can someone show the steps and explain c and d parts of the problem. I'm not really sure what these formulas are or how they work pertaining to the question. please explain this like im 5 years old so confused!Answers
(c) The significance level of the test in this case is the probability of landing the ball in the same space for the 2nd time, in 3 or fewer spins. This can happen in 2 spins or 3 spins.
So we can write 
= P(X=2) + P(X=3), where X is the variable representing the number of spins until the ball lands in the same space for the second time.
Probability of selecting a single space out of the 38 equally likely spaces = 1/38
Probability of selecting a different space other than a designated single space = 1 - 1/38 = 37/38
Probability (X=2) = probability of selecting a single space x probability of the ball landing in the same slot x probability of the ball landing in the same slot = 38 x (1/38) x (1/38)= 
P(X=3) = probability of selecting a single space x probability of the ball landing in the same slot x probability of the ball landing in the same slot x probability of the ball landing in a different x no. of possible combinations among the 3 spaces.
The last part, no. of possible combinations among the 3 spaces can be (same space, different space, same space) or (different space, same space, same space). Hence this is equal to 2.
P(X=3) = 38 x (1/38) x (1/38) x (37/38) x 2
Hence, 
, or 7.76%
(d) The power of the test is the above probability, given that the probability of the ball landing in a particular space is 0.075 and the probability of the ball landing in each of the other spaces is 0.025.
For the following calculations, the ball can either land in the biased space with probability 0.075 or it can land in one of the unbiased spaces with probability 0.025
Probability (X=2) = probability of the ball landing in the biased slot x probability of the ball landing in the biased slot + probability of selecting a single unbiased slot out of the 37 slots x probability of the ball landing in the same slot x probability of the ball landing in the same slot = 0.075 x 0.075 + 37 x 0.025 x 0.025 = 0.00563 + 0.02313 = 0.02875
P(X=3) = probability that the ball lands in the biased slot twice in 3 spins + probability that the ball lands in one of the unbiased slots in 3 spins = probability of ball landing in biased slot x probability of ball landing in biased slot x probability of ball landing in unbiased slot x no.
Of possible combinations (=2 as explained in part (c)) + probability of selecting an unbiased slot out of the 37 unbiased slots x probability of ball landing in that unbiased slot x probability of ball landing in the same unbiased slot x probability of ball landing in any of the other slots x no. of possible combinations (=2 as explained in part (c)) = 0.075 x 0.075 x (1-0.075) x 2 + 37 x 0.025 x 0.025 x (1-0.025) x 2 = 0.01041 + 0.04509 = 0.0555
Power = P(X=2) + P(X=3) = 0.02875 + 0.0555 = 0.08425, or 8.425%