Question 1025224: A roulette wheel has 38 pockets numbered 0,00,and 1 to 36. Of all 38 pockets 18 are red, 18 are black and 2 are green. Each time the wheel is spun, a ball lands in one of the pockets, and each pocket is equally likely.
Question: If you spin the wheel twice, what is the probability that the ball lands in a diffrent color pocket in both spins?
I am having trouble setting up the problem, or if I am even doing the problem correctly.
Answer by mathmate(429)   (Show Source):
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Question:
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A roulette wheel has 38 pockets numbered 0,00,and 1 to 36. Of all 38 pockets 18 are red, 18 are black and 2 are green. Each time the wheel is spun, a ball lands in one of the pockets, and each pocket is equally likely.
Question: If you spin the wheel twice, what is the probability that the ball lands in a diffrent color pocket in both spins?
I am having trouble setting up the problem, or if I am even doing the problem correctly.
Solution:
Each spin is independent of the others, so the probabilities of the three possible outcomes (R,G,B) for each spin are:
P(R) = (18/38)
P(G) = (2/38)
P(B) = (18/38)
The probability of a two step experiment can be obtained as the product of the two outcomes. For example,
P(RR)=(18/38)(18/38)=81/361=0.2244 (approximately)
P(GB)=(2/38)(18/38)=9/361=0.0249 (approximately)
...
Since there are only three colours, the possible outcome (of two spins) with identical colours are RR, GG, BB, the probability of each can be calculated as above.
Since the other outcomes (RG,RB,GR,GB,BR,BG) are all different colours, the sum of these probabilities is the complement of the three identical colours, i.e.
P(same colour)=P(RR)+P(GG)+P(BB)
P(different colours)=1-P(same colour)
Alternatively, (but it's a little more work),
P(different colours)=P(RG)+P(RB)+P(GR)+P(GB)+P(BR)+P(BG)
This set-up should get you the required answer in a few simple steps.
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