Good morning Candice! Let me know if you need my workings

For part a, we simply realise that this event is basically a recurring scenario similar to that of a ball drawn with replacement.

We realise that a ball is thrown each time with a “success” (getting a “6”) probability of 1/6 and a “failure” (not getting a “6”) probability of 5/6.

(Since this event can go on indefinitely, we cannot really draw the complete tree diagram, and at your level, it is not immediately obvious how the sum of all the probabilities of every single possibility is equal to one; this one is A Level stuff about geometric progression)

To achieve a “success” on the third attempt, the first two attempts must have been a “failure” (if a “success” was obtained earlier, the game would have ended on that turn instead).

So, the probability of obtaining a “success” in the third attempt
= P (“failure” on first attempt) x P (“failure” on second attempt) x P (“success” on third attempt)
= 5/6 x 5/6 x 1/6
= 25/216

And in a similar idea, the probability of obtaining a “success” in the fourth attempt is 5/6 x 5/6 x 5/6 x 1/6, or 125/1296.

There are four cases ending by the fourth attempt, so the probability of obtaining a “success” by the fourth attempt is

1/6 + (5/6 x 1/6) + (5/6 x 5/6 x 1/6) + (5/6 x 5/6 x 5/6 x 1/6)

(I have no calculator with me now)

As a side note not related to the question, in the (very unlikely) event that there is no “success” indefinitely, the probability of this happening is

5/6 x 5/6 x 5/6 x ...

or

(5/6)^infinity

and the probability of obtaining a “success” on the nth try is

5/6^(n - 1) x 1/6

because there must have been (n - 1) “failures” preceding the one “success”.

Part b I will do later if you need help

Thank you so much Mr Eric :D
Yes, I understand part (a) now. And I could do part (b) too, hence, you don't have to spend time on part (b) for me. Many thanks !

Have a great day ahead :)