Question

a bag contains 4 green marbles, 6 red marbles, and 5 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 green and the second blue? a $\frac{4}{45}$ b $\frac{1}{14}$ c $\frac{2}{21}$ d $\frac{3}{5}$

Explanation:

Step1: Calculate total marbles

The total number of marbles is $4 + 6+5=15$.

Step2: Calculate probability of first - green

The probability of getting a green marble on the first draw, $P(G_1)$, is $\frac{4}{15}$ since there are 4 green marbles out of 15 total marbles.

Step3: Calculate probability of second - blue

Since the marble is replaced, the total number of marbles remains 15. The probability of getting a blue marble on the second draw, $P(B_2)$, is $\frac{5}{15}=\frac{1}{3}$ as there are 5 blue marbles out of 15 total marbles.

Step4: Calculate joint probability

Since the two draws are independent events, the probability that the first is green and the second is blue is $P(G_1\cap B_2)=P(G_1)\times P(B_2)$. Substitute the values: $\frac{4}{15}\times\frac{1}{3}=\frac{4}{45}$.

Answer:

A. $\frac{4}{45}$