Question 1119066: The serial numbers on $5 bills include three letters followed by seven digits. Assuming the digits are assigned at random, what is the probability that a serial number will contain,
(a)exactly two 5s
(b) at least four 5's
(c) all 5s?
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! 10 digits so for two 5s, it would be 7C2, the number of different places the 5s can occur, multiplied by 0.1^2*0.9^5=0.1240
at least four: do 0, 1,2,3 and subtract from 1. For 0, it would be.9^7=0.4783
For 1 it would be 7*.1*.9^6=0.3720
For 2 it would be 0.1240 from above
For 3 it would be 35 (7C3)*.1^3*.9^4=0.0230
That sum is 0.9973 so the probability of at least four 5s is 0.0027
Can check by doing 4, which is 0.00255 and 5 which would be 0.00017.
Probability all 5s is .1^7, or 1 x 10^(-7)
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