If you are randomly sampling one at a time with replacement for a bag that contains eight blue marbles, seven red marbles, and five green marbles, what is the probability of obtaining

a. A blue marble in one draw from the bag?

b. Three blue marbles in three draws from the bag?

c. A red, a green, and a blue marble in that order in three draws from the bag?

d. At least two red marbles in three draws from the bag?

16. Answer the same questions as in problem 15, except sampling is one at a time without replacement.

Sampling with replacement.

a) 2/5 = 0.4

b) 8/125 = 0.064

c) 7/200 = 0.035

d) 1127/4000 = 0.28175

Sampling without replacement

a) 2/5 = 0.4

b) 14/285 = 0.0491

c) 7/171 = 0.0409

d) 861/3420 = 0.25175

Step-by-step explanation:

Total number of marbles = 8+7+5 = 20

Probability of drawing a Blue marble = P(B) = 8/20 = 2/5

Probability of drawing a Red marble = P(R) = 7/20

Probability of drawing a green marble = P(G) = 5/20 = 1/4

Sampling with replacement.

a) Probability of a blue marble in one draw from the bag = 8/20 = 2/5 = 0.4

b) Probability of three blue marbles in three draws from the bag = (8/20) × (8/20) × (8/20) = 8/125

c) Probability of a red, a green, and a blue marble in that order in three draws from the bag = (7/20) × (5/20) × (8/20) = 280/8000 = 7/200

d) Probability of at least two red marbles in three draws from the bag

This is a sum of the probability of drawing two red marbles out of three and three red marbles from three draws.

Probability of drawing two red marbles out of three draws = 3 × P(R) × P(R) × P(other balls) = (7/20) × (7/20) × (13/20) = 1911/8000

The 3 is there because this can be done in 3 different orders

P(other balls) = 1 - (7/20) = 13/20

Probability of drawing three red marbles in three draws = (7/20) × (7/20) × (7/20) = 343/8000

Probability of at least two red marbles in three draws from the bag = (1911/8000) + (343/8000) = 2254/8000 = 1127/4000

Sampling without replacement

In this sampling method, the sample space and number of marbles decrease as each marble goes out of the bag.

a) Probability of a blue marble in one draw from the bag = 8/20 = 2/5 = 0.4

b) Probability of three blue marbles in three draws from the bag = (8/20) × (7/19) × (6/18) = 336/6840 = 14/285

c) Probability of a red, a green, and a blue marble in that order in three draws from the bag = (7/20) × (5/19) × (8/18) = 280/6840 = 7/171

d) Probability of at least two red marbles in three draws from the bag

This is a sum of the probability of drawing two red marbles out of three and three red marbles from three draws.

Probability of drawing two red marbles out of three draws = 3 × (7/20) × (6/19) × (12/18) = 1512/6840

The 3 is there to multiply because this can be done in three different orders

Probability of drawing three red marbles in three draws = (7/20) × (6/19) × (5/18) = 210/6840

Probability of at least two red marbles in three draws from the bag = (1512/6840) + (210/6840) = 1722/8000 = 861/3420

found it xd it ok yt

-2 is the right