Answer:
1) Answer is D
2) Answer is C
3) Answer is D
Explanation:
Problem 1
Since we are given 12 numbers to choose from, there's a 2/12 chance of selecting a 2. Additionally, there's a 3/11 chance of selecting a number divisible by 4 AFTER selecting a 2. Thus, since the two events are independent, their probabilities are multiplied, so the probability of selecting a 2 and then a number divisible by 4 is (2/12)(3/11) = 6/132 = 1/22, so option D is correct.
Problem 2
There's a 4/12 chance of selecting an odd number and a 6/11 chance of selecting an even number AFTER selecting an odd number (remember that 0 is neither even nor odd). Multiplying the probabilities, the probability of drawing an odd number and then drawing an even number is (4/12)(6/11) = 24/132 = 2/11, so option C is correct.
Problem 3
There's a 2/12 chance of selecting a zero and a 1/11 chance of selecting a second zero. Thus, multiplying their probabilities, the probability of drawing 2 zeroes is (2/12)(1/11) = 2/132 = 1/66, so option D is correct.