A student must pass through 12 sets of traffic lights on his way to university. Suppose that each of the lights is green 39.5% of the time, yellow 5.9% of the time, and red 54.6% of the time. Suppose also that the traffic lights function independently

Question

(b) What is the probability that the student encounters exactly five red lights on his way to university one day?
(c) What is the probability that the student encounters at least three green lights on his way to university one day?
(d) In a five-day school week, what is the probability that the student encounters exactly six yellow lights on his way to university?

Answer 1

a) prob(red) = .546
prob(not red) = .454
prob(exactly 5red) = C(12,5) (.546)^5 (.454)^7
= appr .153

b)
prob (green) = .395
prob(not green) = .605

prob(at least 3 green) = prob(3green) + prob(4 green) + ... + prob(12 green)
=1 - prob(0 not green) - prob(1 not green) - prob(2 not green)

= 1 - C(12,0) (.605)^0 (.395)^12 - C(12,1) (.605) (.395)^11 - C(12,2) (.605)^2 (.395)^10
= 1 - ....

you do the button-pushing