Database Reference
In-Depth Information
#balls). Multiplying gives (#red/balls)
2
(1-#red/balls)
3
, generally (#red/
#balls)
i
(1-#red/#balls)
n-i
.
The only thing that remains is to take into account the 10 different
orderings. This comes from first considering each of the balls to be
individual numberings and then looking at the number of ways to order 5
numbered balls. There are 5×4×3×2×1 = 120 ways to do that. This is usually
represented as 5!, which is called a
factorial function
. But, the red and
non-red balls are indistinguishable from each other, so this method results
in overcounting. By the same argument, there are 2! ways of ordering the 2
red balls and 3! ways of ordering the 3 non-red balls. Each of the final 10
orderings is really being overcounted by 2!×3! orderings, so dividing gives
5!/(2!×3!) = 10 orderings. In other words, there are “5 choose 2” possible
combinations. This generally is stated as “n choose k,” which is the number
of different ways that k objects can be chosen from a group of n objects. This
number of possible combinations is given by the formula n!/(k!×(n-k)!).
Expectation and Variance
What about the average number of red balls drawn if the experiment where
5 balls are drawn is repeated an infinite number of times? A statistician asks
the question, “What are the expected number of red balls?” and calls the
answerthe
expectation
.Instatisticsliterature,thisisusuallywrittenasE[X]
(with square brackets), and X is known as a
random variable
.
Thedefinitionofanexpectationisquitesimple.Forexamplesliketheearlier
one, where there are discrete events such as drawing 2 balls, the expectation
is simply the sum of each possible value of X multiplied by the probability of
that value of X, as follows:
More generally, the expectation is actually defined for any function of X,
written as E[f(x)] and having the same form for computation:
