a) 0.1563
b) 0.1667
Step-by-step explanation:
a)
Inside the bag of the problem, we have:
r = 5 (number of red marbles)
g = 8 (number of green marbles)
b = 3 (number of blue marbles)
In this part, we replace the first marble before drawing the second.
Here we want to find the probability of selecting a red, then a green marble.
The total number of marbles is
n = r + g + b = 5 + 8 + 3 = 16
So, the probability of drawing a red marble at the 1st attempt is
[tex]p(r)=\frac{r}{n}=\frac{5}{16}[/tex]
Then, the first marble is replaced back into the bag before making the second draw.
The probability of drawing a green marbles at the second draw is therefore
[tex]p(g)=\frac{g}{n}=\frac{8}{16}=\frac{1}{2}[/tex]
Since the two events are independent, the combined probability of selecting a red, then a green marble is:
[tex]p(rg)=p(r)\cdot p(g)=\frac{5}{16}\cdot \frac{1}{2}=\frac{5}{32}=0.1563[/tex]
b)
In this second experiment, we do not replace the first marble before drawing the second one.
As in part a), the probability of drawing a red marble at the 1st draw is
[tex]p(r)=\frac{r}{n}=\frac{5}{16}[/tex]
Now, the situation changes. In fact, we do not put back the red marble into the bag. Therefore, at the second draw, the number of balls inside the bag is
[tex]n'=n-1=16-1=15[/tex]
And so ,the probability of drawing a green marbles at the second draw is
[tex]p(g)=\frac{g}{n'}=\frac{8}{15}[/tex]
As before, the two events are independent, so the combined probability of selecting a red, then a green marble is:
[tex]p(rg)=p(r)\cdot p(g)=\frac{5}{16}\cdot \frac{8}{15}=\frac{5}{30}=\frac{1}{6}=0.1667[/tex]
