To solve this problem, we'll apply Bayes' Theorem. The goal is to find the probability that the card drawn is blue, given that it's a 3.

Step 1: Define Events

• Let $B$ be the event that the card is blue.
• Let $R$ be the event that the card is red.
• Let $E$ be the event that the card is a 3.

Step 2: Write Down Given Information

• The total number of cards is 32 (since 8 blue cards are removed from the total 40).
• There are 12 blue cards and 20 red cards remaining.
• Each pack has exactly one card labeled "3".

Step 3: Bayes' Theorem Formula

Bayes' Theorem is given by: $P(B \mid E) = \frac{P(E \mid B) \cdot P(B)}{P(E)}$

Step 4: Calculate the Probabilities

• $P(E \mid B)$: The probability of drawing a 3 given that the card is blue. Since there is exactly one 3 in the blue cards, and 12 blue cards total: $P(E \mid B) = \frac{1}{12}$
• $P(E \mid R)$: The probability of drawing a 3 given that the card is red. Since there is exactly one 3 in the red cards, and 20 red cards total: $P(E \mid R) = \frac{1}{20}$
• $P(B)$: The probability of drawing a blue card, which is: $P(B) = \frac{12}{32} = \frac{3}{8}$
• $P(R)$: The probability of drawing a red card, which is: $P(R) = \frac{20}{32} = \frac{5}{8}$

Step 5: Total Probability of Drawing a 3, $P(E)$

The total probability of drawing a 3, $P(E)$, is calculated by considering both possibilities (drawing a blue 3 and a red 3): $P(E) = P(E \mid B) \cdot P(B) + P(E \mid R) \cdot P(R)$ $P(E) = \left(\frac{1}{12}\right) \cdot \left(\frac{3}{8}\right) + \left(\frac{1}{20}\right) \cdot \left(\frac{5}{8}\right)$ $P(E) = \frac{1 \times 3}{12 \times 8} + \frac{1 \times 5}{20 \times 8}$ $P(E) = \frac{3}{96} + \frac{5}{160} = \frac{1}{32} + \frac{1}{32} = \frac{1}{16}$

Step 6: Apply Bayes' Theorem

Finally, calculate $P(B \mid E)$: $P(B \mid E) = \frac{P(E \mid B) \cdot P(B)}{P(E)}$ $P(B \mid E) = \frac{\frac{1}{12} \cdot \frac{3}{8}}{\frac{1}{16}} = \frac{\frac{3}{96}}{\frac{1}{16}} = \frac{3}{6} = \frac{1}{2}$

Final Answer

The probability that the card drawn is blue, given that it’s a 3, is $\frac{1}{2}$.

Would you like more details on this solution or have any other questions?

Here are 5 related questions:

1. How does Bayes' Theorem apply to real-world decision-making scenarios?
2. What is the significance of conditional probability in statistics?
3. How would the probability change if more or fewer blue cards were removed?
4. Can we use Bayes' Theorem for continuous probability distributions?
5. What if we drew two cards, and one was a 3? How would the probability calculation change?

Tip: When dealing with problems involving conditional probability, always clarify what you know and what you need to find out. This helps in correctly identifying the events and applying the right formulas.