To calculate the probability that both balls drawn are red, we follow these steps:

Step 1: Total number of balls

The bag contains:

• 3 red balls
• 4 white balls

Thus, the total number of balls = $3 + 4 = 7$.

Step 2: Probability of drawing two red balls without replacement

The probability that both balls are red can be calculated using conditional probabilities since the draws are without replacement.

1. Probability of drawing a red ball on the first draw: $P(\text{First red}) = \frac{\text{Number of red balls}}{\text{Total balls}} = \frac{3}{7}.$

2. Probability of drawing a red ball on the second draw (given the first was red): After drawing one red ball, there are:

   • 2 red balls left
   • 6 total balls left

   Therefore: $P(\text{Second red | First red}) = \frac{2}{6} = \frac{1}{3}.$

3. Combined probability of both draws being red: Since both events are sequential and depend on each other, multiply the probabilities: $P(\text{Both red}) = P(\text{First red}) \times P(\text{Second red | First red}).$ Substituting the values: $P(\text{Both red}) = \frac{3}{7} \times \frac{1}{3} = \frac{3}{21} = \frac{1}{7}.$

Final Answer:

The probability that both balls drawn are red is: $\boxed{\frac{1}{7}}$

Do you need further clarification or a breakdown of any step?

Here are 5 related questions to expand your understanding:

1. What is the probability that one ball is red and the other is white?
2. How does the probability change if the draws are with replacement?
3. What is the probability that both balls are white?
4. How many ways can you draw 2 balls from the bag, regardless of color?
5. How does increasing the number of red balls affect the probability of drawing two reds?

Tip: For probability problems, always check if events are dependent (without replacement) or independent (with replacement). This affects how probabilities are calculated.