Question

Andrew has a bag that contains 8 blue marbles, 6 red marbles, and 10 green marbles. He selects a marble, replaces it in the bag, and then selects another. What is the probability that the first marble is blue and the second marble is red?

Answer 1

Answer:
`(1)/(12)`

Explanation:

Given : Andrew has a bag that contains 8 blue marbles, 6 red marbles, and 10 green marbles.

Total marbles = 8+6+10=24

Probability of drawing first marble is blue :

`P(B)=\frac{\text{Number of blue marbles }}{\text{Total marbles }}\\\\=(8)/(24)=(1)/(3)`

Since, he replaces the marbles so the total marbles remains same in bag.

The probability of that second marble is red :-

`P(R)=(6)/(24)=(1)/(4)`

Since replacement occurs, then both events remains independent.

Thus, the that the first marble is blue and the second marble is red will be :-

`P(B)* P(R)=(1)/(3)*(1)/(4)=(1)/(12)`

Hence, the required probability =
`(1)/(12)`