In a recent year it was reported that approximately 300 mill

In a recent year, it was reported that approximately 300 million golf balls were lost in the United States. (I don\'t lose that many!) Assume that the number of golf balls lost in an 18-hole round of golf is distributed as a Poisson random variable with the mean of 4 lost balls. (That sounds about right for me.)

a. State the assumptions that need to be made to claim the number of balls lost is distributed as a Poisson random variable. Is it reasonable that the distribution is Poisson?

Given the assumptions in (a), what is the probability that:

b. no balls will be lost on an 18-hole round of golf?

c. 3 or fewer balls will be lost on an 18-hole round of golf?

d. 5 or more balls will be lost on an 18-hole round of golf?

e. What happens to all those golf balls? (A rhetorical question)

Solution

a)

The probability of observing a single event over a small interval is approximately proportional to the size of that interval.

The probability of two events occurring in the same narrow interval is negligible.

The probability of an event within a certain interval does not change over different intervals.

The probability of an event in one interval is independent of the probability of an event in any other non-overlapping interval.

From these, yes, it is reasonable to assume that we have a Poisson random variable here.

b)

Note that the probability of x successes out of n trials is          
          
P(x) = u^x e^(-u) / x!          
          
where          
          
u = the mean number of successes =    4      
          
x = the number of successes =    0      
          
Thus, the probability is          
          
P (    0   ) =    0.018315639 [ANSWER]

*********************

c)

Using a cumulative poisson distribution table or technology, matching          
          
u = the mean number of successes =    4      
          
x = the maximum number of successes =    3      
          
Then the cumulative probability is          
          
P(at most   3   ) =    0.43347012 [ANSWER]

**********************

d)

Note that P(at least x) = 1 - P(at most x - 1).          
          
Using a cumulative poisson distribution table or technology, matching          
          
u = the mean number of successes =    4      
          
x = our critical value of successes =    5      
          
Then the cumulative probability of P(at most x - 1) from a table/technology is          
          
P(at most   4   ) =    0.628836935
          
Thus, the probability of at least   5   successes is  
          
P(at least   5   ) =    0.371163065 [ANSWER]