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Grandad · #3 November 8th, 2009, 02:04 AM

Quote:

Originally Posted by ntrantrinh

...
2. A small town has a network of 115 residential streets, all containing approximately the same number of residents. If a canvasser randomly selects 20 people from the phone book to promote a product, what is the probability that at least two of the people live on the same street?

(Given Answer: 0.827)

In addition to the assumption that all the streets contain approximately the same number of people, we must also assume that this number is fairly large so that the numbers in each street remain approximately equal after some people have been chosen.

Then we can say that when the first person has been chosen, the probability that the second one chosen is in a different street is $\frac{114}{115}$ .

The probability that the third one chosen is from a street different from the first two is $\frac{113}{115}$ , and so on ...

So the probability that all $20$ are from different streets is

$\frac{114\times113\times112\times...\times96}{115^{19}}=\frac{114!}{95!115^{19}}\approx 0.173$

So the probability that at least two are from the same street $= 1 - 0.173 = 0.827$ .

Alternative method
With repetitions allowed, there are $\frac{115^{20}}{20!}$ ways of selecting $20$ streets from $115$ . With no repetitions there are $\binom{115}{20}=\frac{115!}{20!95!}$ ways.

So the probability of selecting 20 different streets $= \frac{115!20!}{20!95!115^{20}}= \frac{114!}{95!115^{19}}$

Grandad