Thank you so much for your advice, Mr J.

For part (b), does the cumbersome method of calculation mean 23C1 (0.0001) (0.9999)^22 + 23C2 (0.0001)^2 (0.9999)^21 + …. + 23C22 (0.0001)^22 (0.9999)?

Yes. If you try it you'll get the same answer as 1 - 0.9999²³

Note that you can choose to put a 23 C 23 in front of 0.9999²³ if you want to, but it's not really needed since 23 C 23 just equals 1.

i.e there is only one way to not win any prizes.

Thanks Mr J. I got it now :)
The analogy really helped!

I hope you have a good day ahead. Stay safe and take care :D

Welcome.

The analogy however, would be too cumbersome for real-life application (10000 differently numbered balls is a lot to manufacture and handle!) so how the numbers are picked for 4D (or equivalent games in the world) is something like this :


There are 4 slots, one for each digit.
The digits are picked one at a time.

For each slot, there are 10 balls numbered from 0 to 9. A machine will spin these 10 balls until one comes out of its opening and falls into the slot.


Similarly for Toto, 6 numbers and 1 additional number are picked out of 49, but for this there is no replacement so the machine only releases 7 numbers, 1 at a time.

Thank you so much for your great advice, Mr J.
Really appreciate :)