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October 22, 2009a rental agency, which leases heavy equipment by the day, has found that one expensive piece of equipment is lease, on the average, only one day in 5. if rental on day is independent of rental on any other day, find the probability distribution of Y, the number of days between a pair of rentals
Let Y be the number of trials until the first success. Y is a sum of Bernoulli trials in a way similar to a Binomial. The difference here is that for a binomial you are looking at the number of success in n trials. The Geometric is looking for the number of trials before the first success.
Y has the Geometric Distribution with success probability p then:
Y ~ Geometric(p)
P(Y = y) = p * (1 - p) ^ (y - 1) for y = 1, 2, 3, 4, ….
P(Y = y) = 0 otherwise.
As you can see, the probability mass function is derived by looking at having y - 1 failures and then 1 success.
The Expectation or Mean of the Geometric, i.e., how many trails you expect before the first success is 1/p.
The Variance of the Geometric is (1 - p) / p^2
We cannot display the complete probability mass function but for the values y = 1 to y = 10 we have:
Y ~ Geometric( 0.2 )
E(Y) = 5
Var(Y) = 20
P(Y = 1 ) = 0.2
P(Y = 2 ) = 0.16
P(Y = 3 ) = 0.128
P(Y = 4 ) = 0.1024
P(Y = 5 ) = 0.08192
P(Y = 6 ) = 0.065536
P(Y = 7 ) = 0.0524288
P(Y = 8 ) = 0.04194304
P(Y = 9 ) = 0.03355443
P(Y = 10 ) = 0.02684355
….
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Horaayy..there are 2 comment(s) for me so far ;)
Viewing rental of the piece as a "success", Y is a geometric random variable with parameter 0.2. Its distribution is
p_Y (k) = (0.8)^k (0.2), k=0,1,2,…
This is a bit different from the usual interpretation of p_Y(k); it is usually "the probability that k independent Bernoulli trials are required until we get a success", in which case k would have to be at least 1. In the present case, k can of course be 0, as indicated above.
References :
Let Y be the number of trials until the first success. Y is a sum of Bernoulli trials in a way similar to a Binomial. The difference here is that for a binomial you are looking at the number of success in n trials. The Geometric is looking for the number of trials before the first success.
Y has the Geometric Distribution with success probability p then:
Y ~ Geometric(p)
P(Y = y) = p * (1 - p) ^ (y - 1) for y = 1, 2, 3, 4, ….
P(Y = y) = 0 otherwise.
As you can see, the probability mass function is derived by looking at having y - 1 failures and then 1 success.
The Expectation or Mean of the Geometric, i.e., how many trails you expect before the first success is 1/p.
The Variance of the Geometric is (1 - p) / p^2
We cannot display the complete probability mass function but for the values y = 1 to y = 10 we have:
Y ~ Geometric( 0.2 )
E(Y) = 5
Var(Y) = 20
P(Y = 1 ) = 0.2
P(Y = 2 ) = 0.16
P(Y = 3 ) = 0.128
P(Y = 4 ) = 0.1024
P(Y = 5 ) = 0.08192
P(Y = 6 ) = 0.065536
P(Y = 7 ) = 0.0524288
P(Y = 8 ) = 0.04194304
P(Y = 9 ) = 0.03355443
P(Y = 10 ) = 0.02684355
….
References :