anonymous
  • anonymous
binomial geometric hypergeometric For each one, Explain the conditions in which we would use it. Justify the formula used. Give an example of a situation in which it could be used
Mathematics
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schrodinger
  • schrodinger
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anonymous
  • anonymous
@Michele_Laino
anonymous
  • anonymous
@ganeshie8
anonymous
  • anonymous
Hey there ! Please help me!

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anonymous
  • anonymous
Do you know Laino?
anonymous
  • anonymous
Or it is out of your area of experience?
Michele_Laino
  • Michele_Laino
when i was at my university, I used the binomial distribution only
anonymous
  • anonymous
binominal distribution is used when a system gives equal chance of something to happen right?
anonymous
  • anonymous
For example the flipping of the coins
anonymous
  • anonymous
Do you know the formula to binominal distribution?
Michele_Laino
  • Michele_Laino
not necessarily. For example, if I have a dice, and i throw that dice say 500 time, then I measure the probability to get the number 3, then the requested probability is: 1/6, whereas the probability to get another number is 5/6
Michele_Laino
  • Michele_Laino
so total probability to get a 3 say 200 times, is: \[\Large P\left( 3 \right) = \left( {\begin{array}{*{20}{c}} {500} \\ {200} \end{array}} \right){\left( {\frac{1}{6}} \right)^{200}}{\left( {\frac{5}{6}} \right)^{300}}\]
Michele_Laino
  • Michele_Laino
that is a possible application for a binomial distribution
anonymous
  • anonymous
The formula provided explains what happens when one rolls a dice for 300 times where he gets a 3?
Michele_Laino
  • Michele_Laino
that formula, gives us the probability to get the number 3, 200 times, if I throw my dice 500 times
anonymous
  • anonymous
oh ok. So the formula in itself is already justified.
anonymous
  • anonymous
You were the brightest pickle in the jar:) Others are just peeking and going away lol
Michele_Laino
  • Michele_Laino
thanks! I try my best!!
anonymous
  • anonymous
binomial -- we have N independent trials with success rate p, and we're interested in the number of successes k out of these N trials; equivalently, we have some objects, and p is the proportion of objects that are special, and we're interested in the number of special objects k we get after picking N such objects if we sample objects *with replacement* (so the proportion p is constant) geometric -- we have N independent trials with success rate p, and we're interested in the number of trials k we need to get *one* success hypergeometric -- we have T objects, and M of them are special, and we're interested in the number of special objects k we get after picking n such objects if we sample *without replacement* (so the proportion changes from M/T with each object we pick)

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