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#1 |
76038 Posts |
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Hi. I have a math problem and in desperate need for help.
Imagine that you roll a dice 9 times. Let X stand for the number of sixes you get on those rolls. What is the probability distribution of X? What is the probability of you getting maximum 1 six? What is the probability of you getting at least 3 sixes? I'm thankful for all help :) |
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#2 | ||
Nov 2003
22·5·373 Posts |
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We can recommend some, if you like. The term "probability distribution" is nonsense at worst and poorly defined at best. Do you mean density function (pdf), or do you mean cumulative distribution function (cdf)? In either event, the answer to your question will be found in any book on probability. Also, repeat after me: Google is my friend. Quote:
reading. Noone is going to do your homework for you. I recommend Hogg & Craig as a good book. |
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#3 | |
"William"
May 2003
New Haven
2,371 Posts |
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#4 | |
Nov 2003
22×5×373 Posts |
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Teach someone to fish and they eat for a lifetime.... It is better to teach someone how/where to find answers for themselves rather than just handing them the answer. |
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#5 | |
"William"
May 2003
New Haven
2,371 Posts |
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We have very different opinions about what constitutes "handing him the answers." He asked a trivial question and you told him to go read a whole book. I told him what to look up in the index - he still has to read and understand the information. You told him to google, I told him what to google. He still has to find it and understand it. |
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#6 | |
"Lucan"
Dec 2006
England
2·3·13·83 Posts |
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Probabilty of the first 3 throws being a six and the remaining 6 throws not being a six is (1/6)3 *(5/6)6 There are 9!/(6!3!) ways of placing the 3 sixes (hence "Binomial" distribution). Last fiddled with by davieddy on 2011-07-14 at 15:28 |
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#7 | |
"Forget I exist"
Jul 2009
Dumbassville
20C016 Posts |
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#8 | |
"Lucan"
Dec 2006
England
2·3·13·83 Posts |
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I suspect the excercise comes from a chapter entitled "Binomial Distribution". BTW I was being intentionally helpful, in support of WBLipp. This isn't the homework forum, and the poster sounds as if (s)he is beyond "help" anyway. David |
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#9 |
Dec 2010
Monticello
5×359 Posts |
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This problem can also be solved with combinatorics....keeping the die rolls in order, there are 6^9 possible sequences of 9 rolls of six sided die...if you count the number of them with one six, you have the odds of a single six....
That is, suppose the six is in the first position...there are then 8^5 ways for the remaining 8 rolls to fall and leave you with exactly one six.... This should give an idea of what to do.... And RDS: Your effort is appreciated. Thank you. |
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#10 |
Oct 2007
Manchester, UK
2×5×137 Posts |
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#11 |
Bamboozled!
"๐บ๐๐ท๐ท๐ญ"
May 2003
Down not across
22·3·941 Posts |
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