+82 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
+129898 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%
+129898 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%
