Negative Binomial and Geometric Distributions

• Suppose, instead of performing a fixed or given number of trials, one performs independently a Bernoulli trial repeatedly until a desired number of successes is obtained, and then stop.
• Then, the question is that how many trials are required to get the desired number of successes?
• Negative binomial experiments: the $k^{th}$ success occurs on the $x^{th}$ trial.
• Negative binomial random variable: the number $X$ of trials to produce $k$ success in a negative binomial experiment.
• Negative binomial distribution: If repeated independent trials can result in a success with probability $p$ and a failure with probability $q = 1 - p$, then the probability distribution of the random variable $X$, the number of the trial on which the $k^{th}$ success occurs, is
  $$b^*(x;k,p)=\left( \begin{array}{c} x-1\\ k-1\\ \end{array}\right)p^kq^{x-k},~x=k,k+1,k+2,\ldots$$

• Example 5.17: Suppose that team $A$ has probability 0.55 of winning over the team $B$ and both teams $A$ and $B$ face each other in an NBA 4-out-of-7 championship series.

  i

  What is the probability that team $A$ will win the series in six games?

  $$b^*(x;k,p)=$$

  $$b^*(6;4,0.55)=$$

  $$\left( \begin{array}{c} 5\\ 3\\ \end{array}\right)*0.55^4*(1*0.55)^{6-4}=0.1853$$

  ii

  What is the probability that team $A$ will win the series?

  $$b^*(4;4,0.55)+b^*(5;4,0.55)+b^*(6;4,0.55)+b^*(7;4,0.55)$$

  $$= 0.0915 + 0.1647 + 0.1853 + 0.1668 = 0.6083$$

  iii

  If both teams face each other in a regional play-off series and the winner is decided by winning three out of five games, what is the probability that team $A$ will win a play-off?

  $$b^*(3;3,0.55)+b^*(4;3,0.55)+b^*(5;3,0.55)$$

  $$= 0.1664 + 0.2246 + 0.2021 = 0.5931$$

• Why name negative binomial? The binomial coefficient is defined even when $n$ is negative (or is not an integer).
  $$\left( \begin{array}{c} n\\ k\\ \end{array}\right)=\frac{n(n-1)(n-2)\ldots(n-k+1)}{k!}$$
  $$(p+q)^n =\sum_{k=0}^\infty \left( \begin{array}{c} n\\ k\\ \end{array}\right)p^kq^{n-k}$$

• Each term in the expansion of $p^k(1-q)^{-k}$ corresponds to the value of $b^*(x;k,p)$ for $x=k,k+1,k+2,\ldots$
  $$1=p^k*p^{-k}=p^k*(1-q)^{-k}=p^k*\sum_{x=0}^\infty \left( \begin{array}{c} -k\\ x\\ \end{array}\right)(-q)^x$$

• Consider a binomial experiment to get a first success. This implies that for the case of that we encountered the number of failures prior to the first success.
• Geometric Distribution: If repeated independent trials can result in a success with probability $p$ and a failure with probability $q = 1 - p$, then the probability distribution of the random variable $X$, the number of the trial on which the first success occurs, is
  $$g(x;p)=b^*(x;1,p)=pq^{x-1},x = 1,2,3,\ldots$$

• Example: Consider a problem of log in into a communication network. It is know that the probability of success rate, $p= 0.35$ during busy hours.

  iv

  Probability that one is able to log into the network at $3^{rd}$ trial:

  $$P (Y = 3) = (0.65)^2 (0.35) = 0.1479.$$

  v

  Probability that one is able to log into within 3 trials:

  $$P (Y \leq 3) = P (Y = 1)+P (Y = 2)+P (Y = 3)$$

  $$= 0.35+0.2275+0.1479 =0.7254$$

  vi

  The average number of trials to get into the network:

  $$E (Y) = 1/0.35 =2.85.$$

  That is, it takes about 3 times on average

• Example 5.18: In a certain manufacturing process it is known that, on the average, 1 in every 100 items is defective.
• What is the probability that the fifth item inspected is the first defective item found?
  $$g(x;p)=$$
  $$g(5;0.01)=$$
  $$0.01x0.99^4 = 0.0096$$

• Example 5.19: At ``busy time'' a telephone exchange is very near capacity, so callers have difficulty placing their calls.
• It may be of interest to know the number of attempts necessary in order to gain a connection.
• Suppose that let $p = 0.05$ be the probability of a connection during busy time.
• We are interested in knowing the probability that 5 attempts are necessary for a successful call.
  $$P(X=x)=g(5;0.05)=0.05*0.95^4=0.041$$

• Theorem 5.4:
  
• In the system of telephone exchange, trials occurring prior to a success represent a cost.
• A high probability of requiring a large of number of attempts is not beneficial to the scientists or engineers.