Insights Blog
-- Browse All Articles --
Physics Articles
Physics Tutorials
Physics Guides
Physics FAQ
Math Articles
Math Tutorials
Math Guides
Math FAQ
Education Articles
Education Guides
Bio/Chem Articles
Technology Guides
Computer Science Tutorials
Forums
Intro Physics Homework Help
Advanced Physics Homework Help
Precalculus Homework Help
Calculus Homework Help
Bio/Chem Homework Help
Engineering Homework Help
Trending
Featured Threads
Log in
Register
What's new
Search
Search
Search titles only
By:
Intro Physics Homework Help
Advanced Physics Homework Help
Precalculus Homework Help
Calculus Homework Help
Bio/Chem Homework Help
Engineering Homework Help
Menu
Log in
Register
Navigation
More options
Contact us
Close Menu
JavaScript is disabled. For a better experience, please enable JavaScript in your browser before proceeding.
You are using an out of date browser. It may not display this or other websites correctly.
You should upgrade or use an
alternative browser
.
Forums
Homework Help
Precalculus Mathematics Homework Help
Probability of At Least 1 Child Receiving 6 or 7 Candies
Reply to thread
Message
[QUOTE="haruspex, post: 5004809, member: 334404"] At the risk of confusing you further, there is at least one more way of looking at it. Many combinatorial problems involve arrange items into groups under constraints, but counting the distinct arrangements rather than discussing probability. A neat trick is used. In my last example, four objects into two cells, consider 4+2-1=5 things in a line. These represent the four objects and the dividing line between the two cells. The number of ways of partitioning the objects into the cells is the same as the number of ways of choosing which 'thing' is the dividing line: 5. When we apply the 'at least one in each cell' rule it drops to three. As a basis for some probability question about those distributions, they are equally likely. So we now have three different interpretations yielding three different probability distributions: 2/7, 3/7, 2/7; 1/4, 1/2, 1/4; 1/3, 1/3, 1/3. This last view gives you likely the easiest version of your original problem. For (a) I get 0.9 (almost exactly). The given answer just looks wild. I'd love to know how anyone came up with it. [/QUOTE]
Insert quotes…
Post reply
Forums
Homework Help
Precalculus Mathematics Homework Help
Probability of At Least 1 Child Receiving 6 or 7 Candies
Back
Top