drtdrums Posted July 11, 2013 Share Posted July 11, 2013 I've had a couple questions like this in the past and received help promptly, so I figured I'd try again. Thanks in advance! Physiological Condition 1 has an epidemiological occurrence of 0.00026/1. Physiological Condition 2 has an epidemiological occurrence of 0.034/1. Physiological Condition 3 has an epidemiological occurrence of 0.00008/1. What is the epidemiological probability, stated as a percentage, that one person has all three conditions? Link to comment Share on other sites More sharing options...
dfitzo53 Posted July 11, 2013 Share Posted July 11, 2013 For the probability of more than one independent event just multiply them all together. Link to comment Share on other sites More sharing options...
drtdrums Posted July 11, 2013 Author Share Posted July 11, 2013 For the probability of more than one independent event just multiply them all together. Thanks! Lol. Now I remember. Trying to help a friend's kid with his math, and it's been so long I can't remember anything. Link to comment Share on other sites More sharing options...
pointyfootball Posted July 11, 2013 Share Posted July 11, 2013 .00000007%? Link to comment Share on other sites More sharing options...
skinfaninny Posted July 11, 2013 Share Posted July 11, 2013 You can calculate the probability of two or more independent events by multiplying their individual probabilities. P(A and B and C) = P(A) x P( B ) x P( C ) .00026 x .034 x .00008 = .0000000007072 Link to comment Share on other sites More sharing options...
CrypticVillain Posted July 11, 2013 Share Posted July 11, 2013 .00000007%? I think you are missing two 0s... You can calculate the probability of two or more independent events by multiplying their individual probabilities. P(A and B and C) = P(A) x P( B ) x P( C ) .00026 x .034 x .00008 = .0000000007072 The full answer reminds me of an old teacher I had... She would have given me half credit if I didn't put the full answer because I didn't round up. Because of that I almost missed out on getting straight A's by 2 points but my mother went up and raised all kinds of hell lol.... Link to comment Share on other sites More sharing options...
drtdrums Posted July 11, 2013 Author Share Posted July 11, 2013 Would stating the correct probability as "1 in _____" format be: 1 in 1,414,027,149.32 ? Link to comment Share on other sites More sharing options...
skinfaninny Posted July 11, 2013 Share Posted July 11, 2013 .00000007%? I think you are missing two 0s... >You can calculate the probability of two or more independent events by multiplying their individual probabilities. P(A and B and C) = P(A) x P( B ) x P( C ) .00026 x .034 x .00008 = .0000000007072 The full answer reminds me of an old teacher I had... She would have given me half credit if I didn't put the full answer because I didn't round up. Because of that I almost missed out on getting straight A's by 2 points but my mother went up and raised all kinds of hell lol.... I had a teacher like that. I learned to never round off the answer unless it specifically states you can. You'll never get partial credit for giving the complete answer. Would stating the correct probability as "1 in _____" format be: 1 in 1,414,027,149.32 ? Correct. To get the 1 in xxx format, just divide 1 by the probability. Link to comment Share on other sites More sharing options...
skinfaninny Posted July 11, 2013 Share Posted July 11, 2013 Gawd I hate statistics! My worst class in college by far. My professor used to work for NASA on the Apollo project as a statistician. I always said that he was probably the sole cause of Apollo 13. His lectures were all over the place and he kept covering the same things using different terminology which confused all of us. My notes were a mess I ended up skipping lectures and doing all of my studying from the textbook. By doing that I brought my grade from a B- to an A. Of course after that I did the same thing for the rest of my classes, only showing up for labs and exams, and raised my GPA by over 1 point. I think I have ADD and just couldn't focus in class. Link to comment Share on other sites More sharing options...
TD_washingtonredskins Posted July 11, 2013 Share Posted July 11, 2013 Is your "friend" a hot chick? If so, I hope these answers help. Link to comment Share on other sites More sharing options...
CrypticVillain Posted July 11, 2013 Share Posted July 11, 2013 Gawd I hate statistics! My worst class in college by far. My professor used to work for NASA on the Apollo project as a statistician. I always said that he was probably the sole cause of Apollo 13. His lectures were all over the place and he kept covering the same things using different terminology which confused all of us. My notes were a mess I ended up skipping lectures and doing all of my studying from the textbook. By doing that I brought my grade from a B- to an A. Of course after that I did the same thing for the rest of my classes, only showing up for labs and exams, and raised my GPA by over 1 point. I think I have ADD and just couldn't focus in class. Man, I can't even do that at my school. If we miss a certain amount of classes we get an automatic F. That is university wide. Link to comment Share on other sites More sharing options...
pointyfootball Posted July 11, 2013 Share Posted July 11, 2013 .00000007%? I think you are missing two 0s... Didn't OP want it as percentage? Link to comment Share on other sites More sharing options...
CrypticVillain Posted July 11, 2013 Share Posted July 11, 2013 .00000007%? I think you are missing two 0s... Didn't OP want it as percentage? Oh, I didn't notice that. So you multiplied it times 100 then. My bad. I am in summer mode right now lol Link to comment Share on other sites More sharing options...
Larry Posted July 11, 2013 Share Posted July 11, 2013 Of course, observing that were all assuming that these three probabilities are independent. Often when we're talking about medical conditions and a single person, they aren't. But then, if they aren't independent, then you can't answer the question. Link to comment Share on other sites More sharing options...
pointyfootball Posted July 11, 2013 Share Posted July 11, 2013 Of course, observing that were all assuming that these three probabilities are independent. Often when we're talking about medical conditions and a single person, they aren't. But then, if they aren't independent, then you can't answer the question. Ok Sheldon. Link to comment Share on other sites More sharing options...
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