1. simple probability question

Could someone help me with this question ; it would be much appreciated

Ivan and Jo are chess players. In any game the probabilities of Ivan beating Jo, Jo beating Ivan or the game resulting in a draw are 0.6. 0.1 or 0.3 respectively. They play a series of three games. Calculate the probability that:
a) The three games are drawn

b) They win alternate games with Ivan winning the first one
c) Exactly 2 of the 3 games are drawn
d) Joe does not win a game

I'm not sure if this is right but for
a) i just did 0.3 ^ 3
and for
b) 0.6 * 0.1 * 0.6

So yeah, if anyone could assist me on these, it would be great

2. a) $P(\text{Three games are draw}) = P(\text{first game draw}) \times P(\text{second game draw}) \times P(\text{third game draw})$

So, yes, it's right!

b) $P(\text{alternate wins, Ivan wins first}) = P(\text{Ivan wins}) \times P(\text{Jo wins}) \times P(\text{Ivan wins})$

Yes, right.

c) $P(\text{2 games draw}) = P(DD\bar{D}) + P(D\bar{D}D) + P(\bar{D}DD)$

where D represents the even that the game is a draw.

d) $P(\text{Jo does not win}) = P(\text{Jo loses all game}) + P(\text{Jo loses 1 game and draws 2})+ P(\text{Jo loses 2 games and draw 1}) + P(\text{Jo draws all game})$