- #1

chimpfunkz

- 10

- 0

I had a brain teaser that I was hoping you could solve:

"In order to guarantee all emergency call are received, the phone company wants all phone nubers with 9 follower by 2 1's to be routed to 9-1-1 emergency center. This will ensure that any real emergency victims will still get help they require regardless of whether or not they hit extra keys. The digits 9-1-1 do not need to be consecutive but they do need to be in that order (e.g. 198-8141 would get routed to 9-1-1 but 514-9313 would not be since 9 is only followed by one 1.) How many 7 digit phone numbers would not be able to be used, since they would be routed to the 9-1-1 emergency center? assume all 7 digit phone numbers are available, from 000-0000 to 999-9999."

What do you think the answer is?

I worked it as a binomial, so I did

sum sum (i bin n), n=2 to i, i=2 to 6

but this gave me 99, which is definitely incorrect.

I also tried to do it as a binomial probability, and did it as

sum ( sum (i bin n) (.1)^n (.9)^(i-n))), n=2 to i, i=2 to 6

which gave me approximately .2860250, which is 2, 860 250 out of the 10000000 possible combinations

Am I right?

"In order to guarantee all emergency call are received, the phone company wants all phone nubers with 9 follower by 2 1's to be routed to 9-1-1 emergency center. This will ensure that any real emergency victims will still get help they require regardless of whether or not they hit extra keys. The digits 9-1-1 do not need to be consecutive but they do need to be in that order (e.g. 198-8141 would get routed to 9-1-1 but 514-9313 would not be since 9 is only followed by one 1.) How many 7 digit phone numbers would not be able to be used, since they would be routed to the 9-1-1 emergency center? assume all 7 digit phone numbers are available, from 000-0000 to 999-9999."

What do you think the answer is?

I worked it as a binomial, so I did

sum sum (i bin n), n=2 to i, i=2 to 6

but this gave me 99, which is definitely incorrect.

I also tried to do it as a binomial probability, and did it as

sum ( sum (i bin n) (.1)^n (.9)^(i-n))), n=2 to i, i=2 to 6

which gave me approximately .2860250, which is 2, 860 250 out of the 10000000 possible combinations

Am I right?

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