Download The Geometric Distribution (Waiting Times)

The Geometric Distribution (Waiting Times)
Can we answer the following questions:
1. What is the probability that you can roll a sum of seven on a pair of dices in fewer than three
rolls?
2. If a traffic light flows at a given rate, what is the expected length of time that a pedestrian must
wait before crossing the street?
The probability model that enables us to answer these questions is the
A geometric distribution has a specified number of independent trials with two possible outcomes,
success or failure. The random variable is the number of unsuccessful outcomes before a success
occurs.
Note that the number of trials prior to the first success is called the waiting period/time before
success.
If X is a random variable representing the waiting time, then the Geometric Distribution is given by
𝑃 (𝑋 = 𝑥 ) = 𝑞 𝑥 𝑝
where p is the probability of success on each trial and q=1-p is the probability of failure on each
trial and x=0,1,2,...
𝑞
The expected value of a random variable X that is geometrically distributed is (𝑋 ) = .
𝑝
Example 1: In a repeated rolling of a pair of dices,
a) what is the probability that the first roll of doubles occurs on the third roll?
b) what is the expected waiting time (the number of rolls) before you roll doubles?
Solution: p=6/36=1/6 and q=1-p=5/6.
Let the random variable, X, be the number of rolls prior to the first double being rolled.
a) If the first double is to occur on the third roll then x=2.
5 2 1
25
𝑃 (𝑋 = 2) = ( ) ( ) =
6
6
216
Therefore, the probability of the first doubles occurring on the third roll is equal to 25/216.
b) The expected waiting time is 𝐸 (𝑋 ) =
𝑞
𝑝
=
5
6
1
6
= 5. On average we would expect five rolls on non-
doubles before doubles would appear.
Example 2: Jamaal has a success rate of 68% for scoring on free throws in basketball. What is the
expected waiting time before he misses the basket on a free throw?
Solution: The random variable, X, is the number of trials before Jamaal misses on a free throw.
Note that a success is Jamaal failing to score. Thus, q=0.68 and p=0.32.
E(X)=q/p=2.1
The expectation is that Jamaal will score on 2.1 free throws before missing.
Example 3: In a gambling game a player tosses a coin until a head is uppermost. He then receives
$2n, where n is the number of tosses.
a) what is the probability that the player receives $8.00 in one play of the game?
b) If the player must pay $5.00 to play, what is the win/loss per game?
Solution:
We have Bernoulli trials with p=1/2. Let X be the random variable representing the number of
tosses before a head.
a) To win $8.00 the first win must occur on the fourth toss. Therefore, x=3.
1 3 1
1
P (X = 3) = ( ) ( ) =
2
2
16
Therefore, the probability of winning $8.00 is 1/16.
b) The expected waiting time before a head turns up is E(X)=(1/2)/(1/2)=1.
We can then conclude that we can expect to toss a head on the second toss, which would win
$4.00. The expected gain per game is $4-$5=-$1 (a loss per game of $1.00).
Example 3: What is the probability of rolling a sum of seven in fewer than three rolls of a pair of
dice?
Solution: Let the random variable X be the number of trials before a success. The number of trials
before a success is either 0 or 1. There are six sums of seven so, p=6/36=1/6 and q=5/6.
5 0
1
5 1
1
11
6
6
6
6
36
So, 𝑃(𝑋 < 2) = 𝑃(𝑋 = 0) + 𝑃(𝑋 = 1) = ( ) ( ) + ( ) ( ) =
Therefore, the probability of a seven being rolled before the third is 11/36.
Example 4: Suppose that an intersection you pass on your way to school has a traffic light that is
green for 40 s and then amber or red for a total of 60 s.
a) What is the probability that the light will be green when you reach the intersection at least once
a week?
b) What is the expected number of days before the light is green when you reach the intersection?
Solution: a) Each trial is independent with p=40/100=0.4 and q=1-p=0.40.
There are five school days in a week. To get the green light on one of those five days, your waiting
time must be four days or less.
So, P(X<4)=P(X=0)+P(X=1)+P(X=2)+P(X=3)=0.60x0.4+0.61x0.4+0.62x0.4+0.63x0.4=0.8704
The probability of the light being green when you reach the intersection at least once a week is
0.87 (approx. 87%.)
b) E(X)=0.60/0.40=1.5. Therefore, the expected time before catching a green light is 1.5 days.