A bag contains two red marbles three green marbles and four blue marbles if we choose a marble then another marble without putting the first one back in the bag what is the probability that the first marble will be green in the second will be red

Guest Feb 6, 2020

#1**+2 **

**A bag contains two red marbles three green marbles and four blue marbles if we choose a marble then another marble without putting the first one back in the bag what is the probability that the first marble will be green in the second will be red**

\(\text{There are ${\color{red}2} + {\color{green}3} + {\color{blue}4} = 9\ $marbles } \\ \text{first one drawn out is ${\color{green}\text{green}}$ the probability is: $\dfrac{\text{all } {\color{green}\text{green}} \text{ marbles}}{\text{all marbles}} = \dfrac{{\color{green}3}}{9}$ } \\ \text{There are now ${\color{red}2} + {\color{green}2} + {\color{blue}4} = 8\ $marbles (no replacement) } \\ \text{second one drawn out is ${\color{red}\text{red}}$ the probability is: $\dfrac{\text{all } {\color{red}\text{red}} \text{ marbles}}{\text{all marbles now}} = \dfrac{{\color{red}2}}{8}$ } \)

The probability that the first marble will be green in the second will be red is \(\dfrac{{\color{green}3}}{9}\times \dfrac{{\color{red}2}}{8} =\dfrac{1}{12} \)

heureka Feb 6, 2020