Question

A bowl has 4 green marbles 3 red marbles and 2yellow marbles what is the probability that you are going to select a red marble and a yellow marble. You replace the marble before another marble is selected

Answer 1

ANSWER

2/27

EXPLANATION

There are a total of 9 marbles in the bowl. The probability of drawing a red marble is,

$P(red)=\frac{\#red.marbles}{\#total.marbles}=\frac{3}{9}=\frac{1}{3}$

Then you draw another marble, but you put the first back in the bowl, so the total number of marbles is the same. The probability of drawing a yellow marble is,

$P(yellow)=\frac{\#yellow.marbles}{\#total.marbles}=\frac{2}{9}$

The probability of drawing a red marble and a yellow marble is,

$P(red.and.yellow)=P(red)\cdot P(yellow)=\frac{1}{3}\cdot\frac{2}{9}=\frac{2}{27}$