### Video Transcript

Students at a school must wear a
sweatshirt or a blazer and are allowed to wear both. In one class of 32 students, 12
students wear blazers, and four of the students who wear a blazer also wear a
sweatshirt. Let π΄ be the event of randomly
selecting a student from the class who wears a blazer and let π΅ be the event of
randomly selecting a student from the class who wears a sweatshirt. There are four parts to this
question. Find the probability of π΄, find
the probability of π΅, find the probability of π΄ intersection π΅, and find the
probability of π΄ union π΅.

In all four cases, weβre asked to
give our answer as a fraction in its simplest form. We could work out the first and
third parts directly from the question. We are told that 12 out of the 32
students wear blazers; therefore, the probability of π΄ is 12 out of 32. As both the numerator and
denominator are divisible by four, this simplifies to three out of eight or
three-eighths. We are also told that four of the
students who wear a blazer also wear a sweatshirt. This means that the probability of
π΄ intersection π΅ is four out of 32, as this is the probability of selecting a
student who wears a blazer and a sweatshirt. Once again, we can divide the
numerator and denominator by four, giving us one-eighth.

A key word in the question is
βmust,β as weβre told that students must wear a sweatshirt or a blazer. We can use this fact to write down
the probability of π΄ union π΅. As all of the students must wear a
sweatshirt or a blazer or both, the probability of π΄ union π΅ is 32 over 32, which
is equal to one. It is certain that a randomly
selected student will be wearing at least one of a sweatshirt or a blazer.

This leaves us with the second part
of the question, calculating the probability of π΅, the event of randomly selecting
a student who wears a sweatshirt. We can answer this by using the
addition rule of probability, which states that the probability of π΄ union π΅ is
equal to the probability of π΄ plus the probability of π΅ minus the probability of
π΄ intersection π΅. Substituting in the values we
already know, we have one is equal to three-eighth plus the probability of π΅ minus
one-eighth. The right-hand side simplifies to
two-eighths plus the probability of π΅, which in turn is equal to a quarter plus the
probability of π΅. Subtracting one-quarter from both
sides of this equation gives us the probability of π΅ is equal to
three-quarters.

We now have answers to all four
parts of this question. They are three-eighths,
three-quarters, one-eighth, and one, respectively. We could also represent the
information on a Venn diagram, where the numbers shown are the number of students in
each section. There were 12 students that wear
blazers, 24 students that wear sweatshirts, and four students that wear both. The three numbers sum to give us a
total of 32 students.