Math Is Fun Forum

Hello everyone. We have a problem in school with a lot of different probability problems, including the following, which I'm not 100% sure about:

Penalties

A match between team A and team B has to be decided on penalties. Every team has to kick 5 times. We assume that be every try there's a 70% chance of making a goal. Team A starts. Find the probability that after 5 tries to each team, team A has won, team B has won or the penalties end equal, when:
a) No team has missed and team B has to kick for the last time.
b) No team has missed and team B has to shoot for the 2nd last time.
c) A has missed the first shot.
d) Find more of interest on your own.

The way I think about it is this:

Definition:
S = Score
F = Fail

a) team A has a 30% chance of winning and thus team B has a 30% chance of losing. There's a 70% chance of the penalties ending equal.

b) There are several ways the match can end:
Team A score is: SSSS-
Team B score is: SSS--

Assuming team A wins:
B: 4. S 5. F
A:        5. S

B: 4. F (5. S or 5. F - doesn't matter)
A:         5. S

So the probability of team A winning would be the probability of each of these, i.e.

and

Assuming team B wins:
B: 4. S 5. S
A:        5. F

Probability:

Assuming the match ends equal:
B: 4. S 5. S
A:        5. S

B: 4. S 5. F
A:        5. F

Probabilities:

and

c)
There are LOTS of combinations here:

Assuming team A wins:
1.
A: 1. M 2. S 3. S 4. S 5. S - 4 scores
B: 1. M 2. M 3. S 4. S 5. S - 3 scores 2 misses

Team B is a binomial, team A is not:

2.
A: 1. M 2. M 3. S 4. S 5. S - 3 scores 1 miss
B: 1. M 2. M 3. M 4. S 5. S - 2 scores 3 misses

Both teams are binomials:

...and so forth.

Assuming team B wins:
1.
A: 1. M 2. S 3. S 4. S 5. S
B: 1. S 2. S 3. S 4. S 5. S

...continues as above.

Equal:
The same.

d)
Both teams have shot twice, but team A has missed both of their kicks. What's the probability team A will win?

and

Team A has shot three times and missed one. Team B has missed all three. What's the probability the match will end equal?

The problem is as follows:
"A card is drawn from a deck of cards, the card is noted and thereafter returned to the deck. This experiment is repeated 12 times. Find the probability for the following events:
A: There are exactly 9 court cards"

There are 3 more events but more or less the same as A, so I omitted them. The formulation of the problem is originally also not in english, I've translated it.

This is what I've got so far:

Total court cards per deck: 12 <=> probability of choosing a court card is 12/52.
Thus the probability for choosing a court card 9 times must be (12/52)^9.

I'm unsure about what to do next. Should I multiply it by (40/52)^3 since that's the probability of the 3 remaining picks from the deck?

Thanks for helping!