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

Total marbles = \(9 + 3+8=20\)

Step2: Calculate probability of first - marble being blue

Probability of blue \(P(B)=\frac{8}{20}\)

Step3: Calculate probability of second - marble being green

Since the marble is replaced, total marbles remain 20. Probability of green \(P(G)=\frac{9}{20}\)

Step4: Calculate the combined probability

Since the two events (first - draw and second - draw) are independent, we use the formula \(P(A\cap B)=P(A)\times P(B)\). So \(P = \frac{8}{20}\times\frac{9}{20}=\frac{72}{400}=\frac{9}{50}\)

Answer:

A. \(\frac{9}{50}\)