1
$\begingroup$

I am a bit confused whether I should use a hypergeometric or a multinomial distribution when I encounter a questions having more than two X_i and I kinda remember that for multinomial distribution the sum of the theta for all the X_i would be equal to 1.

I have two questions as the examples while one of them is for hypergeometric and the other is for multinomial distribution.

  1. A quality control engineer inspects a random sample of two hand-held calculators from each incoming lot of size 18 and accepts the lot if they are both in good work- ing condition; otherwise, the entire lot is inspected with the cost charged to the vendor. What are the probabilities that such a lot will be accepted without further inspection if it contains (a) 4 calculators that are not in good working condition; (b) 8 calculators that are not in good working condition; (c) 12 calculators that are not in good working condition?

This above one is hypergeometric distribution

  1. The probabilities are 0.40, 0.50, and 0.10 that, in city driving, a certain kind of compact car will average less than 28 miles per gallon, from 28 to 32 miles per gallon, or more than 32 miles per gallon. Find the probability that among 10 such cars tested, 3 will average less than Special Probability Distributions 28 miles per gallon, 6 will average from 28 to 32 miles per gallon, and 1 will average more than 32 miles per gallon.

The above one is multinomial distribution.

I know they are both not in the same situations, but the two distributions have similar forms of formulas, and that was why I got confused and hope to clarify more and help myself make the right judgement.

Thank you guys in advance for any help!

Regards,

$\endgroup$
3
$\begingroup$

One important difference is that the hypergeometric distribution assumes sampling without replacement, and the multinomial assumes sampling with replacement. A second important difference is that there are two categories for the (regular) hypergeometric distribution and there may be $k \ge 2$ categories for the multinomial distribution (see the Comment by @whuber).

Let's look at a specific example of each. I will illustrate the respective computations without answering all of the cases listed in your question.

Hypergeometric. The finite population from which you are sampling is a lot of $T =18$ calculators. Suppose that $a=4$ or the 18 are defective (and $T-a =14$ are good). The engineer samples $n = 2$ calculators without replacement. The hypergeometric random variable $X$ is the number of defective calculators among the two sampled. You seek $$P(\text{Accept Lot}) = P(X = 0) = \frac{{a \choose 0}{b \choose 2}}{{T \choose 2}} = 0.5948.$$

dhyper(0, 4, 14, 2)
[1] 0.5947712
choose(14, 2)/choose(18,2)
[1] 0.5947712

Sometimes one approximates a hypergeometric probability by a binomial one. That involves falsely assuming that sampling is with replacement. The difference between sampling with and without replacement is small if the sample size $n$ is small compared with the population size $T.$ (A rule of thumb is to use a binomial approximation only if $n/T < 0.1.$ In the current example this condition is not quite met. If $X^\prime \sim \mathsf{Binom}(n = 2,\, p = 4/18),$ then $P(X^\prime = 0) = 0.6049,$ not a horrible approximation of $0.5948,$ but not optimal, especially not when the exact alternative involves such a simple hypergeometric computation.)

dbinom(0, 2, 4/18)
[1] 0.6049383

Multinomial. Suppose you have three mileage categories $(A, B, C)$ for a certain model of car in city driving. The respective probabilities of these categories for any one car are $p_a = .4,$ $p_b = .5,$ $p_c = .1,$ where $p_a + p_b + p_c = 1.$ You select $n = 4$ cars at random. (Selection is either with replacement or the population of this model of car is so large that there is essentially no chance of choosing the same car twice.) The probability of getting exactly one car of type A, two of type B, and one of type C is $$P(A=1,\,B=2,\,C=3) = {4 \choose 1,2,1}p_a^1 p_b^2 p_c^1 = 0.12,$$ where the 'multinomial coefficient' is computed as $\frac{4!}{1!\cdot 2! \cdot 1!}=12.$ There are 12 possible arrangements of the required letters: ABBC, BABC, BBAC, BBCA, ..., CBBA.

12*.4*.5^2*.1
[1] 0.12

The probability that of $n = 10$ such cars selected at random exactly three will be of type A is (precisely) a binomial probability with $n = 10$ and $p_a = .4.$ The result is 0.2150.

dbinom(3, 10, .4)
[1] 0.2149908
$\endgroup$
  • 2
    $\begingroup$ +1. It occurred to me after reading your post that you could tie the two distributions together by discussing the multivariate hypergeometric distribution, which might make the distinctions even clearer. Here's a table: $$\begin{array}{rcc}& \text{2 categories} & \text{n categories} \\ \hline \text{With replacement:} & \text{Binomial} & \color{red}{\text{Multinomial}} \\ \text{Without replacement:} & \color{red}{\text{Hypergeometric}} & \text{Multivariate hypergeometric} \end{array}$$ $\endgroup$ – whuber Sep 30 '18 at 21:10
  • $\begingroup$ @whuber: Thanks, added mention of this to my first paragraph. $\endgroup$ – BruceET Sep 30 '18 at 21:38
  • $\begingroup$ Thank you very much for all the help!! It makes sense to me. Thanks!! :) Much appreciation to both of you! $\endgroup$ – Chen Sep 30 '18 at 23:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.