you have 8 blue marbles, 7 red marbles, and 5 green marbles. What is the probability of obtaining at least 2 red marbles in three draws with replacements?

Question

you have 8 blue marbles, 7 red marbles, and 5 green marbles. What is the probability of obtaining at least 2 red marbles in three draws with replacements?

I have the answer, just can't figure out how to get it.

Thanks!

Answer 1

Seven out of the twenty are red. The probability of NOT getting ANY red in three tries is (13/20)^3 = 0.2746

The probability of getting only one red is 3*(13/20)^2*(7/20) = 0.4436
The probability of getting two or more is 1 - 0.2736 - 0.4436 = 0.2828

Answer 2

I thought I was understanding this but I am really not sure about the symbols.

Does the 3*(13/20)^2*(7/20) translate to 3 times 13/20 times 2 times 7/20?
Sorry I am not up on the symbols...
I will keep trying.

Answer 3

3(13/20)^2(7/20)

means
three times (13/20) squared times (7/20)
in general the probability of getting r successes out of n tries in binomial distribution if probability of success is p is
P(r) = C(n,r) p(r)^r (1-p)^(n-r)

Answer 4

That is even more confusing. I am totally lost on this problem.

I know that I have to multiply the three possibilities, so (7/20)(7/20)(13/20) three times (same numbers, different order) and I get 0.07963. I thought I would add those three numbers or essentially multiply that number by three, however, I did not get the right answer doing it that way.

Thanks for any help. I am totally confused.

Answer 5

To find the probability of obtaining at least 2 red marbles in three draws with replacements, we need to consider both the scenarios where we get exactly 2 red marbles and the scenario where we get all 3 red marbles.

To determine the probability of getting exactly 2 red marbles, we need to consider the total number of ways we can select 2 red marbles out of the 3 draws. Since we are drawing with replacements, the probability of drawing a red marble in any given draw is constant and equal to the ratio of the number of red marbles to the total number of marbles.

So, the probability of drawing exactly 2 red marbles out of 3 draws is calculated as:

P(exactly 2 red marbles) = (number of ways to draw exactly 2 red marbles) * (probability of drawing a red marble) * (probability of drawing a red marble) * (probability of drawing a non-red marble)

The number of ways to draw exactly 2 red marbles out of 3 is calculated using combinations. In this case, we have 7 red marbles, so the number of ways to draw exactly 2 red marbles is given by:

7 choose 2 = 7! / (2! * (7-2)!) = 21

Substituting the values into the formula, we have:

P(exactly 2 red marbles) = 21 * (7/20) * (7/20) * (13/20)

To find the probability of getting all 3 red marbles, we use a similar approach. The number of ways to draw all 3 red marbles out of 3 is simply 1, as there is only one way to select all 3 marbles. Therefore, the probability of drawing all 3 red marbles is:

P(all 3 red marbles) = 1 * (7/20) * (7/20) * (7/20)

Finally, to find the probability of obtaining at least 2 red marbles, we sum up the probabilities of the two scenarios:

P(at least 2 red marbles) = P(exactly 2 red marbles) + P(all 3 red marbles)

Now you can calculate the probabilities based on the formula provided above.