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, andthen choosing a red marble?t

The probability for choosing green and red is [tex]\frac{1}{12}[/tex]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, [tex]\text { probability of an event }=\frac{\text { number of favourable outcomes }}{\text { total number of outcomes }}[/tex]So, total possible outcomes = 1 red + 1 yellow + 1 blue + 1 green = 4[tex]\text { Now, probability for green marble }=\frac{1 \text { green marble }}{4 \text { marbles }}=\frac{1}{4}[/tex]And now, total outcomes will be only 3 as we are not replacing the picked marble.[tex]\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}[/tex]Hence, the probability for choosing green and red is [tex]\frac{1}{12}[/tex]