To solve the problem of finding the probability of drawing a black ball in the next draw after three draws from a bag containing 5 black and 3 red balls, we will analyze the situation step by step.
### Step 1: Understand the Initial Setup
The bag contains:
- 5 black balls
- 3 red balls
- Total = 5 + 3 = 8 balls
### Step 2: Calculate the Probability of Drawing a Black Ball
When drawing a ball, we need to consider the possible outcomes after three draws. The probability of drawing a black ball can change depending on how many black and red balls are drawn in the first three draws.
### Step 3: Analyze Different Scenarios
1. **All three draws are black balls:**
- Probability of first black: \( \frac{5}{8} \)
- Probability of second black: \( \frac{4}{7} \)
- Probability of third black: \( \frac{3}{6} \)
- Combined probability:
\[
P(\text{3 black}) = \frac{5}{8} \times \frac{4}{7} \times \frac{3}{6} = \frac{60}{336} = \frac{5}{28}
\]
2. **All three draws are red balls:**
- Probability of first red: \( \frac{3}{8} \)
- Probability of second red: \( \frac{2}{7} \)
- Probability of third red: \( \frac{1}{6} \)
- Combined probability:
\[
P(\text{3 red}) = \frac{3}{8} \times \frac{2}{7} \times \frac{1}{6} = \frac{6}{336} = \frac{1}{56}
\]
3. **Two black and one red:**
- There are 3 ways to arrange this (BBR, BRB, RBB).
- Probability for one arrangement (e.g., BBR):
\[
P(BBR) = \frac{5}{8} \times \frac{4}{7} \times \frac{3}{6} = \frac{60}{336}
\]
- Total for two black and one red:
\[
P(2B, 1R) = 3 \times \frac{60}{336} = \frac{180}{336} = \frac{15}{28}
\]
4. **One black and two red:**
- There are also 3 ways to arrange this (BRR, RBR, RRB).
- Probability for one arrangement (e.g., BRR):
\[
P(BRR) = \frac{5}{8} \times \frac{3}{7} \times \frac{2}{6} = \frac{30}{336}
\]
- Total for one black and two red:
\[
P(1B, 2R) = 3 \times \frac{30}{336} = \frac{90}{336} = \frac{15}{56}
\]
### Step 4: Calculate the Probability of Drawing a Black Ball Next
Now we can find the probability of drawing a black ball next based on the scenarios:
1. **If all three drawn were black:**
- Remaining: 2 black, 3 red
- Probability of black next: \( \frac{2}{5} \)
2. **If all three drawn were red:**
- Remaining: 5 black, 0 red
- Probability of black next: \( 1 \)
3. **If two black and one red:**
- Remaining: 3 black, 2 red
- Probability of black next: \( \frac{3}{5} \)
4. **If one black and two red:**
- Remaining: 4 black, 1 red
- Probability of black next: \( \frac{4}{5} \)
### Step 5: Combine the Probabilities
Using the law of total probability, we combine these results based on their respective probabilities:
\[
P(B) = P(B|3B) \cdot P(3B) + P(B|3R) \cdot P(3R) + P(B|2B,1R) \cdot P(2B,1R) + P(B|1B,2R) \cdot P(1B,2R)
\]
Calculating this gives us:
\[
P(B) = \left(\frac{2}{5} \cdot \frac{5}{28}\right) + \left(1 \cdot \frac{1}{56}\right) + \left(\frac{3}{5} \cdot \frac{15}{28}\right) + \left(\frac{4}{5} \cdot \frac{15}{56}\right)
\]
After calculating each term, we find:
\[
P(B) = \frac{2}{28} + \frac{1}{56} + \frac{9}{28} + \frac{12}{56}
\]
Finding a common denominator and simplifying gives us the final probability.
### Final Answer
After performing the calculations, the final probability of drawing a black ball in the next draw is \( \frac{5}{8} \) or \( 0.625 \).
