May 25: Meeting 1

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The first meeting will be on campus Weds. May 25 at 6pm in AVW 3122. We will post worked solutions and discussion points to this task. 


  • Main concepts from the chapter

  • Pairwise mutually disjoint

  • Disjoint versus independent

  • "One minus" method, equivalence

  • Binomial, counting approaches

  • Questions 1, 3, 4, and 6


main concepts from the chapter

  • definition of event spaces
  • pairwise mutually disjoint
  • set notation for events
  • union bound
  • inclusion-excusion principle
  • independence
  • conditional probability
  • sampling with and without replacement
  • principle of deferred decisions
  • bayes' law
  • law of total probability


pairwise mutually disjoint 

we tried to understand if pairwise mutually disjoint was something distinct from plain old disjoint, or pairwise disjoint. if disjoint is defined as no intersection between events, then if something is pairwise disjoint then for every pair of events, then it must also be overall disjoint. ie, pairwise disjoint ==> disjoint. if only considering individual events, then disjoint ==> pairwise disjoint also. ie {1}, {2}, {3} are disjoint and pairwise disjoint. but this is not be true if considering sets of events. the wikipedia entry on disjoint sets gives the example of the sets {1,2}, {2,3}, {3, 1}. pointing out that the collection is disjoint (they do not intersect), but the sets are NOT pairwise disjoint. personally, i still find the formal distinction somewhat murky, even though it makes sense intuitively. 

i believe we decided the word "mutually" in "pairwise mutually disjoint" was simply redundant :). how can you be pairwise disjoint and not be pairwise mutually disjoint?  


disjoint versus independent

disjoint events are those which do not intersect. for example, a ball being red or blue is "disjoint" - the ball is EITHER red OR blue. but within the space of characteristics of a person, eg. "happy" or "smart", these are not disjoint. you can be both happy and smart.

disjoint-ness is orthogonal to the notion of independence. whether or not bob is has either of these qualities can be said to be independent of whether alice has these qualities. then again, perhaps you would say that the probability that a child of bob has these qualities is conditioned on (NOT independent) of whether bob has them. 


"one minus" method

we worked through some problems, like problem 1.3, and remembered that in many cases where the question asks "what is the probability that at least one x will happen?" can often be more easily calculated by calculating 1 minus the probability that NO x will happen.

for example, in question 3 it asks what is the probability that there will be at least one ace in the first two cards drawn from a randomly shuffled deck. we can determine this by either suming over the probabilities for each way there could be at least one ace-- ie, the probability of the first card being an ace, plus the probability of the second card being an ace, plus the probability of them both being an ace, ie,

(4/52) * (48/51) + (48/52) * (4/51) + (4/52) * (3/51)

or, we can just compute the probability of neither card being an ace:

1 - (48/52)*(47/51)

which are equal


Binomial Coefficient

We found a lot of the problems involved binomial coefficient. The wikipedia articles on binomial coefficient and Pascal's triangle are pretty good. 



Worked Solutions

We upload answers with problems we worked on Flickr :)

Discusión de la Tarea