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? a 1/13 b 3/26 c 5/7 d 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? a 1/13 b 3/26 c 5/7 d 3/28

Answer

Explanation:

Step1: Calculate total marbles

Let's assume there are 4 green, 3 red and 7 blue marbles. Total marbles = 4 + 3+7 = 14.

Step2: Calculate probability of first - blue

The probability of picking a blue marble first, $P(\text{blue}_1)=\frac{7}{14}=\frac{1}{2}$.

Step3: Calculate probability of second - red

Since the first marble is put back, the total number of marbles remains 14. The probability of picking a red marble second, $P(\text{red}_2)=\frac{3}{14}$.

Step4: Calculate combined probability

Since the two events are independent, the probability that the first is blue and the second is red is $P = P(\text{blue}_1)\times P(\text{red}_2)=\frac{1}{2}\times\frac{3}{14}=\frac{3}{28}$.

Answer:

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