a bag contains 4 green marbles, 3 red marbles, and 7 blue marbles. one marble is taken from the bag and put…

a bag contains 4 green marbles, 3 red marbles, and 7 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 red?\na $\frac{1}{13}$\nb $\frac{3}{26}$\nc $\frac{5}{7}$\nd $\frac{3}{28}$

a bag contains 4 green marbles, 3 red marbles, and 7 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 red?\na $\frac{1}{13}$\nb $\frac{3}{26}$\nc $\frac{5}{7}$\nd $\frac{3}{28}$

Answer

Explanation:

Step1: Calculate probability of first - blue

The total number of marbles is $4 + 3+7=14$. The number of blue marbles is 7. The probability of getting a blue marble on the first draw, $P(B_1)$, is $\frac{7}{14}=\frac{1}{2}$ since the marble is replaced.

Step2: Calculate probability of second - red

The number of red marbles is 3. The probability of getting a red marble on the second draw, $P(R_2)$, is $\frac{3}{14}$ as the marble was replaced and total number of marbles is still 14.

Step3: Calculate joint - probability

Since the two events are independent (because of replacement), the probability that the first is blue and the second is red is $P(B_1)\times P(R_2)$. So $P=\frac{1}{2}\times\frac{3}{14}=\frac{3}{28}$.

Answer:

D. $\frac{3}{28}$