# Probability of an event [closed]

How should I go about solving the following problem?

If you randomly type a 6-digit number on a note, what is the probability that you can see the same number if you flip your note upside down? How would you explain your answer to a 6-year-old?

The only piece I have been able to calculate is that there are 9x10x10x10x10x10 numbers.

Is this correct? and/or what is the correct way to get the answer to this problem?

Thanks a lot for your help!

For the number to be same when seen upside down:

1. It can only contain either 0,1,8,6,9.

2. For first and last digit :
For it to be a 6 digit number first digit and the last digit cannot be 0.
If first digit is 6 then last should be 9 and vice-versa.
If first digit is other than 6 or 9 (ie. 1,8) then they both should have same digit.

3. For second and fifth digit / For third and fourth digit :
If first digit is 6 then last should be 9 and vice-versa.
If first digit is other than 6 or 9 (ie. 1,8) then they both should have same digit.

Total Number of 6 digit numbers = 9 * 10 * 10 * 10 * 10 * 10 ([9] as first digit cannot be 0)

Numerator = 4 * 5 * 5 ([4] as first and last digit cant be 0)

So,
Probability should be = Numerator / Total number of 6 digit numbers

• Thanks a lot for the answer and an explanation. I think there could be one modification in point 1 of point 2 For it to be a 6 digit number first digit and the last digit cannot be 0 - The last digit can be 0 for it to be a six-digit number. What do you think? Commented Aug 19, 2020 at 13:48
• If the last digit is 0 , then when we reverse it ,the resulting number will not be a 6 digit number. Also for it to be same when seen upside down its first and last digit should be same. Hope it helps.
– Shiv
Commented Aug 20, 2020 at 5:31
• Yes, very much. Thanks a lot! Commented Aug 21, 2020 at 17:39