Question

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

Answer 1

@Michele_Laino

Answer 2

@ganeshie8

Answer 3

Hey there ! Please help me!

Answer 4

Do you know Laino?

Answer 5

Or it is out of your area of experience?

Answer 6

when i was at my university, I used the binomial distribution only

Answer 7

binominal distribution is used when a system gives equal chance of something to happen right?

Answer 8

For example the flipping of the coins

Answer 9

Do you know the formula to binominal distribution?

Answer 10

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

Answer 11

so total probability to get a 3 say 200 times, is:

\[\Large P\left( 3 \right) = \left( {\begin{array}{*{20}{c}}
{500} \\
{200}
\end{array}} \right){\left( {\frac{1}{6}} \right)^{200}}{\left( {\frac{5}{6}} \right)^{300}}\]

Answer 12

that is a possible application for a binomial distribution

Answer 13

The formula provided explains what happens when one rolls a dice for 300 times where he gets a 3?

Answer 14

that formula, gives us the probability to get the number 3, 200 times, if I throw my dice 500 times

Answer 15

oh ok. So the formula in itself is already justified.

Answer 16

You were the brightest pickle in the jar:) Others are just peeking and going away lol

Answer 17

thanks! I try my best!!

Answer 18

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)