AP Statistics Curriculum 2007 Distrib Multinomial
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 \(r_1=r_{red} = 0\), \(r_2=r_{green} = 2\), and \(r_3=r_{blue} = 3\).
- For any particular trial, the probability of drawing a red, green, or blue marble is 2/9, 3/9, and 5/9, respectively. Hence, \(p_1=p_{red} = 2/9\), \(p_2=p_{green} = 1/3\), and \(p_3=p_{blue} = 5/9\).
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 (See this Binomial Calculator)
\[{n\choose i}=\frac{n!}{k!(n-k)!}\]
\[{n\choose i_1,i_2,\cdots, i_k}= \frac{n!}{i_1! i_2! \cdots i_k!}\]
- 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 (See this Binomial distribution calculator)
\[p=P(X=r)={n\choose r}p^r(1-p)^{n-r}, \forall 0\leq r \leq n\] \[p=P(X_1=r_1 \cap X_1=r_1 \cap \cdots \cap X_k=r_k | r_1+r_2+\cdots+r_k=n)={n\choose i_1,i_2,\cdots, i_k}p_1^{r_1}p_2^{r_2}\cdots p_k^{r_k}, \forall r_1+r_2+\cdots+r_k=n\]
Example
Suppose we study N independent trials with results falling in one of k possible categories labeled \(1,2, cdots, k\). Let \(p_i\) be the probability of a trial resulting in the \(i^{th}\) category, where \(p_1+p_2+\cdots++p_k =1\). Let \(N_i\) be the number of trials resulting in the \(i^{th}\) 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={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)=?
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 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 this one 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 \[P(A) \approx {8 \over 1,000}=0.008\] .
References
- SOCR Home page: http://www.socr.ucla.edu
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