For #2 it should be 1/2 since each time you toss a coin there's half a chance of getting tails and 1/2 a chance of getting heads. If it asks the probability of getting 3 tails in 4 consecutive tosses than it should be 1/2 * 1/2 * 1/2 * 1/2= 1/16

Hmm I don't think that's right. The way you're doing it the probability of getting 3 H/1 T is same as 2 H/ 2 T, or even 4 H (cause each would be 1/2). 1/16 would be the probability of getting exactly HHHT, but what about HHTH (another 1/16) or HTHH (another 1/16) or THHH (another 1/16). So it comes out to a total of 4/16 or 1/4.

But if you don't wanna count like that, (which you really cant for like one of the problems I saw asking this with 9 coins. You do this:

Total possible outcomes is 2^4 =16, right? Now how many ways can you choose 3 Heads out of 4 trials? Since it doesnt matter if you get HHHT or HHHT (lol), it's a combination 4C3 = 4!/3!1! = 4

So the probability of getting 3 heads out of the 4 trials is 4/16=1/4.

I think there was another way too, which can be used if the probability of the two events are not equal (what I said wouldn't work if the probability of getting a head was 2/3 and tail 1/3).

I think it had something like nCr*p^r*q^n-r (where p was the probability of the event u want and q was the complementary event I think). Don't quote me on this one though. This is what I remember right now, it can be wrong (likely). (but it does give the right answer here, 4C3=4, p=1/2, q=1/2, r=3, n=4, so you get 4*(1/2^3)*(1/2^1)=4/16, 1/4, so maybe that is the right one)