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68% of cars on a toll-road have EZ-pass. 5 cars are approaching the toll-gates. What is the probability that:

a. None have E-Z pass?
b. All have E-Z pass?
c. All but 2 have E-Z pass?
d. At least 2 have E-Z pass?
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4 Answers

This is a binomial distribution problem. I highly recommend researching this distribution online or in your textbook. Here is the pmf. P(N=n) = p(n) = k C n q^n*(1-q)^(k-n) where k is the total number of "trials," n is the number of "successes," and q is the probability of success in a single trial. k C n is the number of combinations of n objects chosen from k objects. In this problem the trials are the cars at the toll booth, k = 5, q=0.68.
a) n= 0 p(0) = (5 C 0)*0.68^0*0.32^5=0.0034
b) n= 5 p(5) = (5 C 5)*0.68^5*0.32^0=0.1454
c) n= 3 p(3) = (5 C 3)*0.68^3*0.32^2=0.3220
d) n= 2, 3, 4, or 5 We can get this from n is not 0 or 1 or 1-p(0)-p(1). We already have p(0)=0.0034
p(1)= (5 C 1)*0.68^1*0.32^4=0.0357
so the answer is 1-.0034-.0357=.9609

You will need the binomial distribution B(5,0.68;k) with probabilities P=C(5,k)*0.68k*0.325-k , where C(5,k) is the binomial coefficient.
a) k=0,  P=C(5,0)*0.680*0.325=0.325=0.00335.
If you have a statistics calculator like the TI-84, use the function binomialpdf(5,0.68,0).
b) k=5, use binomialpdf(5,0.68,5), get P=C(5,5)*0.6805*0.320=0.145
c) k=3, use binomialpdf(5,0.68,3), get P=0.322
d) k≥2, P(k≥2)=1-P(k<2)=1-0.039=0.961
On the TI-84, use 1-binomialcdf(5,0.68,1). (This is the cumulative distribution, it adds up
all values 0..k.)
a) (1-.68)^5 = .32^5
b) .68^5
c) 5C2 (.68)^2 (.32)^3, binomial probability
d) 1 - ∑{i=0,1)5Ci (.68)^(5-i) (.32)^i, using complement set
Given P(a) = 68%
P(/a) = 1 - P(a) = 1-0.68 = 32%
P(b) = 1 x P(a) = 68%
P(c) = P(a) x (1-P(/c)) = 40.8%
P(d) = P(a) x (1-P(/d)) = 27.2%


Tutors. Please confirm. i had to review to determine what was wrong.  Thanks!