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
(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?
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
To solve these questions, we'll need to use the concept of probability and the principles of independent events. Here's how we can approach each question:
(b) To find the probability that the student encounters exactly five red lights, we can use the binomial distribution formula. Each set of traffic lights can be considered a trial, and the probability of encountering a red light (success) is 54.6%.
The formula for the probability of exactly k successes in n trials is:
P(X = k) = (n C k) * p^k * (1 - p)^(n - k),
Where (n C k) is the combination formula n choose k, p is the probability of success, and (1 - p) is the probability of failure.
In this case, n (the number of trials) is 12, k (the number of successes) is 5, and p (the probability of encountering a red light) is 54.6%.
Therefore, the probability of encountering exactly five red lights is:
P(X = 5) = (12 C 5) * (0.546)^5 * (1 - 0.546)^(12 - 5).
Calculating this expression will give you the desired probability.
(c) To find the probability that the student encounters at least three green lights, we need to consider all possible combinations of green lights. The student can encounter 3, 4, 5, 6, 7, 8, 9, 10, 11, or 12 green lights.
For each case, we calculate the probability using the binomial distribution formula as described in part (b). Then, we sum up the probabilities for all these cases.
P(at least 3 green lights) = P(X = 3) + P(X = 4) + ... + P(X = 12),
Calculate each of these probabilities using the formula described in part (b), and sum them up.
(d) To find the probability that the student encounters exactly six yellow lights over a five-day school week, we need to consider the number of yellow lights out of the total of 5*12 = 60 lights (12 lights per day).
Using the binomial distribution formula, each yellow light on a particular day has a 5.9% chance of occurring, and each day is treated as an independent trial.
The formula for the probability of exactly k successes in n trials can be applied here, with n = 60, k = 6, and p (the probability of encountering a yellow light on any given day) = 5.9%.
P(X = 6) = (60 C 6) * (0.059)^6 * (1 - 0.059)^(60 - 6).
Evaluate this expression to obtain the probability of encountering exactly six yellow lights.
Remember to use a calculator or statistical software to perform the calculations involving combinations and exponents.