Answer:
The data we have is:
the probability that a civil servant own a car is 1/6
Two civil servants are selected at random.
a) The probability that each own a car.
Ok, here we have two events:
Person 1 haves a car.
Person 2 haves a car.
The probability of each of those events is the same, 1/6.
P1 = 1/6
P2 = 1/6
Now, the probability of both events happening at the same time is equal to the product of the individual probabilities.
P = P1*P2 = (1/6)*(1/6) = 1/36.
b) Only one has a car.
Suppose that Person 1 has the car and Person 2 has not a car.
The probability for person 1 is 1/6.
And for person 2, is the negation of having a car:
So if we have prob of having a car = 1/6
Then the probability of not having a car is = 1 - 1/6 = 5/6.
Then we have:
P1 = 1/6
P2 = 5/6.
The joint probability is:
P = P1*P2 = (1/6)*(5/6) = 5/36.
But we also have the case where person 1 does not have a car, and person 2 does have one, then we have a permutation, and the actual probability is two times the obtained above.
P = 2*(5/36) = 10/36
