Item 3 Find the probability of randomly selecting a red, then a green marble from a bag of 5 red, 8 green, and 3 blue marbles when (a) you replace the first marble before drawing the second, and (b) you do not replace the first marble. Then, compare the probabilities. Round your answers to four decimal places. a. The probability of selecting a red, then a green marble when you replace the first marble before drawing the second is about . b. The probability of selecting a red, then a green marble when you do not replace the first marble before drawing the second is about . You are about times more likely to pick a red, then a green when you do not replace the first marble.

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

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

Since the two events are independent, the combined probability of selecting a red, then a green marble is:

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

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

And so ,the probability of drawing a green marbles at the second draw is

As before, the two events are independent, so the combined probability of selecting a red, then a green marble is:

do you mean add both of the equations together

because if you do the answer is 10x-5

step-by-step explanation: