Word problom involving the probability of a union Fran the Fearless has a well-known act at a local zoo. As part of the act, Fran handles a scary animal that she picks at random out of the Box of Terror. The following animals are in the box: 6 poisonous snakes, 7 nonpoisonous snakes, 10 poisonous spiders, and 6 nonpoisonous spiders. What is the probability that the animal Fran picks is poisonous or is a spider? Do not round intermediate computations, and round your answer to the nearest hundredth.

To find the probability that the animal Fran picks is poisonous or a spider, we can use the principle of the probability of a union. First, let's determine the total number of animals in the Box of Terror: Total animals = 6 poisonous snakes + 7 nonpoisonous snakes + 10 poisonous spiders + 6 nonpoisonous spiders = 29 animals. Next, we need to find the number of animals that are either poisonous or spiders. 1. **Poisonous animals**: - Poisonous snakes = 6 - Poisonous spiders = 10 - Total poisonous = 6 + 10 = 16 2. **Spider animals**: - Nonpoisonous spiders = 6 - Poisonous spiders = 10 - Total spiders = 6 + 10 = 16 Using the formula for the probability of a union: \[ P(A \cup B) = P(A) + P(B) - P(A \cap B) \] Here, \(P(A)\) is the probability of picking a poisonous animal, \(P(B)\) is the probability of picking a spider, and \(P(A \cap B)\) is the probability of picking a poisonous spider. - \(P(A)\) = Number of poisonous animals / Total number of animals = \(16/29\). - \(P(B)\) = Number of spiders / Total number of animals = \(16/29\). - \(P(A \cap B)\) = Number of poisonous spiders / Total number of animals = \(10/29\). Now substitute the values: \[ P(A \cup B) = \frac{16}{29} + \frac{16}{29} - \frac{10}{29} \] \[ P(A \cup B) = \frac{16 + 16 - 10}{29} = \frac{22}{29} \] Finally, calculate the probability: \[ P(A \cup B) = \frac{22}{29} \approx 0.7586 \] So, rounding to the nearest hundredth, the probability is approximately **0.76**.