Question 1178072
Let's solve this problem step by step.

**Understanding the Problem**

* We have a box with 3 red balls and 4 black balls.
* We draw balls one at a time without replacement.
* We stop when all 4 black balls are drawn.
* We want to find the expected number of red balls left in the box.

**Defining the Random Variable**

Let R be the random variable representing the number of red balls left in the box when all 4 black balls are drawn.

**Possible Scenarios**

To determine the expected value of R, we need to consider the possible scenarios and their probabilities:

* The last black ball could be drawn at the 4th, 5th, 6th, or 7th draw.

**Calculating Probabilities**

1.  **Last black ball at the 4th draw (all black balls drawn first):**
    * This means the first 4 draws are all black.
    * Probability: (4/7) * (3/6) * (2/5) * (1/4) = 1/35
    * Red balls left: 3

2.  **Last black ball at the 5th draw:**
    * This means 3 black balls and 1 red ball are drawn in the first 4 draws, and the 5th draw is a black ball.
    * Probability: [C(4, 3) * C(3, 1) / C(7, 4)] * (1/3) = [4 * 3 / 35] * (1/3) = 4/35
    * Red balls left: 2

3.  **Last black ball at the 6th draw:**
    * This means 3 black balls and 2 red balls are drawn in the first 5 draws, and the 6th draw is a black ball.
    * Probability: [C(4, 3) * C(3, 2) / C(7, 5)]*(1/2) = [4 * 3 / 21] * (1/2) = 2/7 = 10/35
    * Red balls left: 1

4.  **Last black ball at the 7th draw:**
    * This means 3 black balls and 3 red balls are drawn in the first 6 draws, and the 7th draw is a black ball.
    * Probability: [C(4, 3) * C(3, 3) / C(7, 6)] * (1/1) = [4 * 1 / 7] * (1) = 4/7 = 20/35
    * Red balls left: 0

**Calculating Expected Value**

E(R) = (3 * 1/35) + (2 * 4/35) + (1 * 10/35) + (0 * 20/35)
E(R) = (3/35) + (8/35) + (10/35) + 0
E(R) = 21/35
E(R) = 3/5 = 0.6

**Therefore, the expected number of red balls left in the box is 0.6.**