Answer:
Here we have four individual events.
First selection, second selection, third selection and fourth selection.
Where initially, the options are 16 Americans and 13 Canadians.
Now, to see if the events are dependent or independent.
Suppose that in the first selection, an American is selected.
Then the second selection has 15 Americans and 13 Canadians as options.
Now, if in the first selection, a Canadian is selected.
Then in the second selection, the options will be 16 Americans and 12 Canadians.
Then the events are not independent.
Now we want to find the probability of randomly selecting a four-person committee consisting entirely of Canadians from a pool of 16 Americans and 13 Canadians
The probability of randomly selecting a Canadian is equal to the quotient between the number of Canadians, and the total number of postulants.
For the first selection, we have 13 Canadians, and 29 people in total, then the probability is:
P1 = 13/29.
Suppose that a Canadian is selected, now there are 12 Canadians and 28 people in total, then the probability for the second selection is:
P2 = 12/28
with the same reasoning as above, the probabilities for the third and fourth selections will be:
P3 = 11/27
P4 = 10/26.
The joint probability will be equal to the product of the individual probabilities, this is:
P = P1*P2*P3*P4 = (13/29)*(12/28)*(11/27)*(10/26) = 0.03
