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For a certain airport containing three runways it is known that in the ideal setting the following are the probabilities that the individual runways are accessed by a randomly arriving commercial jet: runway 1 = 1/2, runway 2 = 1/8, runway 3 = 3/8. What is the probability that 6 randomly arriving airplanes are distributed in the following fashion?

Runway 1:   2 airplanes
Runway 2:   1 airplanes
Runway 3:   3 airplanes

yuhki  Nov 21, 2014

Best Answer 

 #1
avatar+93038 
+15

I'll give this one a shot

The probabilty that 2 planes use runway 1 is (1/2)^2

The probability that 1 plane uses runway 2 is (1/8)^1

The probability of three planes u8se runway 3 is (3/8)^3

And we have 6! arrangements of these probabilities....but some are indistinguishable from others. So, the number of distinguishable arrangements is 6! / [ 2! * 3!] = 60

So the probability is .... 60*(1/2)^2*(1/8)^1*(3/8)*3 = 0.098876953125 = about 9.8%

 

CPhill  Nov 21, 2014
 #1
avatar+93038 
+15
Best Answer

I'll give this one a shot

The probabilty that 2 planes use runway 1 is (1/2)^2

The probability that 1 plane uses runway 2 is (1/8)^1

The probability of three planes u8se runway 3 is (3/8)^3

And we have 6! arrangements of these probabilities....but some are indistinguishable from others. So, the number of distinguishable arrangements is 6! / [ 2! * 3!] = 60

So the probability is .... 60*(1/2)^2*(1/8)^1*(3/8)*3 = 0.098876953125 = about 9.8%

 

CPhill  Nov 21, 2014
 #2
avatar+94203 
0

That sounds like it could be right Chris.  

 

I don't have any better ideas.  3 more points for you   :))

Melody  Nov 22, 2014

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