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 $\frac{1}{13}$ b $\frac{3}{26}$ c $\frac{5}{7}$ d $\frac{3}{28}$

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? a $\frac{1}{13}$ b $\frac{3}{26}$ c $\frac{5}{7}$ d $\frac{3}{28}$

Answer

Answer:

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

Explanation:

Step1: Calculate total number of marbles

$4 + 3+7=14$

Step2: Calculate probability of first - blue marble

The probability of getting a blue marble first, $P(\text{blue})=\frac{7}{14}=\frac{1}{2}$ since there are 7 blue marbles out of 14 total marbles.

Step3: Calculate probability of second - red marble

Since the marble is replaced, the total number of marbles remains 14. The probability of getting a red marble second, $P(\text{red})=\frac{3}{14}$ as there are 3 red marbles out of 14 total marbles.

Step4: Calculate combined probability

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