This problem involves calculating a conditional probability. We are given a scenario where balls are drawn from a box without replacement, and we need to find the probability of an event occurring given that another event has already happened.

Initial Ball Count in the Box

Initially, the box contains:

• 4 white balls (W)
• 3 red balls (R)

The total number of balls in the box at the start is $4 + 3 = 7$.

Understanding the Conditional Event

We are given specific information: "the first removed ball is white". This is the condition that we know has occurred. This changes the state of the box for the second draw.

Updated Ball Count After First Draw

Since the first ball removed was white, we now have one less white ball and one less total ball in the box.

After the first white ball is removed, the box contains:

• $4 - 1 = 3$ white balls
• 3 red balls

The new total number of balls in the box is $3 + 3 = 6$.

Calculating the Probability of the Second Ball Being Red

Now, we need to find the probability that the second ball removed is red, given the updated state of the box. This is a direct probability calculation from the remaining balls.

The probability is calculated as the ratio of the number of favorable outcomes (drawing a red ball) to the total number of possible outcomes (total remaining balls).

Probability (Second ball is Red | First ball is White) = $\frac{\text{Number of red balls remaining}}{\text{Total number of balls remaining}}$

Substituting the numbers:

Probability = $\frac{3}{6}$

Simplifying the fraction:

Probability = $\frac{1}{2}$

So, the probability that the second removed ball is red, given the first removed ball was white, is $\frac{1}{2}$.