Binomial and Multinomial Distribution

• Bernoulli Random Variable: Suppose that we have a random variable $X$ that has just two outcomes (e.g., success/failure) with probability $p$ and $1 - p = q$, respectively.
• We call the random variable Bernoulli random variable. By representing a random variable $X$ to be the number of successes,
  $$X=\left\lbrace \begin{array}{l} 1,~with ~probability~p,\\ 0,~with ~probability~1-p,\\ \end{array}\right.$$

• An experiment that involve the Bernoulli random variable is called the Bernoulli experiment or Bernoulli trial.
• The Bernoulli Probability Distribution: The probability distribution of $X$ is given by
  $$P (X = x) = p(x) = p^xq^{1-x},~ x = 0, 1$$
  where $p$ is called the parameter of Bernoulli probability distribution.

• Since $X$ can be only 0 and 1;
  $$E(X) = 1*p + 0 *q = p, \sigma_X^2 = E(X^2) - [E(X)]^2 = p - p^2 = pq.$$

• Example 5.2: Generalization of the model of tossing a coin: $S = \{Success, Failure\}$ with $P(Success) = p$ and $P(Failure) = 1 - p = q$.
• An experiment often consists of repeated trials, each with two possible outcomes that may be labeled success or failure.
• The properties of the Bernoulli Process
  • The experiment consists of $n$ repeated trials.
  • Each trial is called Bernoulli trial. Each trial results in an outcome that may be classified as a success or a failure.
  • We may choose to define either outcome as a success.
  • The probability of success, denoted by $p$, remains constant from trial to trial.
  • The repeated trials are independent.

• Binomial Distribution. A Bernoulli trial can result in a success with probability $p$ and a failure with probability $q = 1 - p$.
• Then the probability distribution of the binomial random variable $X$, the number of successes in $n$ independent trials, is
  
• where
  $$\left( \begin{array}{c} n\\ x\\ \end{array}\right)$$

  : is the number of sample points that have $x$ successes.

  $$B(r;n,p)=\sum_{x=0}^r b(x;n,p)$$

  : is the binomial sums.

• A random variable with this probability distribution is said to be binomially distributed.
• The values of the binomial sums can be found in Table A.1

Figure: Binomial Probability Sums $B(r;n,p)=\sum_{x=0}^r b(x;n,p)$.

• Example: Suppose a professional basket player tries 5 free throws. The player is known to make 80% successful rate.
• Let $X$ be the number of free throws he will make, then $X\sim b(n = 5, p = 0.8)$.
  

  Table 1: Possible cases of 4 successes (o) and 1 miss (x) among 5 trials with $p = 0.8, q = 0.2$.

  Trial	Possible Event	$p(x)$	Probability
  1	oooox	ppppq	$(0.8)^4(0.2)^1$ = 0.08192
  2	oooxo	pppqp	$(0.8)^4(0.2)^1$ = 0.08192
  3	ooxoo	ppqpp	$(0.8)^4(0.2)^1$ = 0.08192
  4	oxooo	pqppp	$(0.8)^4(0.2)^1$ = 0.08192
  5	xoooo	qpppp	$(0.8)^4(0.2)^1$ = 0.08192

  
  Since there are $$\left( \begin{array}{c} 5\\ 4\\ \end{array}\right)=5$$ cases of making 4 successes among 5 trials,
  $$P (X = 4) = 5(0.8)^4(0.2)^1 = 0.4096.$$

• Example 5.4: The probability that a certain kind of component will survive a given shock test is $\frac{3}{4}$.
• Find the probability that exactly 2 of the next 4 components tested survive.
  $$b(x;n,p)=$$
  $$b(2;4,\frac{3}{4})=$$
  $$\left( \begin{array}{c} 4\\ 2\\ \end{array}\right) \left( \frac{3}{4}\right)^2 \left( \frac{1}{4}\right)^2=$$
  $$\frac{4!}{2!2!}*\frac{3^2}{4^4}=\frac{27}{128}$$

• Where Does the Name Binomial Come From? Binomial distribution corresponds to the binomial expansion of $(q + p)$

since $p+q=1$

• Example 5.5: The probability that a patient recovers from a rare blood disease is 0.4.
• If 15 people are known to have contracted this disease, what is the probability that

  i

  at least 10 survive,

  $$P(X\geq10)=1-P(x<10)=1-\sum_{x=0}^9b(x;15,0.4)=1-0.9662=0.0338$$

  ii

  from 3 to 8 survive, and

  $$P(3\leq X \leq 8)=\sum_{x=3}^8b(x;15,0.4)=\sum_{x=0}^8b(x;15,0.4) -\sum_{x=0}^2b(x;15,0.4)$$

  $$=0.9050-0.0271=0.8779$$

  iii

  exactly 5 survive?

  $$P(X=5)=b(5;15,0.4)=\sum_{x=0}^5b(x;15,0.4)-\sum_{x=0}^4b(x;15,0.4)$$

  $$=0.4032-0.2173=0.1859$$

• Example 5.6: A large chain retailer purchases a certain kind of electronic device from a manufacturer. The manufacturer indicates that the defective rate of the device is 3%

  iv

  The inspector of the retailer randomly picks 20 items from a shipment. What is the probability that there will be at least one defective item among these 20?

