Question

A pond contains 200 fishes of which 40 are marked. A second pond contains 300 fishes of which 50 are marked. One fish is drawn from each of the ponds. What is the probability that the fishes drawn are both marked?

• A. 1/90
• B. 1/60
• C. 1/30
• D. 1/11

Hint:

For independent probability questions, simplify each individual fraction first (e.g., \( 40/200 = 1/5 \) and \( 50/300 = 1/6 \)) before multiplying. This keeps the calculations simple and prevents arithmetic errors.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
This problem involves calculating the joint probability of two independent events.
Because the fish are drawn from two separate ponds, the outcome of the draw from the first pond does not affect the outcome of the draw from the second pond.
Key Formula or Approach:
For two independent events \( A \) and \( B \), the probability of both events occurring is the product of their individual probabilities:
\[ P(A \text{ and } B) = P(A) \times P(B) \]

Step 2: Detailed Explanation:
Let us calculate the individual probabilities:
1. Let Event \( A \) be drawing a marked fish from the first pond.
The first pond contains 200 fish, of which 40 are marked:
\[ P(A) = \frac{\text{Number of marked fish}}{\text{Total number of fish}} = \frac{40}{200} = \frac{1}{5} \]
2. Let Event \( B \) be drawing a marked fish from the second pond.
The second pond contains 300 fish, of which 50 are marked:
\[ P(B) = \frac{\text{Number of marked fish}}{\text{Total number of fish}} = \frac{50}{300} = \frac{1}{6} \]
3. Since the events are independent, multiply the two probabilities to find the joint probability:
\[ P(A \text{ and } B) = P(A) \times P(B) \]
\[ P(A \text{ and } B) = \frac{1}{5} \times \frac{1}{6} = \frac{1}{30} \]
Therefore, the probability of drawing a marked fish from both ponds is \( \frac{1}{30} \).

Step 3: Final Answer:
The probability that both drawn fish are marked is \( \frac{1}{30} \).
Thus, the correct choice is (C).