a bag contains 9 green marbles, 3 red marbles, and 8 blue marbles. one marble is taken from the bag and put…

a bag contains 9 green marbles, 3 red marbles, and 8 blue marbles. one marble is taken from the bag and put back after checking its color. a second marble is then taken out. what is the probability that the first is blue and the second green? a $\frac{9}{50}$ b $\frac{17}{20}$ c $\frac{1}{19}$ d $\frac{18}{95}$

a bag contains 9 green marbles, 3 red marbles, and 8 blue marbles. one marble is taken from the bag and put back after checking its color. a second marble is then taken out. what is the probability that the first is blue and the second green? a $\frac{9}{50}$ b $\frac{17}{20}$ c $\frac{1}{19}$ d $\frac{18}{95}$

Answer

Explanation:

Step1: Calculate total number of marbles

The total number of marbles is $9 + 3+8=20$.

Step2: Calculate probability of first - blue marble

The probability of picking a blue marble first, $P(\text{blue}_1)$, is $\frac{8}{20}$ since there are 8 blue marbles out of 20 total marbles.

Step3: Calculate probability of second - green marble

Since the first marble is put back, the total number of marbles remains 20. The probability of picking a green marble second, $P(\text{green}_2)$, is $\frac{9}{20}$ as there are 9 green marbles out of 20 total marbles.

Step4: Calculate combined probability

Since the two events are independent (because the first marble is replaced), the probability that the first is blue and the second is green is $P = P(\text{blue}_1)\times P(\text{green}_2)=\frac{8}{20}\times\frac{9}{20}=\frac{72}{400}=\frac{9}{50}$.

Answer:

A. $\frac{9}{50}$