Question

A university requires its biology majors to take a course called BioResearch. The prerequisite for this course is that students must have taken either a Statistics course or a computer course. By the time they are juniors, \(52 \%\) of the Biology majors have taken Statistics, \(23 \%\) have had a computer course, and \(7 \%\) have done both. a) What percent of the junior Biology majors are ineligible for BioResearch? b) What's the probability that a junior Biology major who has taken Statistics has also taken a computer course? c) Are taking these two courses disjoint events? Explain. d) Are taking these two courses independent events? Explain.

Short answer

a) 32% ineligible. b) 13.46%. c) No, not disjoint. d) No, not independent.

Step-by-step solution

Identify eligibility group

A student must take either Statistics or a computer course (or both) to be eligible for the BioResearch course. We need to find the percentage of students who have done neither to determine the ineligible percentage.

Use principle of inclusion-exclusion

The formula for students who have taken either course is: \( P(A \cup B) = P(A) + P(B) - P(A \cap B) \). Here, \( P(A) = 0.52 \), \( P(B) = 0.23 \), \( P(A \cap B) = 0.07 \). So \( P(A \cup B) = 0.52 + 0.23 - 0.07 = 0.68 \).

Calculate ineligible percentage

Since 68% are eligible, the percentage of junior Biology majors ineligible for BioResearch is \(100\% - 68\% = 32\%\).

Calculate conditional probability

The probability that a junior Biology major who has taken Statistics has also taken a computer course is given by \( P(B|A) = \frac{P(A \cap B)}{P(A)} \). Thus, \( P(B|A) = \frac{0.07}{0.52} = 0.1346 \), or about \(13.46\%\).

Check if events are disjoint

Disjoint events cannot occur simultaneously, meaning \( P(A \cap B) = 0 \). Since \( P(A \cap B) = 0.07 \), these events are not disjoint.

Check for independence of events

Two events, A and B, are independent if \( P(A \cap B) = P(A) \times P(B) \). Here, \( 0.07 eq 0.52 \times 0.23 \), which means these events are not independent.

Key concepts

Conditional Probability

When we talk about conditional probability, we refer to the likelihood of an event occurring given that another event has already happened. It's like asking, "If I know this one thing happened, what are the chances this other thing happens too?" In mathematical terms, the conditional probability of event B occurring, given that event A has already happened, is given by the formula:\[ P(B|A) = \frac{P(A \cap B)}{P(A)} \]

In our exercise, we wanted to know the chance that a Biology major has taken a computer course, given that they have taken Statistics.

To solve this, we used the formula, where \( P(A \cap B) = 0.07 \) and \( P(A) = 0.52 \), resulting in \( P(B|A) = 13.46\% \).

This calculation tells us that there is a 13.46% chance that a student who took Statistics also took a computer course.

Inclusion-Exclusion Principle

The Inclusion-Exclusion Principle is a helpful tool for finding the probability of the union of two events, A and B. It ensures that we don't double-count the probability of both A and B happening. The formula for this is:\[ P(A \cup B) = P(A) + P(B) - P(A \cap B) \]

This principle is used to calculate the number of students eligible for the BioResearch course.

We know that 52% of students took Statistics and 23% took a computer course, but 7% took both. We need to subtract the 7% who are counted twice, leading to \( P(A \cup B) = 68\% \).

Therefore, 68% of students took at least one of the required courses.

This method helps manage overlapping data, providing a clearer picture of multiple conditions being met.

Independent Events

Events are independent if the occurrence of one event does not affect the probability of the other event happening. Essentially, knowing that one event occurred doesn't give us any additional information about whether the other event will happen. We check independence by the condition:\[ P(A \cap B) = P(A) \times P(B) \]

In our scenario, we tested whether taking Statistics and a computer course occurs in such a way.

We found that \( P(A \cap B) = 0.07 \) isn't equal to \( P(A) \times P(B) = 0.52 \times 0.23 \), showing the courses are not independent.

This result tells us that the probability of taking one course affects the likelihood of having taken the other, showing a dependence between these events.

Disjoint Events

Disjoint events, also known as mutually exclusive events, are scenarios where two events cannot occur simultaneously. If events are disjoint, their intersection's probability is zero:\[ P(A \cap B) = 0 \]

In the given problem, we confirmed whether having completed Statistics and a computer course are disjoint.

Since \( P(A \cap B) = 0.07 \), the events are not disjoint, meaning students can and do take both courses.

This clearly shows that both courses can overlap for some students, indicating possible concurrence and not exclusivity.