AP Statistics Curriculum 2007 Distrib Multinomial

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The multinomial experiments (and multinomial distribtuions) directly extend the their [[AP_Statistics_Curriculum_2007_Distrib_Binomial |bi-nomial counterparts]].  
The multinomial experiments (and multinomial distribtuions) directly extend the their [[AP_Statistics_Curriculum_2007_Distrib_Binomial |bi-nomial counterparts]].  
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* Examples of Multinomial experiments
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===Multinomial experiments===
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** Rolling a hexagonal Die 5 times: Where the outcome space is the colection of 5-tuples, where each element is a value such that: <math>1\leq value\leq 6</math>.
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A multinomial experiment is an experiment that has the following properties:
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* The experiment consists of '''k repeated trials'''.
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* Each trial has a '''discrete''' number of possible outcomes.
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* On any given trial, the probability that a particular outcome will occur is '''constant'''.
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* The trials are '''independent'''; that is, the outcome on one trial does not affect the outcome on other trials.
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* The Multinomial random variable (RV): Mathematically, a (k) multinomial trial is modeled by a random variable
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====Examples of Multinomial experiments====
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<center><math>X(outcome) = \begin{cases}x_o,\\
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*  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}''?
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x_1,\\
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\cdots,\\
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x_k.\end{cases}</math></center>
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If <math>p_i=P(X=x_i)</math>, then:
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*To solve this problem, we apply the multinomial formula. We know the following:
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: ''expected value'' of X, <math>E[X]=\sum_{i=1}^k{x_i\times p_i}</math>.  
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** The experiment consists of 5 trials, so k = 5.
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: standard deviation of X, <math>SD[X]=\sqrt{\sum_{i=1}^k{(x_i-E[X])^2\times p_i}}</math>.
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** The 5 trials produce 0 red, 2 green marbles, and 3 blue marbles; so <math>r_1=r_{red} = 0</math>, <math>r_2=r_{green} = 2</math>, and <math>r_3=r_{blue} = 3</math>.
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** For any particular trial, the probability of drawing a red, green, or blue marble is 2/9, 3/9, and 5/9, respectively. Hence, <math>p_1=p_{red} = 2/9</math>, <math>p_2=p_{green} = 1/3</math>, and <math>p_3=p_{blue} = 5/9</math>.
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Plugging these values into the multinomial formula we get the probability of the event of interest to be:
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: <math>P(A) = {5\choose 0, 2, 3}p_1^{r_1}p_2^{r_2}p_3^{r_3}</math>
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: <math>P(A) = {5! \over 0!\times 2! \times 3! }\times (2/9)^0 \times (1/3)^2\times (5/9)^3=0.19052.</math>
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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===
===Synergies between Binomial and Multinomial processes/probabilities/coefficients===

Revision as of 23:50, 4 March 2008

Contents

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

  • The Binomial vs. Multinomial 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}}
  • The Binomial vs. Multinomial Probabilities
p=P(X=r)={n\choose i}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


References




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