Multiple Choice:

1. The probability that the first marble taken out of the box is red is 2/12 = 1/6. The probability that the second marble taken out of the box is red is 1/11. The probability of two red marbles taken out of the box is the product of 1/6 and 1/11. This is 1/66.

2. P(A) = the probability of the first child choosing a flavor.

P(B) = the probability of the second child choosing a flavor different than the one chosen by the first child.

P(C) = the probability of the third child choosing a flavor different than the flavors chosen by the first two children.

P(D) = the probability of the fourth child choosing a flavor different than the flavors chosen by the other three children.

P(A and B and C and D) = P(A)·P(B | A)·P(C | A and B)·P(D | A and B and C).

P(A) = 8/8.

P(B | A) = 7/8.

P(C | A and B) = 6/8.

P(D | A and B and C) = 5/8.

P(A and B and C and D) = 105/256.

3. The probability of choosing an apple is:

(3 baskets/6 baskets)·(2 apples/3 fruits) + (2 baskets/6 baskets)·(3 apples/4 fruits) + (1 basket/6 baskets)·(0 apples/3 fruits) =

1/3 + 1/4

.58.

4. The mean μ = 10·.7 = 7.

The standard deviation σ = √(10·.7·(1 - .7)) = 1.45.

5. 1/66

Free Response:

1. A) p(less than 250,000 newspapers sold) = .01 + .07 + .14 + .25 = .47.

b) The expected amount is $100,000 + $10,000·(.15 + .1 + .04) = $100,000 + $2,900 = $102,900.

c) p( (x>260,000) and (270,000 < x < 280,000) ) = p(270,000 < x <280,000)/p(x>260,000) = .1/(.15 + .1 + .04) = .345.

2. a. The sum of the given probabilities is 0.9, so P(blue) = 0.1

b. The sum of the given probabilities is 0.7, so P(blue) = 0.3.

c. P(plain M&M is red, yellow, or orange) = 0.2 + 0.2 + 0.1 = 0.5. P(peanut M&M is red, yellow, or orange) = 0.1 + 0.2 + 0.1 = 0.4.

3. An individual light remains lit for 3 years with probability 1 − 0.02; the whole string remains lit with probability (1− 0.02)20 = (0.98)20 = 0.6676.