probability

Question

6 blue marbles and 8 green marbles are placed in a jar. Two marbles are then drawn in succession without replacement. What is the probability that the first marble drawn is green, given that the second marble drawn is blue?

Answer 1

-First, calculate the total number of marbles by adding the number of green and blue marbles (6 + 8 = 14).

-To determine the probability of the first marble being green, divide the number of green marbles (8) by the total number of marbles in the jar (14). --> 8 ÷ 14 = ?

Note: For this question, it is irrelevant that the second marble is blue (because there are still more blue and green marbles left in the jar and the removal of the second marble has no effect on the color of marble you previously withdrew from the jar).

Answer 2

Let A be the event green of the first pick

Let B be the event blue on the second pick

 

P(A|B) = p(A∩B)/P(B)

 

P(A∩B) = (8/14) * (6/13)

 

P(B) = (8/14) * (6/13) + (6/14) * (5/13)

 

P(A|B) = 8*6/[(8*6) + (6*5)] = 48/78 = 12/17