Question

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 } \\)

Explanation:

Step1: Calculate the total number of marbles

The total number of marbles is \(4 + 3+7=14\).

Step2: Calculate the probability of drawing a blue marble first

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

Step3: Calculate the probability of drawing a red marble second

Since the first marble is put back, the total number of marbles remains 14. The probability of drawing a red marble second, \(P(\text{red})\), is \(\frac{3}{14}\) (since there are 3 red marbles out of 14 total).

Step4: Calculate the combined probability

Since the two events (drawing blue first and red second) are independent (because the first marble is replaced), we use the formula \(P(A\cap B)=P(A)\times P(B)\). So \(P(\text{blue then red})=\frac{1}{2}\times\frac{3}{14}=\frac{3}{28}\)

Answer:

D. \(\frac{3}{28}\)