2 Answers

Janovak · Answer 1 · 2020-06-06T19:40:04+0000

Answer:

The probability of drawing a second green marble, given that the first marble is green is:

(3)/(8)

Let A denote the event that first marble is green.

B denote the event that the second marble is green.

A∩B denote the event that both the marbles are green.

Let P denote the probability of an event.

We are asked to find:

P(B|A) i.e. probability of drawing a second green marble, given that the first marble is green.

We know that:

$P(B|A)=(P(A\bigcap B))/(P(A))$

Probability of drawing one green marble is 2/5 i.e.

P(A)=(2)/(5)

The probability of drawing two green marbles from the urn without replacement is 3/20 i.e.

$P(A\bigcap B)=(3)/(20)$

Hence, we have:

$P(B|A)=((3)/(20))/((2)/(5))\\\\\\i.e.\\\\\\P(B|A)=(3* 5)/(20* 2)\\\\\\P(B|A)=(3)/(8)$