Help of Probability!?

Question

I absolutely abhor probability questions so I need your help!

1. If you flip three coins simultaneously, what's the probability you'll get three heads for the first time on the fifth toss?
a. 0.125
b. 0.0733
c. 0.763
d. 0.1094
e. 0.5

2. You have a rather strange die: three faces are marked with the letter A, two faces with the letter B, and one face with the letter C. What is the probability that you will get a B in three or fewer rolls?
a. 0.333
b. 0.704
c. 0.556
d. 0.037
e. 0.296

3. Imagine a country where only one of every 5 births is a girl. To increase their chances of having a girl, a family is willing to have many children. What is the probability that the first girl they have is the fourth baby?
a. 0.1024
b. 0.4096
c. 0.2048
d. 0.0512
e. 0.0819

Answer 1

1.
Probability of three heads is (1/2)^3 = 1/8
Probability of not three heads is 1 - 1/8 = 7/8
Probability of not getting heads on the first four tries is
(7/8)^4
So the answer is (7/8)^4 * 1/8 = 7^4 / 8^5 = 2401/32768 = .07
(answer B)

2.
Probability of getting a B is 1/3
Probability of not getting a B is 2/3
Probability of getting B on first roll is 1/3
Probability of getting B on second roll (but not the first) is 2/3 * 1/3 = 2/9
Probability of getting B on third roll (but not the first two) is 2/3 * 2/3 * 1/3 = 4/27
Probability of getting B on at least one of the first three rolls is 9/27 + 6/27 + 4/27 = 19/27 = .7
(answer B)

3.
Probability of getting a girl is .2
Probability of not getting a girl is .8
Probability of not getting a girl on first three tries is .8 * .8 * .8 = .512
Probability of then getting a girl on fourth baby is .512 *.2 = .1024
(Answer A)

Answer 2

1. (1-1/8)^4 * 1/8 = (7/8)^4 * 1/8 = 7^4/8^5 = .07327.. (b)

2. Pr that you don't get it in 3 or fewer rolls is (1-2/6)^3 = .296..
so .7037... (b)

3. (4/5)^3 * (1/5) = 4^3 / 5^4 = .1024 (a)