"Blessed are those that noughtexpect,
for they shall not be disappointed."

John Wolcot

8.2 THE GEOMETRIC DISTRIBUTIONS (Pages 434 -444)

OVERVIEW: The Advanced PlacementStatistics Syllabus states that students need only know how to obtaingeometric probabilities through simulation. The geometric setting issomewhat similar to the binomial setting, the basic difference beingthat the geometric setting does not have a fixed number ofobservations.

The geometric setting:

(1) Each observation is in one of twocategories: success or failure.

(2) The probability of success is the same foreach observation.

(3) Observations are independent. (Knowing theresult of one observation tells you nothing about the otherobservations.)

(4) The variable of interest in the number oftrials required to obtain the first success.

Example:

Question: How many times would youexpect to have to roll a single die in order to get a "6"?

Here is a simulation approach (ten trials) usingrandInt(1,6,15) on the TI-83.

Trial 1: 3 5 2 2 3 6 ... (6 rolls)

Trail 2: 1 3 3 2 1 4 2 1 1 6 ... (10rolls)

Trial 3: 5 1 5 6 ... (4 rolls)

Trial 4: 4 1 4 3 5 ... (5 rolls)

Trial 5: 4 5 3 4 6 ... (5 rolls)

Trial 6: 5 5 1 3 3 2 2 1 3 6 ... (10rolls)

Trial 7: 5 3 4 1 6 ... (5 rolls)

Trial 8: 5 2 4 2 1 6 ... (6 rolls)

Trial 9: 4 5 1 4 6 ... (5 rolls)

Trial 10: 2 5 4 2 1 2 3 4 6 (9rolls)

The mean number of rolls for the 10 trials is6.5.

Now, if p is the probability of success, and q =1-p is the probability of failure, then the probability of success onthe first trial is p, on the second trial it is qp, on the thirdtrial it is q2p, etc. Hence, If X is a variable representing the numberof trials until the first success, the expected value of X is

mx = 1p + 2qp +3q2p +4q3p + ....= p(1 + 2q + 3q2 + 4q3 + ...).

Observe that 1 + 2q + 3q2 + 4q3 + ... =(1+q+q2+q3+ ...)(1+q+q2+q3+ ...)

Also, for 0<q<1, the sum of the infiniteseries 1 + q + q2 + q3 = 1/(1-q).

Hence, mx = p (1+q+q2+q3+ ...)(1+q+q2+q3+ ...) = p[1/(1-q)][1/(1-q)] = p/(1-q)2 = p/p2 = 1/p.

In the die-rolling example, the probability ofrolling a 3 is 1/6. Hence, the expected number of rolls before thefirst success is 1/(1/6) = 6.

In the present California Lottery, one chooses 6numbers from the set 1,2,3,...,49,50,51. The State of Californiathen randomly selects 6 numbers from the set. If you happen to matchthe six number chosen by the State, you win millions of dollars. Your probability of matching six is 1/51C6 = 1/18,009,460. That is,you would expect to have to play 18,009,460 times to get your firstsuccess. If you played once a week, you would expect your firstsuccess after 346,336 weeks, or 6,660 years. Good luck!