Answer: 77.28
Explanation:
To solve this problem, we can use the formula:
P(G) = P(G ∩ NG') + P(G ∩ NG)
where P(G) is the probability that a resident has green eyes, P(G ∩ NG') is the probability that a resident has green eyes and does not wear glasses, and P(G ∩ NG) is the probability that a resident has green eyes and wears glasses. NG' represents the complement of wearing glasses, i.e. not wearing glasses.
We are given that 32% of the residents wear glasses, so the probability of not wearing glasses (NG) is:
NG = 1 - 0.32 = 0.68
We are also given that 11% have green eyes and wear glasses:
P(G ∩ NG) = 0.11
We can use this information to find the probability of having green eyes and not wearing glasses:
P(G ∩ NG') = P(G) - P(G ∩ NG) = 0.11 - P(G ∩ NG')
We are also given that 45% of the residents have neither green eyes nor wear glasses:
P(NG' ∩ NG) = 0.45
We can use this information to find the probability of not having green eyes and wearing glasses:
P(NG ∩ G') = P(G' ∩ NG) = 0.45
We can now set up a system of equations to solve for P(G) and P(G ∩ NG'):
P(G ∩ NG') = P(G) - P(G ∩ NG)
P(G' ∩ NG) = P(NG) - P(NG' ∩ NG)
Substituting the values we have:
P(G ∩ NG') = P(G) - 0.11
0.45 = 0.68P(G' ∩ NG)
Solving for P(G' ∩ NG), we get:
P(G' ∩ NG) = 0.45/0.68 = 0.6628
Substituting this value into the first equation, we get:
P(G ∩ NG') = P(G) - 0.11
P(G) = P(G ∩ NG') + 0.11
P(G) = 0.6628 + 0.11
P(G) = 0.7728
Therefore, the probability that a randomly selected resident of Toronto has green eyes is approximately 0.7728 or 77.28%.