Question

An automobile manufacturing plant produced 38 vehicles today: 15 were trucks, 10 were sedans, and 13 were vans. (each vehicle falls into only one of these categories.) plant managers are going to select two of these vehicles for a thorough inspection. the first vehicle will be selected at random, and then the second vehicle will be selected at random from the remaining vehicles. what is the probability that the first vehicle selected is a sedan and the second vehicle is a van? do not round your intermediate computations. round your final answer to three decimal places.

Answer 1

The probability of selecting a sedan first and then a van from the remaining vehicles at the automobile manufacturing plant is approximately 0.092 when rounded to three decimal places.

Step-by-step explanation:

To calculate the probability that the first vehicle selected is a sedan and the second is a van from the automobile manufacturing plant, we use the basic principles of probability.

First, we find the probability of choosing a sedan. There are 10 sedans out of 38 vehicles, so the probability of selecting a sedan first is:

Probability (Sedan first) = \(\frac{10}{38}\).

Once a sedan has been selected, there are now 37 vehicles left to choose from. Since we want to select a van next, and there are 13 vans available, the probability of choosing a van second is:

Probability (Van second given Sedan first) = \(\frac{13}{37}\).

To find the overall probability of both events occurring in sequence, we multiply the probabilities of each individual event:

Overall Probability = Probability (Sedan first) \(\times\) Probability (Van second given Sedan first)

Overall Probability = \(\frac{10}{38}\) \(\times\) \(\frac{13}{37}\) = \(\frac{130}{1406}\).

After calculating the final probability and rounding to three decimal places, we get that the Overall Probability is approximately 0.092.