Question 1178072: A box has 3 red and 4 black balls. You randomly take one ball at a time without putting it back in. Stop as soon as all the black balls are drawn out of the box. How many red balls do you expect to be left in the box? (Hint: find the expected value of the properly defined random variable)
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! 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.**
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