# AP Statistics Curriculum 2007 Distrib Multinomial

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 Revision as of 19:05, 7 November 2009 (view source)IvoDinov (Talk | contribs)m (→See also: typo)← Older edit Current revision as of 01:35, 17 April 2013 (view source)IvoDinov (Talk | contribs) m (→Examples of Multinomial experiments) (12 intermediate revisions not shown) Line 1: Line 1: ==[[AP_Statistics_Curriculum_2007 | General Advance-Placement (AP) Statistics Curriculum]] - Multinomial Random Variables and Experiments== ==[[AP_Statistics_Curriculum_2007 | General Advance-Placement (AP) Statistics Curriculum]] - Multinomial Random Variables and Experiments== - The multinomial experiments (and multinomial distributions) directly extend the their [[AP_Statistics_Curriculum_2007_Distrib_Binomial |bi-nomial counterparts]]. + The multinomial experiments (and multinomial distributions) directly extend their [[AP_Statistics_Curriculum_2007_Distrib_Binomial |bi-nomial counterparts]]. ===Multinomial experiments=== ===Multinomial experiments=== Line 13: Line 13: *  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}''? *  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: + *To solve this problem, we apply the multinomial formula, we know the following: ** The experiment consists of 5 trials, so k = 5. ** The experiment consists of 5 trials, so k = 5. ** The 5 trials produce 0 red, 2 green marbles, and 3 blue marbles; so $r_1=r_{red} = 0$, $r_2=r_{green} = 2$, and $r_3=r_{blue} = 3$. ** The 5 trials produce 0 red, 2 green marbles, and 3 blue marbles; so $r_1=r_{red} = 0$, $r_2=r_{green} = 2$, and $r_3=r_{blue} = 3$. Line 20: Line 20: Plugging these values into the multinomial formula we get the probability of the event of interest to be: Plugging these values into the multinomial formula we get the probability of the event of interest to be: - : $P(A) = {5\choose 0, 2, 3}p_1^{r_1}p_2^{r_2}p_3^{r_3}$ + : $$P(A) = {5\choose r_1,r_2,r_3}p_1^{r_1}p_2^{r_2}p_3^{r_3}$$. In this specific case, $$P(A) = {5\choose 0,2,3}p_1^{0}p_2^{2}p_3^{3}$$. : $P(A) = {5! \over 0!\times 2! \times 3! }\times (2/9)^0 \times (1/3)^2\times (4/9)^3=0.0975461.$ : $P(A) = {5! \over 0!\times 2! \times 3! }\times (2/9)^0 \times (1/3)^2\times (4/9)^3=0.0975461.$ Line 35: Line 35: * The Binomial vs. Multinomial '''Coefficients''' (See this [http://www.ohrt.com/odds/binomial.php Binomial Calculator]) * The Binomial vs. Multinomial '''Coefficients''' (See this [http://www.ohrt.com/odds/binomial.php Binomial Calculator]) - : ${n\choose i}=\frac{n!}{k!(n-k)!}$ + : ${n\choose i}=\frac{n!}{i!(n-i)!}$ : ${n\choose i_1,i_2,\cdots, i_k}= \frac{n!}{i_1! i_2! \cdots i_k!}$ : ${n\choose i_1,i_2,\cdots, i_k}= \frac{n!}{i_1! i_2! \cdots i_k!}$ Line 44: Line 44: a_1^{i_1} \times a_2^{i_2} \times \cdots \times a_k^{i_k}}[/itex] a_1^{i_1} \times a_2^{i_2} \times \cdots \times a_k^{i_k}}[/itex] - * The Binomial vs. Multinomial '''Probabilities''' (See this [http://socr.ucla.edu/Applets.dir/Normal_T_Chi2_F_Tables.htm Binomial distribution calculator]) + * The Binomial vs. Multinomial '''Probabilities''' (See this [http://socr.ucla.edu/Applets.dir/Normal_T_Chi2_F_Tables.htm Binomial distribution calculator] and the [http://socr.ucla.edu/htmls/dist/Multinomial_Distribution.html SOCR Multinomial Distribution calculator]) : $p=P(X=r)={n\choose r}p^r(1-p)^{n-r}, \forall 0\leq r \leq n$ : $p=P(X=r)={n\choose r}p^r(1-p)^{n-r}, \forall 0\leq r \leq n$ : $p=P(X_1=r_1 \cap X_2=r_2 \cap \cdots \cap X_k=r_k | r_1+r_2+\cdots+r_k=n)=$ : $p=P(X_1=r_1 \cap X_2=r_2 \cap \cdots \cap X_k=r_k | r_1+r_2+\cdots+r_k=n)=$ Line 68: Line 68: ===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={3 ones, 3 twos, 2 threes, and 2 fours}. 