Hi,

as already said I've a lot problem with probabilities and combinations.

I encountered the following problems, they seem very similar to me. So I tried to solve them with the same method. But it didn't work.

The problem are:

1) A four digit safe code does not contain the digits 1 and 4 at all. What is the probability that it has at least one even digit?

2) John wrote a phone number on a note that was later lost. John can remember that the number had 7 digits, the digit 1 appeared in the last three places and 0 did not appear at all. What is the probability that the phone number contains at least two prime digits?

1) I have four slots _ _ _ _ that can contain only {0,2,3,5,6,7,8,9}.

Four of them are even, four are odd.

So the probability of having an even digit is \(4/8=1/2\), as well having an odd digit.

The solution can be calculated as:

Probability of at least one even digit = Probability of one even digit + Probability of two even digits + Probability of three even digits + Probability of four even digits

P1 = \(1/2\)

P2 = \(1/2^2=1/4\)

P3 = \(1/2^3=1/8\)

P4 = \(1/2^4=1/16\)

P = P1+P2+P3+P4 = \(15/16\)

2) I have the following situation _ _ _ _ 1 1 1

Empty slots can contain only {2,3,4,5,6,7,8,9}

Prime digits are {2,3,5,7}, not prime digits are {4,6,8,9}.

So the probability of having a prime digit is \(4/8=1/2\), as well having a non prime digit.

I think I have the same situation of the previous problem: filling 4 empty slots with digits (even or prime) that have the same probability of being in the slot (\(1/2\)) or not being (\(1/2\)).

So I created the solution with the same pattern of the previous problem:

Probability of at least two prime digits = Probability of two prime digits + Probability of three prime digits + Probability of four prime digits

P2 = \(1/2^2=1/4\)

P3 = \(1/2^3=1/8\)

P4 = \(1/2^4=1/16\)

P = P2+P3+P4 = \(7/16\)

My solution of problem 2 is not correct, while the one of problem 1 is correct. Why?

Kudos and beers for your help!!!