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#21
11-30-2005, 05:56 PM
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Re: mutual exclusivity
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ok, my take on explaining this graphically and easily [img]/images/graemlins/smile.gif[/img] 1. Draw a square 2. Draw a vertical line down the middle of the square. Left half is A, right half is A' 3. Draw a horizontal line in the square (I drew it about 1/3 up from the bottom so it would look like about 35%). Below the horizontal line is B, above is B'. 4. Your square is now divided into 4 rectangular regions. 5. Finding your answer... -The top right box is P(A' and B') and equals 20%. -If the top right box is 20% that means the bottom right box is 30% (remember the right two boxes are A' and must equal 50%) -If the bottom right box is 30% then the bottom left box must be 5% (remember the bottom two boxes are B and must add to 35%). -But the bottom left box is P(A and B). Congrats you now know P(A and B) is 5%! Pokerscott [/ QUOTE ] There is an issue with this drawing which didn’t prevent you from getting the correct answer, but nevertheless it is important to discuss. Aside from the numbers, the drawing implies that A and B are independent, which they are not. The vertical line down the middle makes A occupy 1/2 of the total square, meaning that its probability is 1/2, and since B goes all the way across horizontally, it gets cut in half by the A/A'line, so that the fraction of B that overlaps A is also 1/2. That means that P(A) = P(A | B), or in words, the probability of being in A is the same as the probability of being in A given that you are in B. This means that A and B are independent by definition since it means that being in B has no bearing on the chance that we are in A. It is equivalent to saying that P(A and B) = P(A)*P(B), which can also be taken as the definition of independence, and this relation can also be seen from your drawing since P(A)*P(B) is the width of A multiplied by the width of B, which is the area of P(A and B). You wrote the number 0.05 in this box because that was the amount left over once you put the numbers in the other boxes, but there is no way that you could make the area of that box 0.05 if, as you say, the vertical line goes down the middle, and the other line is 35% from the bottom. The area is 0.5*0.35. In fact, no matter where you draw the vertical line, and no matter where you draw a horizontal line, as long as the lines go all the way across, you can never make P(A and B) anything other than P(A)*P(B), and they will always represent independent events. This aspect of your drawing doesn't impact your solution, and it's OK as long as we only interpret the drawing as a table of numbers. However, when you draw the line for B 35% from the bottom, that implies that the areas are part of the model, and that A and B are independent, which they are not. This could be corrected by using a circle instead of a square, and then drawing a diameter for A/A', and another chord for B/B' that does not get bisected by A/A'. The main reason I mention this is because many people have difficulty grasping the concept of independence. You have made a drawing which illustrates this concept beautifully, though this was unintentional. Often people confuse the concepts of independence and mutual exclusivity. In a diagram such as this one, mutual exclusivity would be shown by drawing A and B so that they do not overlap, so that P(A intersect B) = P(A and B) = 0. Clearly this bears no relationship to independence, and in fact, events with non-zero probability can never be both independent and mutually exclusive. I don’t know why the title of this thread is “mutual exclusivity”, since that applies to nothing in this problem. 
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#22
11-30-2005, 06:31 PM
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Re: mutual exclusivity
[ QUOTE ]
This aspect of your drawing doesn't impact your solution, and it's OK as long as we only interpret the drawing as a table of numbers. However, when you draw the line for B 35% from the bottom, that implies that the areas are part of the model, and that A and B are independent, which they are not. [/ QUOTE ] No argument here. I probably shouldn't have made hints to in the middle and 1/3 from the bottom. I was just using it as a convenient way to segment the problem so you could easily see what P(A' and B') implied and could use that information to figure out P(A and B). It is pretty obvious that the area inside each square does not correspond to the probability in the square (the left and right bottom boxes are the same size in the drawing but one is 5% and one is 30%!). Anyway, you are right, think of it just as a convenient way of have tabulating the four events. Pokerscott
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