a bag contains the following fourteen marbles. deepak randomly chooses two marbles from the bag, one at a…

a bag contains the following fourteen marbles. deepak randomly chooses two marbles from the bag, one at a time, and replaces the marble after each choice. what is the probability he will choose one green marble and then one red marble? express the probabilities in fraction form. p(green) = p(red) = p(green and red) =
Answer
Explanation:
Step1: Count green marbles
There are 6 green marbles out of 14 total marbles. So $P(\text{green})=\frac{6}{14}=\frac{3}{7}$.
Step2: Count red marbles
There are 2 red marbles out of 14 total marbles. So $P(\text{red})=\frac{2}{14}=\frac{1}{7}$.
Step3: Calculate combined probability
Since the events are independent (due to replacement), $P(\text{green and red}) = P(\text{green})\times P(\text{red})=\frac{3}{7}\times\frac{1}{7}=\frac{3}{49}$.
Answer:
$P(\text{green})=\frac{3}{7}$ $P(\text{red})=\frac{1}{7}$ $P(\text{green and red})=\frac{3}{49}$