Question

A bag contains 7 blue, 3 green, and 3 red marbles. If Agnes randomly draws two marbles from the bag, replacing the first before drawing the second, what is the probability of drawing a green and then a blue marble?

Answer 1

Answer:

$\frac{21}{169}$

Step-by-step explanation:

Given:

Number of blue marbles in a bag = 7

Number of green marbles  in a bag = 3

Number of red marbles  in a bag = 3

Agnes randomly draws two marbles from the bag, replacing the first before drawing the second.

To find:

the probability of drawing a green and then a blue marble

Solution:

Let A denotes the event of drawing a green marble and B denotes the event of drawing a blue marble

$P(A)=\frac{3}{13}\\ P(B)=\frac{7}{13}$

As Agnes randomly draws two marbles from the bag, replacing the first before drawing the second, so A and B are independent events.

So, the probability of drawing a green and then a blue marble

$=P(A)P(B)\\=(\frac{3}{13} )(\frac{7}{13} )=\frac{21}{169}$