Question

A certain type of fish has a 4/5 probability of surviving in a pond, while another fish has a 2/5 probability of survival. What is the probability that exactly one of them survives?

• A. 2/5
• B. 9/16
• C. 14/25
• D. 16/25

Hint:

For "exactly one of two independent events" problems, break it down into two mutually exclusive scenarios: (Event A happens AND Event B does NOT happen) OR (Event A does NOT happen AND Event B happens). Then sum the probabilities of these two scenarios.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
The problem asks for the probability that exactly one of two independent events (fish survival) occurs.

Step 2: Key Formula or Approach:
Let $P(A)$ be the probability that the first fish survives and $P(B)$ be the probability that the second fish survives.
The probability that exactly one of them survives is:
$P(\text{exactly one survives}) = P(A \text{ survives and } B \text{ dies}) + P(A \text{ dies and } B \text{ survives})$.
Since the events are independent:
$P(\text{exactly one survives}) = P(A) \times P(B') + P(A') \times P(B)$.
Where $P(A')$ is the probability that A dies, and $P(B')$ is the probability that B dies.
$P(X') = 1 - P(X)$.

Step 3: Detailed Explanation:
Given:
- Probability of first fish surviving, $P(A) = 4/5$.
- Probability of second fish surviving, $P(B) = 2/5$.
First, calculate the probabilities of them dying:
- Probability of first fish dying, $P(A') = 1 - P(A) = 1 - 4/5 = 1/5$.
- Probability of second fish dying, $P(B') = 1 - P(B) = 1 - 2/5 = 3/5$.
Now, calculate the probability that exactly one of them survives:
\[ P(\text{exactly one survives}) = P(A)P(B') + P(A')P(B) \]
\[ = \left(\frac{4}{5}\right) \times \left(\frac{3}{5}\right) + \left(\frac{1}{5}\right) \times \left(\frac{2}{5}\right) \]
\[ = \frac{12}{25} + \frac{2}{25} \]
\[ = \frac{12 + 2}{25} = \frac{14}{25} \]

Step 4: Final Answer:
The probability that exactly one of them survives is 14/25.