Editing Negative binomial distribution (section)

====Selling candy====
Pat Collis is required to sell candy bars to raise money for the 6th grade field trip.  Pat is (somewhat harshly) not supposed to return home until five candy bars have been sold.  So the child goes door to door, selling candy bars. At each house, there is a 0.6 probability of selling one candy bar and a 0.4 probability of selling nothing.

''What's the probability of selling the last candy bar at the'' {{mvar|n}}-th ''house?''

Successfully selling candy enough times is what defines our stopping criterion (as opposed to failing to sell it), so {{mvar|k}} in this case represents the number of failures and {{mvar|r}} represents the number of successes.  Recall that the {{math|NB(''r'', ''p'')}} distribution describes the probability of {{mvar|k}} failures and {{mvar|r}} successes in {{math|''k'' + ''r''}} {{math|Bernoulli(''p'')}} trials with success on the last trial.  Selling five candy bars means getting five successes.  The number of trials (i.e. houses) this takes is therefore {{math|1=''k'' + 5 = ''n''}}.  The random variable we are interested in is the number of houses, so we substitute {{math|1=''k'' = ''n'' − 5}} into a {{math|NB(5, 0.4)}} mass function and obtain the following mass function of the distribution of houses (for {{math|''n'' ≥ 5}}):

:<math> f(n) = {(n-5) + 5 - 1 \choose n-5} \; (1-0.4)^5 \; 0.4^{n-5} = {n-1 \choose n-5} \; 3^5 \; \frac{2^{n-5}}{5^n}. </math>

''What's the probability that Pat finishes on the tenth house?''

:<math> f(10) = \frac{979776}{9765625} \approx 0.10033. \, </math>

''What's the probability that Pat finishes on or before reaching the eighth house?''

To finish on or before the eighth house, Pat must finish at the fifth, sixth, seventh, or eighth house. Sum those probabilities:
:<math> f(5) = \frac{243}{3125} \approx 0.07776 \, </math>
:<math> f(6) = \frac{486}{3125} \approx 0.15552 \, </math>
:<math> f(7) = \frac{2916}{15625} \approx 0.18662 \, </math>
:<math> f(8) = \frac{13608}{78125} \approx 0.17418 \, </math>
:<math>\sum_{j=5}^8 f(j) = \frac{46413}{78125} \approx 0.59409.</math>

''What's the probability that Pat exhausts all 30 houses that happen to stand in the neighborhood?''

This can be expressed as the probability that Pat [[Complementary event|does not]] finish on the fifth through the thirtieth house:
:<math>1-\sum_{j=5}^{30} f(j) = 1 - I_{0.4}(5, 30-5+1) \approx 1 - 0.999999823 = 0.000000177. </math>

Because of the rather high probability that Pat will sell to each house (60 percent), the probability of her ''not'' fulfilling her quest is vanishingly slim.