A lottery game works as follows: The player draws a numbered

1) I got \(\frac{18}{125}\) in the following way

Three draws of 5 balls result in \(5^3=125\)options.
Not to lose the game, a player has to draw odd numbers on the first draw and even numbers on all other draws. Therefore to lose on the third draw, a player has to draw an odd number on the first draw, an even number on the second and then an odd number on the third draw. Hence, the options to draw (odd,even,odd)=\(3\times 2\times 3=18\)
Hence the probability is \(\frac{18}{125}\)

2) I got \(\frac{198}{3125}\) in the following way.

Again, no to lose a game a play have to draw odd number at the 1st draw and then even numbers...
To collect 7 points, there are 6 possible outcomes:
(1,2,2,2)
(1,4,2)
(1,2,4)
(3,2,2)
(3,4)
(5,2)
P(To collect 7 points and lose on the next)=P(to collect 7 points on the 2st and lose on the 3rd)+P(to collect 7 points on the 3rd and lose on the 4th) +P(to collect 7 points on the 4rth and lose on the 5th)
1) To collect 7 points at the 2nd and lose at the 3rd: 2 events (3,4),(5,2) and then any of the tree odd numbers => \(\frac{6}{5^3}\)
2) To collect 7 points on the 3rd and lose on the 4th, 3 events (1,4,2),(1,2,4),(3,2,2) and then any of the tree odd => \(\frac{9}{5^4}\)
3) to collect 7 points on the 4rth and lose on the 5th, one event (1,2,2,2) and the any of the tree odd=> \(\frac{3}{5^5}\)

Probability= \(\frac{6}{5^3} + \frac{9}{5^4} + \frac{3}{5^5}= \frac{198}{3125}\)

Thanks for the question. The second part was very interesting.