# AP Statistics Curriculum 2007 Distrib Multinomial

(Difference between revisions)
 Revision as of 06:26, 6 March 2008 (view source)IvoDinov (Talk | contribs)m (→Synergies between Binomial and Multinomial processes/probabilities/coefficients)← Older edit Revision as of 06:50, 6 March 2008 (view source)IvoDinov (Talk | contribs) Newer edit → Line 51: Line 51: : P(3 by Air, 3 by Bus, 1 by Auto, 2 by Train) = ? : P(3 by Air, 3 by Bus, 1 by Auto, 2 by Train) = ? : P(2 by air) = ? : P(2 by air) = ? - ===SOCR Multinomial Examples=== ===SOCR Multinomial Examples=== + Suppose we row 10 loaded hexagonal (6-face) dice 8 times and we are interested in the probability of observing the event A={2 ones, 1 three, 2 fours and 3 sixes}. Assume the dice are loaded to the small outcomes according to the following probabilities of the 6 outcomes (''one'' is the most likely and ''six'' is the least likely outcome). +
+ {| class="wikitable" style="text-align:center; width:75%" border="1" + |- + | ''x'' || 1 || 2 || 3 || 4 || 5 || 6 + |- + | ''P(X=x)'' || 0.286 || 0.238 || 0.19 || 0.143 || 0.095 || 0.048 + |} +
+ + : ''P(A)=?'' + + Of course, we can compute this number exactly as: + + : $P(A) =$ + + However, we can also find a pretty close empirically-driven estimate using the [[SOCR_EduMaterials_Activities_DiceSampleExperiment | SOCR Dice Sample Experiment]]. + + For instance, running the [http://socr.ucla.edu/htmls/SOCR_Experiments.html SOCR Dice Sample Experiment] 1,000 times with number of dice n=10, and the loading probabilities listed above, we get an output like the one shown below. + +
[[Image:SOCR_Activities_DiceSampleExperiment_Chui_051807_Fig1.jpg|400px]]
+ + Now, we can actually count how many of these 1,000 trials generated the event ''A'' as an outcome. Then the relative proportion of these outcomes to 1,000 will give us a fairly accurate estimate of the exact probability we computed above + : $P(A) \approx {s \over 1,000}=$ . +

## General Advance-Placement (AP) Statistics Curriculum - Multinomial Random Variables and Experiments

The multinomial experiments (and multinomial distribtuions) directly extend the their bi-nomial counterparts.

### Multinomial experiments

A multinomial experiment is an experiment that has the following properties:

• The experiment consists of k repeated trials.
• Each trial has a discrete number of possible outcomes.
• On any given trial, the probability that a particular outcome will occur is constant.
• The trials are independent; that is, the outcome on one trial does not affect the outcome on other trials.

#### Examples of Multinomial experiments

• Suppose we have an urn containing 9 marbles. Two are red, three are green, and four are blue (2+3+4=9). We randomly select 5 marbles from the urn, with replacement. What is the probability (P(A)) of the event A={selecting 2 green marbles and 3 blue marbles}?
• To solve this problem, we apply the multinomial formula. We know the following:
• The experiment consists of 5 trials, so k = 5.
• The 5 trials produce 0 red, 2 green marbles, and 3 blue marbles; so r1 = rred = 0, r2 = rgreen = 2, and r3 = rblue = 3.
• For any particular trial, the probability of drawing a red, green, or blue marble is 2/9, 3/9, and 5/9, respectively. Hence, p1 = pred = 2 / 9, p2 = pgreen = 1 / 3, and p3 = pblue = 5 / 9.

Plugging these values into the multinomial formula we get the probability of the event of interest to be:

$P(A) = {5\choose 0, 2, 3}p_1^{r_1}p_2^{r_2}p_3^{r_3}$
$P(A) = {5! \over 0!\times 2! \times 3! }\times (2/9)^0 \times (1/3)^2\times (5/9)^3=0.19052.$

Thus, if we draw 5 marbles with replacement from the urn, the probability of drawing no red , 2 green, and 3 blue marbles is 0.19052.

### Synergies between Binomial and Multinomial processes/probabilities/coefficients

${n\choose i}=\frac{n!}{k!(n-k)!}$
${n\choose i_1,i_2,\cdots, i_k}= \frac{n!}{i_1! i_2! \cdots i_k!}$
• The Binomial vs. Multinomial Formulas
$(a+b)^n = \sum_{i=1}^n{{n\choose i}a^1 \times b^{n-i}}$
$(a_1+a_2+\cdots +a_k)^n = \sum_{i_1+i_2\cdots +i_k=n}^n{ {n\choose i_1,i_2,\cdots, i_k} a_1^{i_1} \times a_2^{i_2} \times \cdots \times a_k^{i_k}}$
$p=P(X=r)={n\choose r}p^r(1-p)^{n-r}, \forall 0\leq r \leq n$
$p=P(X_1=r_1 \cap X_1=r_1 \cap \cdots \cap X_k=r_k | r_1+r_2+\cdots+r_k=n)={n\choose i_1,i_2,\cdots, i_k}p_1^{r_1}p_2^{r_2}\cdots p_k^{r_k}, \forall r_1+r_2+\cdots+r_k=n$

### Example

Suppose we study N independent trials with results falling in one of k possible categories labeled 1,2,cdots,k. Let pi be the probability of a trial resulting in the ith category, where $p_1+p_2+\cdots++p_k =1$. Let Ni be the number of trials resulting in the ith category, where $N_1+N_2+\cdots++N_k = N$.

For instance, suppose we have 9 people arriving at a meeting according to the following information:

P(by Air) = 0.4, P(by Bus) = 0.2, P(by Automobile) = 0.3, P(by Train) = 0.1
• Compute the following probabilities
P(3 by Air, 3 by Bus, 1 by Auto, 2 by Train) = ?
P(2 by air) = ?

### SOCR Multinomial Examples

Suppose we row 10 loaded hexagonal (6-face) dice 8 times and we are interested in the probability of observing the event A={2 ones, 1 three, 2 fours and 3 sixes}. Assume the dice are loaded to the small outcomes according to the following probabilities of the 6 outcomes (one is the most likely and six is the least likely outcome).

 x 1 2 3 4 5 6 P(X=x) 0.286 0.238 0.19 0.143 0.095 0.048
P(A)=?

Of course, we can compute this number exactly as:

P(A) =

However, we can also find a pretty close empirically-driven estimate using the SOCR Dice Sample Experiment.

For instance, running the SOCR Dice Sample Experiment 1,000 times with number of dice n=10, and the loading probabilities listed above, we get an output like the one shown below.

Now, we can actually count how many of these 1,000 trials generated the event A as an outcome. Then the relative proportion of these outcomes to 1,000 will give us a fairly accurate estimate of the exact probability we computed above

$P(A) \approx {s \over 1,000}=$ .