i)

P(first card is not '8')
= P(the card drawn is any of the other 7 cards)
= ⅞

7 cards out of 8 cards don't have '8' on them. So 7 outcomes out of total of 8 possible outcomes fit the required probability.

ii)

Using a numbered example first :

If they were asking for 5 draws needed to obtain the number '8' ,

this means the first 4 draws were not '8' and the 5th draw is the '8' (that's why you had 5 draws because you only succeeded on the 5th try)

So for the first 4 draws, probability of each non-8 draw = ⅞

(i.e you draw any of the other 7 cards)


The 5th draw is the '8' and probability for this is ⅛ (only 1 card out of 8 cards is '8')


So required probability

P(1st draw is not 8) × P(2nd draw is not 8) × P(3rd draw is not 8) × P(4th draw is not 8) × P(5th draw is 8)

= ⅞ × ⅞ × ⅞ × ⅞ × ⅛

= (⅞)⁴ × ⅛

So for exactly n draws, the first (n - 1) draws are not '8'. Only the nth draw (which is the last draw) is '8'

Required probability

= ⅞ × ⅞ × ⅞ .... × ⅞ × ⅛

There are (n - 1) '⅞' in the product and only one ⅛

= (⅞)ⁿ⁻¹ × ⅛


This is also equal to :
7ⁿ⁻¹ / 8ⁿ

iii)

'At least n draws to get '8' can have these equivalent meanings :


①
'n or more draws to get '8' '

②
'no '8' for the first (n - 1) draws.


For ①, the '8' can be drawn on the nth draw or on later draws.

It includes the possibilities of getting '8' on the nth draw, (n+1)th, (n+2)th, (n+3)th, ... all the way to infinity.

So what you are adding up are all these probabilities :

Probability = (⅞)ⁿ⁻¹ (⅛) + (⅞)ⁿ (⅛) + (⅞)ⁿ+¹ (⅛) + (⅞)ⁿ+² (⅛) ... ...



For ②, once we reach (n-1) draws with no '8', the subsequent draw/draws that may get '8' have been accounted for already.


The probability is simply = (⅞)ⁿ⁻¹


Note that the probabilities for ① and ② are equal .

① forms a geometric series with a sum to infinity that is equal to ②'s probability.



Side note :

Recall from part ii) that the probability for exactly n draws is (⅞)ⁿ⁻¹ (⅛)


We see a ⅛. What is the other ⅞ ? It represents those (n+1)th, (n+2)th, (n+3)th, .... and draws that go all the way to infinity


Probability to get '8' on (n+1)th draw or later
= (⅞)ⁿ⁻¹ (⅞) = ⅞ⁿ


Add the two together , the result is still going to be (⅞)ⁿ⁻¹