Question

A jar contains 3 blue marbles, 6 red marbles, 1 green marble, and 2 black marbles. A marble is chosen at random from the jar and replaced. Then another marble is chosen at random. Find the probability of the first marble being red and the second marble being green.

Answer 1

Answer: 0.0416

Step-by-step explanation:

A jar contains 3 blue marbles, 6 red marbles, 1 green marble, and 2 black marbles

Total possible outcomes will be:

3+6+1+2 = 12 marbles

Probability of picking red

P(red) = 6/12 = 0.5

P(green) = 1/12 = 0.0833

probability of the first marble being red and the second marble being green will be

P(red) ×P(green) = 0.5×0.0833

= 0.0416

Answer 2

The probability of picking red first and green second [with replacement] = 1/24

Explanation:

Extracting the key information from the question:

*** 1 jar contains 3 blue, 6 red, 1 green and 2 black.

*** 1 marble is chosen at random from the jar and replaced.

*** Another marble is chosen at random.

*** We are required to find the probability that the first is red and the second is green.

First we calculate the total marbles in the jar.

3 blue + 6 red + 1 green + 2 black = 12 marbles.

Possibility of picking a red marble = number of red marbles/total number of marbles.

P(red) = 6/12

Again, probability of picking green P(green) = number of green marbles/total number of marbles.

P(green) = 1/12

Since the pick is with replacement, the probability that the first is red and the second is green =

P(red,green) = 6/12 × 1/12 = 6/144 = 1/24