Question

Roger has a bag of marbles. There are 6 red, 4 blue, 3 white, and 7 green marbles in the bag. If he draws one marble, replaces it, and then draws another, find the following probabilities.
P(red, red) *
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This is a required question
P(not white, green) *
0 points
P(blue or white, red) *
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Find the answer for P(red, red) if Roger does NOT replace the marble before selecting the second one. *
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Answer 1

Using the probability concept, we have that the probabilities are:

• P(red, red): 9%.

• P(not white, green): 29.75%.

• P(blue or white, red): 10.5%.

• P(red, red), without replacement: 7.89%.

What is a probability?

• A probability is given by the number of desired outcomes divided by the number of total outcomes.

P(red, red):

• In each trial, there are 20 marbles.

• In each trial, 6 are red.

Hence:

$p = \frac{6}{20} \times \frac{6}{20} = \frac{36}{400} = 0.09 = 9\%$

P(not white, green):

• There are 17 marbles that are not white.

• There are 7 marbles that are green.

Hence:

$p = \frac{17}{20} \times \frac{7}{20} = 0.2975 = 29.75\%$

P(blue or white, red):

• There are 7 marbles that are blue or white.

• There are 6 marbles that are red.

Hence:

$p = \frac{7}{20} \times \frac{6}{20} = 0.105 = 10.5\%$

P(red, red), without replacement:

• Initially, there are 6 red marbles out of 20.

• Then, without replacement, for the second marble, there will be 5 red out of 19.

Then:

$p = \frac{6}{20} \times \frac{5}{19} = 0.0789 = 7.89\%$

You can learn more about the probability concept at https://brainly.com/question/15536019