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Conditions for negative binomial models

What is the negative binomial distribution?

  • The negative binomial distribution models the number of trials needed to reach a fixed number of successes, r

    • For example, how many times will you have to roll a dice until it lands on a ‘6’ for the third time

  •  There is no one standard form of notation for the negative binomial distribution

    • But for a random variable X that has the negative binomial distribution you could write either:

      • X tilde NB open parentheses r comma space p close parentheses or X tilde Negative space straight B open parentheses r comma space p close parentheses space

    • X is the number of trials that will be required to reach a total of bold italic r successes

    • p is the fixed probability of success in any one trial

What are the conditions for using a negative binomial model?

  • A negative binomial model can be used for an experiment that satisfies the following conditions:

    • The experiment consists of an indefinite number of successive trials

    • The outcome of each trial is independent of the outcomes of all other trials

    • There are exactly two possible outcomes for each trial (success and failure)

    • The probability of success in any one trial (p) is constant 

  •  Note that these conditions are very similar to the conditions for the binomial distribution

    • But for a binomial distribution the number of trials (bold italic n) is fixed

      • And you count the number of successes

    • While for a negative binomial distribution the number of successes (bold italic r) is fixed

      • And you count the number of trials it takes to reach that number of successes

When might the conditions not be satisfied?

  • If asked to criticise a negative binomial model, you may be able to question whether the trials are really independent

    • For example, someone may be repeating an activity until they achieve the rth success

      • The trials may not be independent because the person gets better from practising the activity

      • This also means the probability of success, p, is not constant

    • In order to proceed using the model, you will have to assume trials are independent

Examiner Tips and Tricks

  • Replace the word “trials” with the context (e.g. “flips of a coin”) when commenting on conditions and assumptions

Negative binomial probabilities

What are the probabilities for the negative binomial distribution?

  • If X space tilde space Negative space straight B open parentheses r comma space p close parentheses, then X has the probability function:

    • Error converting from MathML to accessible text.

    • the random variable X is the number of trials needed to get bold italic r successes

    • p is the constant probability of success in one trial

    • straight P open parentheses X equals x close parentheses is the probability that the r th success will occur on the x th trial

  • Note that that is the product of

    • the binomial probability of getting bold italic r bold minus bold 1 successes in bold italic x bold minus bold 1 trials,  ,

    • and the probability of getting a success in the bold italic x bold th trialp

  • open parentheses table row cell x minus 1 end cell row cell r minus 1 end cell end table close parentheses is the binomial coefficient

    • i.e., open parentheses table row cell x minus 1 end cell row cell r minus 1 end cell end table close parentheses equals C presuperscript open parentheses x minus 1 close parentheses end presuperscript subscript open parentheses r minus 1 close parentheses end subscript

  • Also note that there is no greatest possible value of x

    • It could require any number of trials to reach the r th success

    • However for any given r, straight P open parentheses X equals x close parentheses gets closer and closer to zero as x gets larger

Where does the formula come from?

  • Consider rolling a fair dice and wanting to know the probability that it would take 12 rolls for the dice to land on ‘6’ a total of 3 times

    • This is the same as saying that the third ‘6’ occurs on the 12th roll

  • You can model this situation using the random variable X tilde Negative space straight B open parentheses 3 comma 1 over 6 close parentheses

    • ‘success’ here is defined as ‘roll a 6’

    • r equals 3 because we’re interested in the number of trials required to reach a total of 3 successes

    • p equals 1 over 6 is the probability of rolling a ‘6’ on a fair dice

    • X is the number of trials that will be required to reach 3 successes

  • The probability you are looking for is therefore straight P open parentheses X equals 12 close parentheses

  • For the third ‘6’ to occur on the 12th roll, the following things need to happen:

    • ‘6’ must occur exactly 2 times in the first 11 rolls

      • It doesn’t matter which rolls those two ‘6’s occur on

      • The probability of that happening is straight P open parentheses Y equals 2 close parentheses, for Y tilde straight B open parentheses 11 comma fraction numerator space 1 over denominator 6 end fraction close parentheses

      • So that part of the answer is a binomial probability

    • Then a ‘6’ must also occur on the 12th roll

      • The probability of that happening is 1 over 6

    • So the probability of both those things happening is

straight P open parentheses Y equals 2 close parentheses cross times 1 over 6 equals open parentheses open parentheses table row 11 row 2 end table close parentheses open parentheses 1 over 6 close parentheses squared open parentheses 5 over 6 close parentheses to the power of 9 close parentheses cross times 1 over 6 equals open parentheses table row 11 row 2 end table close parentheses open parentheses 1 over 6 close parentheses cubed open parentheses 5 over 6 close parentheses to the power of 9 equals 0.0493489...

  • That same logic can be used to find the general formula for negative binomial probabilities

Is there a connection between negative binomial probabilities and geometric probabilities?

  • Note that when r equals 1,the negative binomial probability function becomes:

  • That is the same as the geometric distribution probability function

  • The geometric distribution is the ‘special case’ of the negative binomial distribution when r equals 1

Examiner Tips and Tricks

  • Make sure you are clear about what xr, p and  straight P open parentheses X equals x close parentheses refer to in the formula!

  • Read the question carefully to determine whether binomial, geometric or negative binomial probabilities are required

Worked Example

Emanuel is playing in a chess tournament, where his probability of winning any one game is 0.55.  Find the probability that:

a) his first win is in the third game he plays

negb-probs-we-a

b) he wins exactly 4 of his first 7 games

negb-probs-we-b

c) he wins for the fourth time in his seventh game

negb-probs-we-c

d) he wins for the fourth time in his seventh game, given that he won his first game

negb-probs-we-d

e) his fourth win occurs in or before his seventh game.

negb-probs-we-e

f) Criticise the model used in this question.

negb-probs-we-f

 

Negative binomial mean & variance

What are the mean and variance of the negative binomial distribution?

  • If X space tilde space Negative space straight B open parentheses r comma space p close parentheses, then

    • The mean of X is  straight E open parentheses X close parentheses equals mu italic equals r over p

    • The variance of X is  Var open parentheses X close parentheses equals sigma to the power of italic 2 italic equals fraction numerator r stretchy left parenthesis 1 minus p stretchy right parenthesis over denominator p to the power of italic 2 end fraction

  • You need to be able to use these formulae to answer questions about the negative binomial distribution.

Examiner Tips and Tricks

  • If a question gives you the mean and/or variance with one known parameters (r or p), form an equation to find the other

Worked Example

Croesus is tossing a biased coin for which the probability of the coin landing on ‘heads’ is p.  The random variable X represents the number of times he needs to flip the coin until it has landed on heads four times.  Given that the mean of X is 10, find:

a) the value of p

negb-mean-var-we-a

b) the standard deviation of X

negb-mean-var-we-b

c) the probability that the fourth ‘head’ will occur on the seventh toss.

negb-mean-var-we-c

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