Probability problem-middle school level A spinner is divided into 15 equal sections, 5 red , 4 blue, 3 green, and 3 yellow.
A) If the spinner is spun once, what is the probability that it will stop on a green section?
 Is it 3/15, 20%?
B) If the spinner is spun twice what is the probability that it will stop on a green section both times?
I am not sure.. Is it 3/15+3/15 therefore 6/15?
c) If the spinner is spun twice what is the probability that it will stop on a green section once and a blue section once?
...I am lost.
D)If you gather data from 100 rounds in which the spinner is spun twice, in how many rounds would you expect to land on a green section once and a blue section once (in either order)?
Thanks!
 A: A) Yes, exactly right.
B) When you're calculating probability for multiple occasions, you multiply. That is, if $P(G)$ is the probability that you spin green, the probability of it occurring twice is $P(G)^2$ (at least in your case).
C) Like above case, you're going to have to find $P(G)$ and $P(B)$ (spin blue). You can either spin green then blue, or vice versa. What can you conclude?
D) This is just telling you to to apply the probability you calculated in C to real life. You'd expect it to occur (probability)*100 times.
A: A) $1/5$.
B) $1/5$ times $1/5$ which is $1/25$.
A: A) Yes
B) Consider, are two greens in a row more likely (than a green and then something else)? If an event is 1/5 and you want the odds of it occurring twice in a row, you multiply, 1/5 * 1/5 = 1/25. For a coin, heads is 1/2. To get two heads in a row? 1/4. Ten in a row? 1/1024 
C) a Bit tougher. Green is 1/5, blue 4/15 so GB is 4/75. But BG is 4/15 * 1/5 or another 4/75 and these add to 8/75. Every color pair will add to 100/100. 
D) This is just the ratio 8/75 = 10.667/100 or 10-2/3 times per hundred experiments. 