  Denote by $X$ the number of defective devices among the 20; $b(x;20,0.03)$

  $$P(X\geq 1)=1-P(X=0)=1-b(0;20,0.03)=1-0.03^0 0.97^{20-0}=0.4562$$

  v

  Suppose that the retailer receives 10 shipments in a month and the inspector randomly tests 20 devices per shipment. What is the probability that there will be 3 shipments containing at least one defective device?

  Denote by $Y$ the number of shipments containing at least one defective item; $b(y;10,0.4562)$

  $$P(Y=3)=\left( \begin{array}{c} 10\\ 3\\ \end{array}\right)0.4562^3(1-0.4562)^{10-3}=0.1602$$

• Theorem 5.2:
  
• Example 5.7: Find the mean and variance of the binomial random variable of Example 5.5 ( $n = 15, p = 0.4$), and then use Chebyshev's theorem to interpret the interval $\mu\pm 2\sigma$
• Solution: Example 5.5 was a binomial experiment with $n=15$ and $p=0.4$
  $$\mu=np=15*0.4=6$$
  $$\sigma^2=npq=15*0.4*0.6=3.6 \Rightarrow \sigma=1.897$$
  The interval
  $$\mu\pm2\sigma=6\pm2*1.897 \Rightarrow 2.206 ~to~ 9.794$$
  has a probability of at least $\frac{3}{4}$.

• Example 5.8: It is conjectured that an impurity exists in 30% of all drinking wells in a certain rural community.
• In order to gain some insight on this problem, it is determined that some tests should be made. It is too expensive to test all of the many wells in the area, so 10 were randomly selected for testing.

  vi

  Using the binomial distribution, what is the probability that exactly three wells have the impurity assuming that the conjecture is correct?

  $$b(x;10,0.3)=P(X=3)=B(3;10,0.3)-B(2;10,0.3)$$

  $$=0.6496-0.3828=0.2668$$

  vii

  What is the probability that more than three wells are impure?

  $$P(X>3)=1-B(3;10,0.3)=1-0.6496=0.3504$$

• Example 5.9: Consider the situation of Example 5.8. The ``30% are impure'' is merely a conjecture put forth by the area water board.
• Suppose 10 wells are randomly selected and 6 are found to contain the impurity. What does this imply about the conjecture? Use a probability statement.
• Solution:
  $$P(X=6)=\sum_{x=0}^6 b(x;10,0.3)-\sum_{x=0}^5 b(x;10,0.3)$$
  $$=0.9894-0.9527=0.0367$$
  For values of $b(x;10,0.3)$, see Table A.1 from text book.

• As a result, it is unlikely (3.6%chance) that 6 wells would be found impure if only 30% of all are impure. This casts considerable doubt on the conjecture and suggests that the impurity problem is much more severe.

• If the number of outcomes, $k$ is more than two, it is referred to as multinomial. Suppose we have $k$ possible outcomes ($k > 2$) in an experiment.
• Multinomial Distribution. If a given trial can result in the $k$ outcomes $E_1,E_2, \ldots, E_k$ with probabilities $p_1,p_2, \ldots, p_k$, then the probability distribution of the random variables $X_1,X_2, \ldots, X_k$ representing the number of occurrences for $E_1,E_2, \ldots, E_k$ in $n$ independent trials is
  
• $\sum_{i=1}^kx_i=n$ and $\sum_{i=1}^k p_i=1$
  $$\left( \begin{array}{c} n\\ x_1,x_2, \ldots, x_k\\ \end{array}\right)=\frac{n!}{x_1!x_2!\ldots x_k!}$$
  is the number of ways that yielding $x_1$ outcomes for $E_1$, $x_2$ outcomes for $E_2$, $\ldots$, $x_k$ outcomes for $E_k$.

• Example: A local gas station sells three types of gasoline; regular, premium and super.
• According to past sales, About 60% customers fuelregular, 30% put premium, and rest 10% buy super.
• For next 10 customers, find the probability that 5 buy regular gas, 4 fuel premium, and 1 buy super gas.
• Solution:

  Let $X_i$ be the number of customers buy $i^{th}$ product, $i = 1, 2, 3$. Then

  $$f(5,4,1;\frac{60}{100},\frac{30}{100},\frac{10}{100},10)=\le... ...right)\frac{60}{100}^5\frac{30}{100}^4\frac{10}{100}^1=0.07936$$

• Example 5.10: For a certain airport containing three runways it is known that in the ideal setting the following are the probabilities that the individual runways are accessed by a randomly arriving commercial jet:

  viii

  Runway 1: $p_1=2/9$

  ix

  Runway 2: $p_2=1/6$

  x

  Runway 3: $p_3=11/18$

• What is the probability that 6 randomly arriving airplanes are distributed in the following fashion? Runway 1: 2 airplanes, Runway 2: 1 airplanes, Runway 3: 3 airplanes.
• Solution: Using the multinomial distribution , we have
  $$f(2,1,3;\frac{2}{9},\frac{1}{6},\frac{11}{18},6)=\left( \be... ...\ \end{array}\right)\frac{2}{9}^2\frac{1}{6}^1\frac{11}{18}^3$$
  $$=\frac{6!}{2!1!3!}*\frac{2}{9}^2*\frac{1}{6}^1*\frac{11}{18}^3=0.1127$$