I can only help you with the (a) part.
(a) We calculate the number of ways to get a successful choice of 3 balls.
There are 15C3 = [(15)(14)(13)]/[(3)(2)(1)] = 455 ways to choose any 3 balls.
From those 455, we will subtract the number of unsuccessful choices,
i.e., those containing a pair of consecutive choices.
There are 14 pairs of consecutively numbered balls: {1,2},{2,3},...,{14,15}
For each of those pairs of balls we can select a third ball to go with it
in 13 ways. That's (14)(13) = 182 ways.
However, among that 182, we have counted each case of the following 13 choices
twice each:
{1,2,3},{2,3,4},...{13,14,15}
To show that, take for example the choice {6,7,8}. It was counted once when
we put 8 with the pair {6,7} and again when we put 6 with the pair {7,8}.
So we subtract to get 182-13 = 169 unsuccessful choices.
We subtract these 169 ways from the 455 to get 455-169=286 possible successful
choices.
So the probability that Mr. Lewis makes a successful choice is 286/455 = 22/35
(b) I suspect I do not not understand the game well enough to do the (b) part.
As I see it, there is not enough information. We are not told how much money he
wins or how much it costs to place a bet. As I see it, the problem will be
quite different if he walks into the casino with millions of dollars in his
pocket than if he walks in with only enough money to play once, and wins enough
each time to continue playing 12 times. So it looks to me as if we would also
need to know how much he wins per successful bet. As I said, I suspect I do not
understand the game well enough to answer the (b) part. Perhaps another tutor
who understands the game will jump in here. Otherwise post again, explaining
how the game operates a little better.
Edwin