Question

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 $\frac{18}{95}$

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 getting a blue marble on the first draw is $P(\text{blue})=\frac{8}{20}$ since there are 8 blue marbles out of 20 total marbles.

Step3: Calculate probability of second - green marble

Since the marble is replaced, the probability of getting a green marble on the second draw is $P(\text{green})=\frac{9}{20}$ as there are 9 green marbles out of 20 total marbles.

Step4: Calculate the combined probability

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

Answer:

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