Let's break down the problem step by step:

Given:

• The box contains:
  2 red balls
  2 white balls
  1 yellow ball

Thus, there are a total of 5 balls in the box.

We are drawing 4 balls, and we want to calculate the probability that the remaining ball is red.

Total possible outcomes:

The total number of ways to draw 4 balls out of 5 is given by the combination formula $\binom{n}{k}$, where $n$ is the total number of balls and $k$ is the number of balls drawn:

$\text{Total number of outcomes} = \binom{5}{4} = 5$

Favorable outcomes:

To ensure that the remaining ball is red, one of the two red balls must be left behind, meaning only one red ball is drawn among the 4 drawn balls. The other 3 drawn balls will come from the 2 white and 1 yellow balls.

So, we must choose 1 red ball from the 2 red balls, and 3 balls from the remaining 3 (2 white + 1 yellow). The number of favorable outcomes is:

$\text{Favorable outcomes} = \binom{2}{1} \times \binom{3}{3} = 2 \times 1 = 2$

Probability:

The probability is the ratio of favorable outcomes to total outcomes:

$\text{Probability} = \frac{\text{Favorable outcomes}}{\text{Total outcomes}} = \frac{2}{5}$

Conclusion:

The probability that the remaining ball is red is $\frac{2}{5}$.

Would you like more details on any of these steps, or do you have further questions?

Here are 5 related questions to consider:

1. How would the probability change if the box had more balls of each color?
2. What if you were asked for the probability of drawing exactly 1 red ball?
3. How can this problem be modeled using conditional probability?
4. How do combinations work in probability problems like this one?
5. What if 3 balls were drawn instead of 4? How would the probability change?

Tip: In probability, breaking the problem into total outcomes and favorable outcomes helps simplify complex questions.