1. 9999 Notes

Hello to all,

New to the forum.
I have a question you might be able to answer.

You have a box with 9999 notes with a single four digit number written on them.
You have two favourite numbers, e.g. Number 1 and number 7.
You pick one note from the box.

What is the probability that from the 4 digits of the note, 3 of them are always
a combination of your favourite numbers?

For example 1118, 1716, 7771, 1131, 7277, 1717, 7777, 1111 etc.

Thank you!
George.

2. Re: 9999 Notes

Hello George. Given 2 out of 10 favorite digits, the probability that there are k favorite digits out of n on the slip of paper is:

$P(k) = \begin {pmatrix} n \\ k \end{pmatrix}0.2^k 0.8^{n-k}$

where $\begin {pmatrix} n \\ k \end{pmatrix}$ is the combination of k out of n items, which is equal to $\frac {n!}{k!(n-k)!}$

So here you want the probability that either 3 or 4 of the 4 digits are among your two favorites:

$P(3) + P(4) = \begin {pmatrix} 4 \\ 3 \end{pmatrix}0.2^3 0.8^1 + \begin {pmatrix} 4 \\ 4 \end{pmatrix}0.2^4 0.8^0 = 0.0256+0.0016 = 0.0272$

3. Re: 9999 Notes

Hello abaines,

I only need the probability that either 3 of the digits are among the favourites.
In the example 7777, one of the 7s is just another digit that happens to be a 7.

So the probability should be 0.0256, correct?

4. Re: 9999 Notes

Originally Posted by hgg
So the probability should be 0.0256, correct?
No - 0.0256 is the probability of getting precisely three of your favorites and one that is not your favorite - for example it captures the cases of 7772 , 7773, 7774, etc, but not 7771 nor 7777. So you have to add the probabilty that all four are favorites as well.

5. Re: 9999 Notes

Oh, I see. Thanks again!