Question

A bag contains 1 red, 1 yellow, 1 blue, and 1 green marble. What is the probability of choosing a green marble, not replacing it, and
then choosing a red marble?
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Answer 1

The probability for choosing green and red is $\frac{1}{12}$

Solution:

Given that , A bag contains 1 red, 1 yellow, 1 blue, and 1 green marble.

We have to find what is the probability of choosing a green marble, not replacing it, and then choosing a red marble?

Now, we know that,

$\text { probability of an event }=\frac{\text { number of favourable outcomes }}{\text { total number of outcomes }}$

So, total possible outcomes = 1 red + 1 yellow + 1 blue + 1 green = 4

$\text { Now, probability for green marble }=\frac{1 \text { green marble }}{4 \text { marbles }}=\frac{1}{4}$

And now, total outcomes will be only 3 as we are not replacing the picked marble.

$\begin{array}{l}{\text { Then, probability for red marble }=\frac{1 \text { red marble }}{3 \text { marbles }}=\frac{1}{3}} \\\\ {\text { Then overall probability }=\frac{1}{4} \times \frac{1}{3}=\frac{1}{12}}\end{array}$

Hence, the probability for choosing green and red is $\frac{1}{12}$

Answer 2

Probability of choosing a green marble, not replacing it, and  then choosing a red marble is $\frac{1}{12}$

Step-by-step explanation:

We have 1 red, 1 yellow, 1 blue, and 1 green marble in bag

Total Marbles = 4

Probability of choosing green marble= $\frac{1}{4}$

Since green marble is not replaced i.e it is not put back into the bag, so

Total Marbles = 4-1 = 3

Probability of choosing red marble= $\frac{1}{3}$

Now, calculating the probability of choosing a green marble, not replacing it, and  then choosing a red marble

= Probability of choosing green marble * Probability of choosing red marble

= $\frac{1}{4}*\frac{1}{3}$

=$\frac{1}{12}$

So, probability of choosing a green marble, not replacing it, and  then choosing a red marble is $\frac{1}{12}$

Keywords: probability

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• brainly.com/question/4021035
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• brainly.com/question/5751004
• brainly.com/question/6649771

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