A jar contains four marbles, each a different color: red, blue, green and yellow. If you draw two marbles from the jar, one after another, replacing the first before drawing the second, what is the probability of getting

a. two red marbles
b. a red marble on the first draw and a green marble on the second draw
c. at least one red marble and one green marble.
d. no yellow marbles

Would someone be kind enough to help me sort this out. I am stuck on three problems for my math assignment this week.

First you have to know how many marbles of each color are in the jar.

you got one marble of each colour in the jar,

so the prob of any specific colour is 1/4

a) 1/4 x 14 = 1/16
b)1/4 x 1/4 = 1/16
c) poor wording.
you are only drawing 2 marbles, so it has to be R,G or GR
Since you are replacing it would be
1/4 x 1/4 x 2
= 1/8
d) 3/4 x 3/4 = 9/16

thank you for the help

Of course, I'd be happy to help you with your math assignment!

To solve these probability problems, we need to first determine the total number of possible outcomes and then count the number of favorable outcomes. The probability is then calculated by dividing the number of favorable outcomes by the total number of possible outcomes.

Let's break down each question one by one:

a. Probability of getting two red marbles.
In this case, we need to calculate the probability of getting a red marble on the first draw and then getting another red marble on the second draw with replacement.

Since there are four marbles in total, each with an equal chance of being drawn, the probability of selecting a red marble on each draw is 1/4. Since the draws are independent and replacement is done, the probability of getting two red marbles is calculated by multiplying the individual probabilities:
P(Two red marbles) = P(Red on 1st draw) * P(Red on 2nd draw) = (1/4) * (1/4) = 1/16.

b. Probability of getting a red marble on the first draw and a green marble on the second draw.
Similar to the previous question, the probability of drawing a red marble on the first draw is 1/4. However, this time we want to draw a green marble on the second draw, so the probability of that happening is 1/4 as well. Again, we multiply these probabilities together since the draws are independent:
P(Red then Green) = P(Red on 1st draw) * P(Green on 2nd draw) = (1/4) * (1/4) = 1/16.

c. Probability of getting at least one red marble and one green marble.
To find the probability of this event, we can subtract the probability of not getting both a red and a green marble from 1.

The probability of not getting both a red and a green marble can be calculated by finding the probability of getting neither a red nor a green marble. The only scenario where this happens is when we draw two yellow marbles. The probability of drawing a yellow marble on each draw is 1/4, so:
P(Not red and Not green) = P(Yellow on 1st draw) * P(Yellow on 2nd draw) = (1/4) * (1/4) = 1/16.

Therefore, the probability of getting at least one red marble and one green marble is:
P(At least one red and one green) = 1 - P(Not red and Not green) = 1 - 1/16 = 15/16.

d. Probability of getting no yellow marbles.
Since there is only one yellow marble in the jar, the probability of not drawing it on each draw is 3/4 (since there are 3 non-yellow marbles out of a total of 4). Since the draws are independent, we can multiply these probabilities to find the overall probability of not getting a yellow marble on either draw:
P(No yellow marbles) = P(Not yellow on 1st draw) * P(Not yellow on 2nd draw) = (3/4) * (3/4) = 9/16.

I hope this explanation helps! Let me know if you have any further questions.