Answer: [tex]\dfrac{1}{12}[/tex]
Step-by-step 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 :
[tex]P(B)=\dfrac{\text{Number of blue marbles }}{\text{Total marbles }}\\\\=\dfrac{8}{24}=\dfrac{1}{3}[/tex]
Since, he replaces the marbles so the total marbles remains same in bag.
The probability of that second marble is red :-
[tex]P(R)=\dfrac{6}{24}=\dfrac{1}{4}[/tex]
Since replacement occurs, then both events remains independent.
Thus, the that the first marble is blue and the second marble is red will be :-
[tex]P(B)\times P(R)=\dfrac{1}{3}\times\dfrac{1}{4}=\dfrac{1}{12}[/tex]
Hence, the required probability = [tex]\dfrac{1}{12}[/tex]
