Question 1181935
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On spinner A, P(red)=P(blue)=P(yellow)+P(orange) = 1/4<br>
On spinner B, P(yellow)=1/2; P(red)=P(blue)=P(orange)=1/6<br>
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<br>
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<br>
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<br>
ANSWER: Players 2 and 3 have an equal probability of making green; player 1 has a lower probability of making green.<br>
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.<br>
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.<br>