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). + Suppose we roll 10 loaded hexagonal (6-face) dice 8 times and we are interested in the probability of observing the event A={3 ones, 3 twos, 2 threes, and 2 fours}. 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" {| class="wikitable" style="text-align:center; width:75%" border="1" Line 79: Line 79: : ''P(A)=?'' Note that the complete description of the event of interest is: : ''P(A)=?'' Note that the complete description of the event of interest is: - : A={3 ones, 3 twos, 2 threes, 2 fours, and 2 others (5's or 6's!)} + : A={3 ones, 3 twos, 2 threes, 2 fours, and 0 others (5's or 6's!)} ====Exact Solution==== ====Exact Solution==== Line 85: Line 85: * By-hand calculations: * By-hand calculations: - : $P(A) = {10! \over 3!\times 3! \times 2! \times 2! \times 2!} \times 0.286^3 \times 0.238^3\times 0.19^2 \times 0.143^2 \times 0.143^2 =$ + : $P(A) = {10! \over 3!\times 3! \times 2! \times 2! \times 0! \times 0!} \times 0.286^3 \times 0.238^3\times 0.19^2 \times 0.143^2 \times 0.095^0 \times 0.048^0=$ :$=0.00586690138260962656816896.$ :$=0.00586690138260962656816896.$ - * Using the [http://socr.ucla.edu/htmls/dist/Multinomial_Distribution.html SOCR Multinomial Distribution Calculator]: Enter the above given information in the [http://socr.ucla.edu/htmls/dist/Multinomial_Distribution.html SOCR Multimonial distribution applet] to get the probability density and cumulative distribution values for the given outcome ''{3,3,2,2,0}'', as shown on the image below. + * Using the [http://socr.ucla.edu/htmls/dist/Multinomial_Distribution.html SOCR Multinomial Distribution Calculator]: Enter the above given information in the [http://socr.ucla.edu/htmls/dist/Multinomial_Distribution.html SOCR Multimonial distribution applet] to get the probability density and cumulative distribution values for the given outcome ''{3,3,2,2,0}'', as shown on the image below. Note that since the event A does not contain the die outcomes of 5 or 6 , we can reduce the case of 6 outcomes $\{1,2,3,4,5,6 \}$ to a case of 5 outcomes $\{1,2,3,4,other \}$ with corresponding probabilities $\{0.286, 0.238, 0.19, 0.143, 0.143 \}$, where we pool together the probabilities of the die outcomes of 5 and 6, i.e., P(other)=0.095+0.048=0.143.
[[Image:SOCR_EBook_Dinov_Multinimial_102209_Fig2.png|500px]]
[[Image:SOCR_EBook_Dinov_Multinimial_102209_Fig2.png|500px]]
Line 102: Line 102: : $P(A) \approx {8 \over 1,000}=0.008$. : $P(A) \approx {8 \over 1,000}=0.008$. - Note that that this approximation is close to the exact answer above. By the [[AP_Statistics_Curriculum_2007_Limits_LLN | Law of Large Numbers (LLN)]], we know that this SOCR empirical approximation to the exact multinomial probability of interest will significantly improve as we increase the number of trials in this experiment to 10,000. + Note that this approximation is close to the exact answer above. By the [[AP_Statistics_Curriculum_2007_Limits_LLN | Law of Large Numbers (LLN)]], we know that this SOCR empirical compared to the exact multinomial probability of interest will significantly improve as we increase the number of trials in this experiment to 10,000. - ===[[hEBook_Problems_Distrib_Multinomial|Problems]]=== + ===[[EBook_Problems_Distrib_Multinomial|Problems]]=== ===See also=== ===See also=== - [[AP_Statistics_Curriculum_2007_Distrib_Dists| Negative Binomial and other Distributions]] + * [[AP_Statistics_Curriculum_2007_Distrib_Dists| Negative Binomial and other Distributions]] - *[[AP_Statistics_Curriculum_2007_Distrib_Dists#Negative_Multinomial_Distribution_.28NMD.29| Negative Multinomial Distribution]] + * [[AP_Statistics_Curriculum_2007_Distrib_Dists#Negative_Multinomial_Distribution_.28NMD.29| Negative Multinomial Distribution]]

