Question

A bag contains 5 red and 7 blue balls. Find the probability of drawing 2 red balls without replacement.

• A. \( \frac{5}{33} \)
• B. \( \frac{10}{33} \)
• C. \( \frac{25}{144} \)
• D. \( \frac{7}{33} \)

Hint:

For probabilities without replacement: - The total number of items decreases after each draw. - Use conditional probability \(P(A \cap B) = P(A)P(B|A)\).

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
We need to find the probability of a compound event where two red balls are drawn sequentially without putting the first one back.
Step 2: Key Formula or Approach:
We use the multiplication rule for dependent events:
\[ P(A \cap B) = P(A) \times P(B|A) \]
Alternatively, we can use combinations: \( P = \frac{\binom{\text{Red}}{2}}{\binom{\text{Total}}{2}} \).
Step 3: Detailed Explanation:
Total balls in the bag = \( 5 \text{ (Red)} + 7 \text{ (Blue)} = 12 \).
Event 1: Drawing the first red ball.
\[ P(\text{Red}_1) = \frac{\text{Number of red balls}}{\text{Total balls}} = \frac{5}{12} \]
Event 2: Drawing the second red ball (given the first was red).
Remaining red balls = \( 4 \).
Remaining total balls = \( 11 \).
\[ P(\text{Red}_2 | \text{Red}_1) = \frac{4}{11} \]
Total Probability:
\[ P = \frac{5}{12} \times \frac{4}{11} = \frac{20}{132} \]
Simplifying the fraction by dividing by 4:
\[ P = \frac{5}{33} \]
Step 4: Final Answer:
The probability is \( \frac{5}{33} \).