Question

3 A bag contains red and blue marbles, such that the probability of drawing a blue marble is 8 An experiment consists of drawing a marble, replacing it, and drawing another marble. The two draws are independent. What is the probability that both of the marbles drawn are blue? a. b. $ 24% * 39% C. d 0 14% 3 A bag contains red and blue marbles , such that the probability of drawing a blue marble is 8 An experiment consists of drawing a marble , replacing it , and drawing another marble . The two draws are independent . What is the probability that both of the marbles drawn are blue ? a . b . $ 24 % * 39 % C. d 0 14 %​

Answer 1

Answer:

My answer: B

Step-by-step explanation:

The probability mass function of random variable X which represents the number of blue marbles drawn in 2 draws with replacement.

X | P(X)

0 | 0.390625

1 | 0.46875

2 | 0.140625

The probability of drawing exactly one blue marble = 0.46875

Complete Question

A bag contains red and blue marbles, such that the probability of drawing a blue marble is 3/8. an experiment consists of drawing a marble, replacing it, and drawing another marble. The two draws are independent. A random variable assigns the number of blue marbles to each outcome.

To write the Probabilty mass function, we have to establish that in two draws with replacement, the possible number of blue marbles that can be drawn is 0, 1 and 2.

Probability of drawing a blue marble = P(B) = (3/8)

Probability of not drawing a blue marble = P(B') = 1 - (3/8) = (5/8)

- Probability of drawing 0 blue marbles in 2 draws with replacement = P(B') × P(B') = (5/8) × (5/8) = (25/64) = 0.390625

- Probability of drawing one blue marble in 2 draws with replacement

= [P(B) × P(B')] + [P(B') × P(B)]

= (2) × (3/8) × (5/8) = (30/64) = (15/32) = 0.46875

- Probability of drawing two blue marble in 2 draws with replacement = P(B) × P(B)

= (3/8) × (3/8) = (9/64) = 0.140625

The probability mass function of random variable X which represents the number of blue marbles draan in 2 draws with replacement.

X | P(X)

0 | 0.390625

1 | 0.46875

2 | 0.140625

b) Probability of drawing one blue marble in 2 draws with replacement

= [P(B) × P(B')] + [P(B') × P(B)]

= (2) × (3/8) × (5/8) = (30/64) = (15/32) = 0.46875

So, B is the answer.