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The positions on a roulette wheel are divided into three colors: red, black, and green. There are 18 red positions, 18 black positions, and two green positions. If the wheel is fair, the chance that the ball lands in any position is equal to the chance that it lands in any other position, 1/38. What is the probability that in 7 independent spins of the wheel, the ball lands in a red slot four times, in a black slot twice and in a green slot once?

yuhki  Nov 21, 2014

Best Answer 

 #1
avatar+92808 
+10

I'm not sure about this...but I'll take a stab at it

The probability of red or black = 18/38 = 9/19

And the probability of green = 2/38 = 1/19

So the probability of a ball landing in red 4 times = (9/19)^4

And the probability of a ball landing in black twice is (9/19)^2

And the probability of landing in green once is (1/19)

And note that we have 7! arrangements of  these results......but some are indistiguishable from others

So, the totlal distinguishable arrangements is  7! / [4! 2! ] = 105

So the total probability is given by.... 105*(9/19)^4*(9/19)^2*(1/19) = 0.0624265233650037 = about 6.2%

 

CPhill  Nov 21, 2014
 #1
avatar+92808 
+10
Best Answer

I'm not sure about this...but I'll take a stab at it

The probability of red or black = 18/38 = 9/19

And the probability of green = 2/38 = 1/19

So the probability of a ball landing in red 4 times = (9/19)^4

And the probability of a ball landing in black twice is (9/19)^2

And the probability of landing in green once is (1/19)

And note that we have 7! arrangements of  these results......but some are indistiguishable from others

So, the totlal distinguishable arrangements is  7! / [4! 2! ] = 105

So the total probability is given by.... 105*(9/19)^4*(9/19)^2*(1/19) = 0.0624265233650037 = about 6.2%

 

CPhill  Nov 21, 2014

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