Geometric Distribution

jijie14

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Oct 6, 2012
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1. Communication: The rack behind a coat-check counter collapses and 20 coats slip off their numbered hangers. When the first person comes to retrieve one of these coats, the clerk brings them out and holds them up one at a time for the customer to identify.
a) what is the probability that the clerk will find the customer's coat
i) on the first try? ​I got 1/20
ii) on the second try? 1/20
iii) on the third try? 1/20
iv) in fewer than 10 tries? 9 x 1/20

b) What is the expected number of coats the clerk will have to bring our before finding the customer's coat?
the answer is 95 but i have no clue how they got that

c) Explain why you cannot use a geometric distribution to calculate the waiting time.
i said because the events aren't independent


 
Hello, jijie14!

1. The rack behind a coat-check counter collapses and 20 coats slip off their numbered hangers.
When the first person comes to retrieve his/her coat, the clerk brings them out and holds them up
one at a time for the customer to identify.

[A] what is the probability that the clerk will find the customer's coat
(i) on the first try? ​I got 1/20
(ii) on the second try? 1/20
(iii) on the third try? 1/20
(iv) in fewer than 10 tries? 9 x 1/20 . These are correct!

What is the expected number of coats the clerk will bring our before finding the customer's coat?
The answer is 95. . This is wrong!

How can the clerk bring out an average of 95 coats?
. . There are only 20 coats!

Do you know the formula for Expected Value?
 
Hello, jijie14!


How can the clerk bring out an average of 95 coats?
. . There are only 20 coats!

Do you know the formula for Expected Value?

Thats the answer in the back of the book , i dont know how they got 95 and i know that E(x) = q/p but that only applies to independent events
 
Thats the answer in the back of the book , i dont know how they got 95 and i know that E(x) = q/p but that only applies to independent events
As you yourself discovered:

The probability that it will take 1 try is \(\displaystyle \dfrac{1}{20}\)

The probability that it will take 2 tries is \(\displaystyle \left(1 - \dfrac{1}{20}\right) * \dfrac{1}{19} = \dfrac{19}{20} * \dfrac{1}{19} = \dfrac{1}{20}.\)

The probability that it will take 3 tries is \(\displaystyle \left(1 - \dfrac{1}{20} - \dfrac{1}{20}\right) * \dfrac{1}{18} = \dfrac{18}{20} * \dfrac{1}{18} = \dfrac{1}{20}.\)

And so on.

Now apply the definition of expected value:

\(\displaystyle \displaystyle \sum_{i=1}^{20}\left(i * \dfrac{1}{20}\right) = \dfrac{1}{20} * \sum_{i=1}^{20}i = \dfrac{1}{20} * \dfrac{20 * 21}{2} = 10.5.\)

As Soroban pointed out, the book's answer of 95 is clearly wrong so stop worrying about it.
 
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