We have two boxes. The first box contains three green balls and eight red balls. The second contains four green balls and three red balls. Bob selects one of the boxes at random. Then he selects a ball from that box at random. If he has a red ball, what is the probability that he selected a ball from the first box?
Solution
This is a conditional probability problem. The goal is to determine the probability that the ball was selected from the first box, given that a red ball was chosen.
Problem Setup:
Box 1: Contains 3 green balls and 8 red balls (total = 11 balls).
Box 2: Contains 4 green balls and 3 red balls (total = 7 balls).
Process: Bob selects a box at random (probability for each box = 21), and then he randomly selects a ball from the chosen box.
We are tasked with finding P(A1∣B), where:
A1: The event of selecting a ball from Box 1.
B: The event of selecting a red ball.
Using Bayes' Theorem:
P(A1∣B)=P(B)P(B∣A1)P(A1)
Step 1: Compute the components of Bayes' Theorem
Probability of selecting a red ball from Box 1, P(B∣A1):
P(B∣A1)=118(since there are 8 red balls out of 11 in Box 1).
Probability of selecting a red ball from Box 2, P(B∣A2):
P(B∣A2)=73(since there are 3 red balls out of 7 in Box 2).
Total probability of selecting a red ball, P(B):
P(B)=P(B∣A1)P(A1)+P(B∣A2)P(A2)
Substituting values:
= \frac{4}{11} + \frac{3}{14}$$
Finding a common denominator ($$154$$):
$$P(B) = \frac{56}{154} + \frac{33}{154} = \frac{89}{154}.$$
4. **Prior probability of selecting Box 1, $$P(A_1)$$**:
$$P(A_1) = \frac{1}{2}.$$
#### Step 2: Apply Bayes' Theorem
$$P(A_1 | B) = \frac{P(B | A_1)P(A_1)}{P(B)}$$
Substitute the values:
$$P(A_1 | B) = \frac{\left(\frac{8}{11}\right)\left(\frac{1}{2}\right)}{\frac{89}{154}}
= \frac{\frac{4}{11}}{\frac{89}{154}}
= \frac{4}{11} \cdot \frac{154}{89}
= \frac{616}{979}.$$
### Final Answer:
The probability that the red ball was selected from the first box is:
$$P(A_1 | B) = \frac{616}{979} \approx 0.629.$$
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Would you like further clarification or more examples? Here are 5 related questions to deepen your understanding:
1. How does Bayes' Theorem apply to real-world problems like medical testing or spam detection?
2. What changes in the solution if Bob chooses a box with unequal probabilities?
3. How would the probability change if the number of red or green balls in one box was altered?
4. Can you derive $$P(A_2 | B)$$, the probability that the red ball came from Box 2?
5. How can you visualize this problem using a tree diagram?
**Tip:** Always double-check your fractions and common denominators in probability problems to avoid minor calculation errors.
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Math Problem Analysis
Mathematical Concepts
Probability Conditional Probability Bayes' Theorem