Question

A certain university has 10 vehicles available for use by faculty and staff. Six of these are vans and four are cars. On a particular day, only two requests for vehicles have been made. Suppose that the two vehicles to be assigned are chosen in a completely random fashion from among the 10 . a. Let \(E\) denote the event that the first vehicle assigned is a van. What is \(P(E)\) ? b. Let \(F\) denote the probability that the second vehicle assigned is a van. What is \(P(F \mid E)\) ? c. Use the results of Parts (a) and (b) to calculate \(P(E\) and \(F)\) (Hint: Use the definition of \(P(F \mid E) .)\)

Short answer

The probabilities are: \(P(E)\) is 0.6, \(P(F | E)\) is \(\frac{5}{9}\), and \(P(E \: and \: F)\) is \(\frac{1}{3}\).

Step-by-step solution

Calculate \(P(E)\)

The event \(E\) is that the first vehicle assigned is a van. Since there are 6 vans out of 10 vehicles, the probability that a randomly chosen vehicle is a van is given by \(\frac{number \:of \:vans}{total\: number\: of\: vehicles} = \frac{6}{10} = 0.6\). So, \(P(E) = 0.6\).

Calculate \(P(F | E)\)

The event \(F\) is that the second vehicle assigned is a van. \(P(F | E)\) is the probability that the second vehicle assigned is a van given that the first one already assigned was a van. After assigning one van, there are now 5 vans and 9 vehicles left. So, \(P(F | E) = \frac{number\: of\: remaining\: vans}{total\: number\: of\: remaining\: vehicles} = \frac{5}{9}.\) So, \(P(F | E) = \frac{5}{9}\).

Calculate \(P(E \: and \:F)\)

According to the definition of conditional probability, the probability that both events E and F occur, which is denoted as \(P(E \: and \: F)\), is the product of the probability of E and the probability of F given E. So, \(P(E \: and \: F) = P(E) \times P(F | E) = 0.6 \times \frac{5}{9} = \frac{1}{3}. \) So, \(P(E \: and \: F) = \frac{1}{3}\).

Key concepts

Conditional Probability

In statistics, one of the most fundamental concepts is that of conditional probability, which measures the probability of an event occurring given that another event has already happened. The conditional probability can be represented as P(A|B), which reads as 'the probability of A given B'.

To understand this with an example, let's consider the textbook exercise which deals with the assignment of vehicles. If the event E is that the first vehicle assigned is a van, and we want to find the conditional probability that the second vehicle assigned is also a van (event F) given that E has already occurred, we use the formula:
\[P(F | E) = \frac{P(E \: and \: F)}{P(E)}\]
In the solution provided, after one van has already been assigned (event E), there are fewer vans to choose from for the second assignment. The conditional probability P(F|E) takes into account this reduced pool of vehicles. This concept helps in understanding probabilities in more complex, dependent scenarios and is widely used in fields ranging from games of chance to predictive modelling.

Random Assignment

Another important concept in probability is random assignment, which refers to the practice of assigning subjects or objects to groups in a way that each subject has an equal chance of being placed in any group. This principle is crucial for conducting fair and unbiased experiments or selections. In the context of our example, the random assignment principle is applied when choosing vehicles to fulfill requests.

The term 'completely random fashion' as stated in the exercise implies each of the 10 vehicles, which includes 6 vans and 4 cars, has an equal probability of being selected. Random assignment ensures that the selection process doesn't favor vans over cars, or vice versa. It's the randomness that maintains the equality of chances among all vehicles, which is fundamental for calculating event probabilities accurately.

Event Probability

Finally, the concept of event probability signifies the chance of an event occurring. It is a measure that varies between 0 and 1, with 0 indicating an impossibility and 1 indicating a certainty. In our example, the event probability is first calculated for the event that a van is selected as the first vehicle:
\[P(E) = \frac{6}{10} = 0.6\]
This outcome represents the likelihood of event E happening in a single trial.

When the first event's outcome influences the second event, the calculation must consider the changed conditions, as seen when calculating P(F|E). The probability of both events E and F occurring is found by multiplying the probability of E with the conditional probability P(F|E).
Enhancing the students' understanding of these probabilities helps in grasping the underlying mechanism of random processes and is pivotal for studies in fields that range from finance to engineering.