# telephone bills

• June 5th 2007, 09:34 AM
honey
telephone bills
A. To use the telephone banking pay bills the customer has to enter the last three digits of each bill. The numbers 0-9 inclusive can be used. If the numbers happen to be on the same on more one bill the customer has to enter the first three digits as well
What is the probability that a person has 2 bills register with the same last three digits?

so:

2 bills
last three digits is it like so:

2/ 1000 = 0.002

or is it:

1/ 1000 = 0.001

???

B. What is the probability that a person with 10 bills to register has at least 2 bills with the same last three digits?

I am not sure?
• June 6th 2007, 05:15 AM
Pterid
OK. We assume that the last three digits on the bills can be treated as random and "uniformly distributed". This just means that on any bill you pick out, the 1000 possible combinations of 3 digits (running from $000$ up to $999$).

For part A, imagine looking at one bill first - it gives you a specific 3 digits. Then look at the other bill - the probability of those 3 digits being equal to those on the first bill is just:

$P_2 = \frac{1}{1000} = 0.001.$

Part B is a little more complicated. To solve it, I'm going to ask a slightly different question:

"You have 10 bills. What is the probability that no two of them share their last three digits?"

This time, imagine laying all 10 bills out in a row. Starting at one end, read the bills one at a time. Stop whenever you see a "match". If you get to the end without reading the same number twice, then (obviously) no two of them are the same.
• When you look at "bill #1", you get one 3-digit code to hold in your head.
• When you look at "bill #2", the probability of that code not being equal to bill #1 is $999/1000$. And, in this case, you can move on to bill #3.
• When you look at "bill #3", the probability of it not being equal to either bill #1 or bill #2 is $998/1000$ (by the time to get to reading bill #3, you know that bills #1 & 2 are different.)
• Keep going. If you make it to bill #10, you'll have 9 distinct numbers in your head, and the probability that bill #10 isn't equal to any of them is $991/1000$.
So, the overall probability of no matches comes by multiplying these nine probabilities together, giving you

$P(no~match) = (1 - P_{10}) = \frac{999}{1000} \times \frac{998}{1000} \times \ldots \times \frac{991}{1000} \approx 0.956$.

So, the probability of that there is a match (2 or more codes being equal) is $P_{10} = 1 - 0.956 = 0.044 = 4.4\%$.

- - -

I fear I might have made this look more complicated than it really is. I invite anyone with a more concise argument to jump in. ;P
• June 6th 2007, 06:06 AM
honey
well
to be honset you expliend it well there was nothing confusing about it I understand now thanks!:)
• June 6th 2007, 12:47 PM
Jhevon
Quote:

Originally Posted by Pterid
Part B is a little more complicated. To solve it, I'm going to ask a slightly different question:

"You have 10 bills. What is the probability that no two of them share their last three digits?"

This time, imagine laying all 10 bills out in a row. Starting at one end, read the bills one at a time. Stop whenever you see a "match". If you get to the end without reading the same number twice, then (obviously) no two of them are the same.
• When you look at "bill #1", you get one 3-digit code to hold in your head.
• When you look at "bill #2", the probability of that code not being equal to bill #1 is $999/1000$. And, in this case, you can move on to bill #3.
• When you look at "bill #3", the probability of it not being equal to either bill #1 or bill #2 is $998/1000$ (by the time to get to reading bill #3, you know that bills #1 & 2 are different.)
• Keep going. If you make it to bill #10, you'll have 9 distinct numbers in your head, and the probability that bill #10 isn't equal to any of them is $991/1000$.
So, the overall probability of no matches comes by multiplying these nine probabilities together, giving you

$P(no~match) = (1 - P_{10}) = \frac{999}{1000} \times \frac{998}{1000} \times \ldots \times \frac{991}{1000} \approx 0.956$.

So, the probability of that there is a match (2 or more codes being equal) is $P_{10} = 1 - 0.956 = 0.044 = 4.4\%$.

- - -

I fear I might have made this look more complicated than it really is. I invite anyone with a more concise argument to jump in. ;P

i got something different for this one. don't be too alrmed though, i'm not good at probability as everyone here will tell you, so chances are i've made some error and not you (in which case i'd appreciate if you would point out my flaw).

I would do this using Bernoulli Trials.

By the method of Bernoulli Trials, the probability, $P(k)$, of $k$ successes in $n$ trials is given by:

$P(k) = P(k \mbox { } successes) = {n \choose k} p^k q^{n-k}$

where $p$ is the probability of success and $q$ is the probability of failure.

Here, we have 10 bills, and hence 10 trials. We consider it a success if we find a match, that probability is $\frac {1}{1000}$, the probability we fail to find a match is $\frac {999}{1000}$.

We want the probability of finding AT LEAST 2 matches. That is, we want to know what are the chances of finding 2 matches or 3 matches or ...or 10 matches. Or algebraically:

we want $P(k \geq 2) = P(2) + P(3) + P(4) + ... + P(10)$

or simply $P(k \geq 2) = 1 - (P(0) + P(1))$

Therefore, $P(k \geq 2) = 1 - {10 \choose 0} \left( \frac {1}{1000} \right)^0 \left( \frac {999}{1000} \right)^{10} - {10 \choose 1} \left( \frac {1}{1000}\right)^1 \left( \frac {999}{1000} \right)^9$

$\Rightarrow P(k \geq 2) \approx 0.0004476 \mbox { or } 0.045 \%$

Can someone please confirm this, or tell me i'm an idiot... well, you don't have to be that harsh if you don't want to:D
• June 6th 2007, 01:01 PM
Plato
Quote:

Originally Posted by Jhevon
i got something different for this one.
I would do this using Bernoulli Trials.

Bernoulli Trials are independent events!
Are these independent?
• June 6th 2007, 02:47 PM
Jhevon
Quote:

Originally Posted by Plato
Bernoulli Trials are independent events!

ah. i forgot that little fact:o

Quote:

Are these independent?
i don't think so. since whether we have a match or mismatch will depend on what numbers we had before, or will have after.