Parking spaces at a business school are arranged by a random monthly lottery. there are 3 spaces for every 10 students who want one and all of them have an equal chance of getting one. what is the probability of a student getting a parking space for at least four out of the nine months? (as a decimal)

27.034%  
Let's define the function P(x) for the probability of getting a parking space exactly x times over a 9 month period. it would be: 
P(x) = (0.3^x)(0.7^(9-x))*9!/(x!(9-x)!)  
Let me explain the above. The raising of (0.3^x)(0.7^(9-x)) is the probability of getting exactly x successes and 9-x failures. Then we shuffle them in the 9! possible arrangements. But since we can't tell the differences between successes, we divide by the x! different ways of arranging the successes. And since we can't distinguish between the different failures, we divide by the (9-x)! different ways of arranging those failures as well. So P(4) = 0.171532242 meaning that there's a 17.153% chance of getting a parking space exactly 4 times. 
Now all we need to do is calculate the sum of P(x) for x ranging from 4 to 9.

So
 P(4) = 0.171532242
 P(5) = 0.073513818
 P(6) = 0.021003948
 P(7) = 0.003857868
 P(8) = 0.000413343
 P(9) = 0.000019683
  And
 0.171532242 + 0.073513818 + 0.021003948 + 0.003857868 + 0.000413343
+ 0.000019683 = 0.270340902 
 So the probability of getting a parking space at least four out of the nine months is 27.034%

128/3reason being 5 and 1/3 = 16/316/3 • 2/3 = 32/31 and 1/3 = 4/3 32/3 • 4/3 this will equal 128/3

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