampling with replacement'', we will end up with a probability of <math>\frac{1}{13}</math> for a queen draw.  By the same argument, we obtain a probability for drawing a spade as <math>\frac{1}{4}</math>.  The expression <math>P(A\cup B)</math> here can be translated as "the chance of drawing a queen or a spade".  If we assume naively (as I used to) that for this case <math>P(A\cup B) = P(A) + P(B)</math>, we can simply add our probabilities together for "the chance of drawing a queen or a spade" as <math>\frac{1}{13}+\frac{1}{4}</math>.  If we were to gather some data experimentally, we would find that our results would differ from the prediction -- the probability observed would be slightly less than <math>\frac{1}{13}+\frac{1}{4}</math>.  Why?  Because we're counting the queen of spades twice in our expression, once as a spade, and again as a queen.  We need to count it only once, as it can only be drawn with probability of <math>\frac{1}{52}</math>.  [[Still confused?]] <br>
+How can we see this from example?  Well, let's consider drawing from a deck of cards.  I'll define two ''events'': "drawing a Queen", and "drawing a Spade".  It is immediately clear that these are not disjoint events, because you can draw a queen that is also a spade.  There are four queens in the deck, so if we perform the experiment of drawing a card, putting it back in the deck and shuffling (what statisticians refer to as ''sampling with replacement'', we will end up with a probability of <math>\frac{1}{13}</math> for a queen draw.  By the same argument, we obtain a probability for drawing a spade as <math>\frac{1}{4}</math>.  The expression <math>P(A\cup B)</math> here can be translated as "the chance of drawing a queen or a spade".  If we assume naively (as I used to) that for this case <math>P(A\cup B) = P(A) + P(B)</math>, we can simply add our probabilities together for "the chance of drawing a queen or a spade" as <math>\frac{1}{13}+\frac{1}{4}</math>.  If we were to gather some data experimentally, we would find that our results would differ from the prediction -- the probability observed would be slightly less than <math>\frac{1}{13}+\frac{1}{4}</math>.  Why?  Because we're counting the queen of spades twice in our expression, once as a spade, and again as a queen.  We need to count it only once, as it can only be drawn with probability of <math>\frac{1}{52}</math>.  [[Still confused?]] <br>
 '''Proof:''' If <math>A</math> and <math>B</math> are not disjoint, we have to avoid the double counting problem by exactly specifying their union.<br>

Introduction to Statistics: Difference between revisions

wikademia>Zach Garner