A fisherman has three favorite fishing spots. He visits each spot with equal probability. The probability of catching a fish at the first spot is $0.4$, at the second spot is $0.45$, and at the third spot is $0.5$. One day, the fisherman went fishing to one of his favorite spots. Find the probability that he will catch a fish.

1. Problem Description

A fisherman has three favorite fishing spots. He visits each spot with equal probability. The probability of catching a fish at the first spot is $0.4$, at the second spot is $0.45$, and at the third spot is $0.5$. One day, the fisherman went fishing to one of his favorite spots. Find the probability that he will catch a fish.

2. Solution Steps

Let $A_1$ be the event that the fisherman chooses the first fishing spot, $A_2$ be the event that he chooses the second fishing spot, and $A_3$ be the event that he chooses the third fishing spot. Let $B$ be the event that the fisherman catches a fish. We are given the following probabilities:

$P(A_1) = P(A_2) = P(A_3) = \frac{1}{3}$

$P(B|A_1) = 0.4$

$P(B|A_2) = 0.45$

$P(B|A_3) = 0.5$

We want to find the probability $P(B)$, which can be found using the law of total probability:

$P(B) = P(B|A_1)P(A_1) + P(B|A_2)P(A_2) + P(B|A_3)P(A_3)$

Plugging in the given values:

$P(B) = (0.4)(\frac{1}{3}) + (0.45)(\frac{1}{3}) + (0.5)(\frac{1}{3})$

$P(B) = \frac{1}{3}(0.4 + 0.45 + 0.5)$

$P(B) = \frac{1}{3}(1.35)$

$P(B) = 0.45$

3. Final Answer

The probability that the fisherman will catch a fish is $0.45$.