Question

Approximately 300 million golf balls were lost in the U.S. in 2009. Assume that the number of golf balls lost in an 18-hole round is distributed as a Poisson random variable with a mean of 3 balls. What is the probability that:

0 balls will be lost in an 18-hole round?

An individual can play an 18-hole round with a single sleeve of golf balls (3 balls)? (Hint: can you continue to play if you have lost all 3 balls?)

An individual who is going to play 3 rounds of golf on a trip can play all three rounds with the single sleeve of balls?

Answer 1

Define

X: number of golf balls lost in an 18-hole round

Given tha X~Poisson(lambda=3)

Pmf of X is given by

probability that 0 ball will be lost in an 18 hole round

=P(X=0)

=0.05( approx.)

it is assumed that "one ball rule" is in effect ,that is a golfer can play a round with balls from a single sleeve only.

An individual can play an 18-hole round with a single sleeve of golf balls (3 balls) means that the individual has not lost all 3 balls of the single sleeve (the single sleeve from which the individual is using balls) in an 18-hole round.

So, the probability that an individual can play an 18-hole round with a single sleeve of golf balls (3 balls) is equivalent to the probability that maximum 2 balls are lost (he/she can play the the round with one remaining ball from the single sleeve)

i.e. P(X=0)+P(X=1)+P(X=2)

Hence, the probability that an individual can play an 18-hole round with a single sleeve of golf balls (3 balls) = P(X=0)+P(X=1)+P(X=2)

define

X₁ : number of balls lost in first round out of 3 rounds (each round is an 18 hole round)

X₂ : number of balls lost in second round out of 3 rounds (each round is an 18 hole round)

X₃ : number of balls lost in third round out of 3 rounds (each round is an 18 hole round)

X₁,X₂,X₃ are independent events

define

Y : number of balls lost in 3 rounds (each round is 18 hole round)

Y=X₁+X₂+X₃

By the reproductive property of Poisson (lambda) distribution

Y ~ Poisson(3*lambda)

i.e. Y ~ Poisson(9)

An individual who is going to play 3 rounds of golf on a trip can play all three rounds with the single sleeve of balls means all 3 balls from a single sleeve are not lost in 3 rounds, i.e. maximum 2 balls from the single sleeve are lost.

So, the probability that an individual who is going to play 3 rounds of golf on a trip can play all three rounds with the single sleeve of balls is equivalent to the probability that maximum 2 balls are lost in 3 rounds

Hence the probability that an individual who is going to play 3 rounds of golf on a trip can play all three rounds with the single sleeve of balls

=P(Y=0)+P(Y=1)+P(Y=2)