To solve this probability question, let's break it down step by step:

Given:

• There are 3 red marbles and 4 blue marbles.
• A marble is taken at random and replaced, then another marble is taken.
• We need to find the probability that both marbles taken are of the same color.

Total number of marbles:

There are $3$ red and $4$ blue marbles, so the total number of marbles is: $3 + 4 = 7$

Probability of drawing two marbles of the same color:

1. Red-Red (Two red marbles):

   • The probability of drawing a red marble in the first draw is: $\frac{3}{7}$
   • Since the marble is replaced, the probability of drawing a red marble again is: $\frac{3}{7}$
   • Therefore, the probability of both being red is: $\frac{3}{7} \times \frac{3}{7} = \frac{9}{49}$

2. Blue-Blue (Two blue marbles):

   • The probability of drawing a blue marble in the first draw is: $\frac{4}{7}$
   • The probability of drawing a blue marble again (with replacement) is: $\frac{4}{7}$
   • Therefore, the probability of both being blue is: $\frac{4}{7} \times \frac{4}{7} = \frac{16}{49}$

Total probability of drawing two marbles of the same color:

We add the probabilities of both red-red and blue-blue: $\frac{9}{49} + \frac{16}{49} = \frac{25}{49}$

Thus, the probability that the two marbles taken are the same color is $\frac{25}{49}$.

Would you like any further details or clarifications?

Related Questions:

1. What is the probability of drawing two marbles of different colors?
2. How would the solution change if the marbles were not replaced?
3. What is the probability of drawing a red marble followed by a blue one?
4. If the bag had 5 red and 5 blue marbles, what would the probability be for drawing two of the same color?
5. How does the probability change if one more red marble is added to the bag?

Tip:

Always check if the marbles are being replaced or not. With replacement, the probability stays the same for each draw; without replacement, it changes.