- Thread starter
- #1

Suppose a phone number is 123-4567, and that the first three numbers and last 4 numbers are scrambled.

First parts I figured out:

Probability of choosing every number correctly: 1/3!4!

probability of choosing of choosing the first 3 correctly:1/3!

Where I got stuck on, the probability of having the first three correct, and at least 1 of the numbers in the last 4 correct. My initial reaction was that it would be 1/3!* the probability of not getting any correct in the last 4. I just do not know how to calculate this, I initially thought it would be (3/4)*(2/3)*(1/2) which would be 1/4 which would have given me a final answer of 18/3!4!.

Looking at the answer in the back of the book showed me the answer was 15/3!4!. I now know this is the answer by writing down all the possible options, but would love to know how to come by this answer with properties of probability.

Any help would be greatly appreciated!