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

The multinomial experiments (and multinomial distributions) directly extend 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 4/9, respectively. Hence, p1 = pred = 2 / 9, p2 = pgreen = 1 / 3, and p3 = pblue = 4 / 9.

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

$$P(A) = {5\choose r_1,r_2,r_3}p_1^{r_1}p_2^{r_2}p_3^{r_3}$$. In this specific case, $$P(A) = {5\choose 0,2,3}p_1^{0}p_2^{2}p_3^{3}$$.
$P(A) = {5! \over 0!\times 2! \times 3! }\times (2/9)^0 \times (1/3)^2\times (4/9)^3=0.0975461.$

Thus, if we draw 5 marbles with replacement from the urn, the probability of drawing no red, 2 green, and 3 blue marbles is 0.0975461.

• Let's again use the urn containing 9 marbles, where the number of red, green and blue marbles are 2, 3 and 4, respectively. This time we select 5 marbles from the urn, but are interested in the probability (P(B)) of the event B={selecting 2 green marbles}! (Note that 2 < 5)
• To solve this problem, we classify balls into green and others! Thus the multinomial experiment consists of 5 trials (k = 5), r1 = rgreen = 2, r2 = rother = 3. In this case, the probabilities of drawing a green or other marble are 3/9, and 6/9, respectively. Notice now the P(other) is the sum of the probabilities of the other colors (complement of green)! Hence,
$P(B) = {5\choose 2, 3}p_1^{r_1}p_2^{r_2} = {5! \over 2! \times 3! }\times (3/9)^2 \times (6/9)^3=0.329218.$

This probability is equivalent to the binomial probability (success=green; failure=other color), B(n=5, p=1/3).

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

${n\choose i}=\frac{n!}{i!(n-i)!}$
${n\choose i_1,i_2,\cdots, i_k}= \frac{n!}{i_1! i_2! \cdots i_k!}$
• The Binomial vs. Multinomial Formulas
$(a+b)^n = \sum_{i=1}^n{{n\choose i}a^i \times b^{n-i}}$
$(a_1+a_2+\cdots +a_k)^n = \sum_{i_1+i_2\cdots +i_k=n}^n{ {n\choose i_1,i_2,\cdots, i_k} a_1^{i_1} \times a_2^{i_2} \times \cdots \times a_k^{i_k}}$
$p=P(X=r)={n\choose r}p^r(1-p)^{n-r}, \forall 0\leq r \leq n$
$p=P(X_1=r_1 \cap X_2=r_2 \cap \cdots \cap X_k=r_k | r_1+r_2+\cdots+r_k=n)=$
$={n\choose i_1,i_2,\cdots, i_k}p_1^{r_1}p_2^{r_2}\cdots p_k^{r_k}, \forall r_1+r_2+\cdots+r_k=n$

### Expectation and variance

The expected number of times for observing the outcome i, over n trials, is

E(Xi) = npi.

Since each diagonal entry is the variance of a binomially distributed random variable, the variance-covariance matrix is defined by:

Diagonal-terms (variances): VAR(Xi) = npi(1 − pi), for each i, and
Off-diagonal terms (covariances): COV(Xi,Xj) = − npipj, for $i\not= j$.

### 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 roll 10 loaded hexagonal (6-face) dice 8 times and we are interested in the probability of observing the event A={3 ones, 3 twos, 2 threes, and 2 fours}. 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)=? Note that the complete description of the event of interest is:
A={3 ones, 3 twos, 2 threes, 2 fours, and 0 others (5's or 6's!)}

#### Exact Solution

Of course, we can compute this probability exactly:

• By-hand calculations:
$P(A) = {10! \over 3!\times 3! \times 2! \times 2! \times 0! \times 0!} \times 0.286^3 \times 0.238^3\times 0.19^2 \times 0.143^2 \times 0.095^0 \times 0.048^0=$
= 0.00586690138260962656816896.
• Using the SOCR Multinomial Distribution Calculator: Enter the above given information in the SOCR Multimonial distribution applet to get the probability density and cumulative distribution values for the given outcome {3,3,2,2,0}, as shown on the image below. Note that since the event A does not contain the die outcomes of 5 or 6 , we can reduce the case of 6 outcomes {1,2,3,4,5,6} to a case of 5 outcomes {1,2,3,4,other} with corresponding probabilities {0.286,0.238,0.19,0.143,0.143}, where we pool together the probabilities of the die outcomes of 5 and 6, i.e., P(other)=0.095+0.048=0.143.

#### Approximate Solution

We can also find a pretty close empirically-driven estimate using the SOCR Dice Experiment.

For instance, running the SOCR Dice 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. In one such experiment of 1,000 trials, there were 8 outcomes of the type {3 ones, 3 twos, 2 threes and 2 fours}. Therefore, the relative proportion of these outcomes to 1,000 will give us a fairly accurate estimate of the exact probability we computed above

$P(A) \approx {8 \over 1,000}=0.008$.

Note that this approximation is close to the exact answer above. By the Law of Large Numbers (LLN), we know that this SOCR empirical compared to the exact multinomial probability of interest will significantly improve as we increase the number of trials in this experiment to 10,000.