Question 1181935: In the game “Making Green”, a player spins twice. If the player gets blue in one section and yellow in the
other, the player wins, because blue and yellow make green. The player can choose either spinner for each spin.
Three people play the game. Player 1 spins spinner A twice. Player 2 spins spinner A once and spinner B once. Player 3
spins spinner B twice. Who has the best chance to Make Green?
I can't add a picture, but hopefully this makes sense. Spinner A is broken down so each color (red, blue, yellow and orange) takes up 25% of the circle. Spinner B has 50% yellow and the other colors have an equal share of the other 50% (16.6% each).
Found 2 solutions by greenestamps, Solver92311:
Answer by greenestamps(13198)   (Show Source):
You can put this solution on YOUR website!
On spinner A, P(red)=P(blue)=P(yellow)+P(orange) = 1/4
On spinner B, P(yellow)=1/2; P(red)=P(blue)=P(orange)=1/6
Player 1 spins spinner A twice. To get green he needs to get either blue or yellow on the first spin and the other of those two colors on the second.
P(making green for player 1) = (2/4)(1/4) = 2/16 = 1/8
Player 2 spins spinner A once and spinner B once. To get green he needs to get either yellow on A and blue on B, or blue on A and yellow on B.
P(making green for player 2) = (1/4)(1/6)+(1/4)(1/2) = 1/24+1/8 = 4/24 = 1/6
Player 3 spins spinner B twice. To get green he needs to get either blue on the first spin and yellow on the second, or yellow on the first and blue on the second.
P(making green for player 3) = (1/6)(1/2)+(1/2)(1/6) = 1/12+1/12 = 1/6
ANSWER: Players 2 and 3 have an equal probability of making green; player 1 has a lower probability of making green.
NOTE: Don't use the approximation 16.6% for the probability of getting red, blue, or orange on spinner B. Keep those probabilities as the fraction 1/6. You want your calculations to be exact, not approximate.
In this particular problem, since the probabilities turn out to be equal for players 2 and 3, using the approximate percentage 16.6% would make the probabilities for those two players slightly different, giving the wrong answer to the question.
Answer by Solver92311(821)  (Show Source):
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