At a school movie night, 150 students bought tickets: 90 underclassmen, of whom 54 bought popcorn, and 60 upperclassmen, of whom 42 bought popcorn. A ticket buyer is selected at random. What is $P(\text{bought popcorn} \mid \text{upperclassman})$?

A$\frac{7}{10}$
B$\frac{16}{25}$
C$\frac{7}{16}$
D$\frac{7}{25}$

We want to find the conditional probability $P(\text{bought popcorn} | \text{upperclassman})$, which represents the probability that a student bought popcorn, given that they are an upperclassman.

Apply the conditional probability formula

The conditional probability formula is defined as:
\[P(A | B) = \frac{n(A \cap B)}{n(B)}\]
where:

$A$ is the event that a student bought popcorn.
$B$ is the event that a student is an upperclassman.
$n(B)$ is the total number of upperclassmen.
$n(A \cap B)$ is the number of upperclassmen who bought popcorn.

From our data:

$n(B) = 60$
$n(A \cap B) = 42$

Calculate and simplify the probability

Now, we substitute the values into the formula:
\[P(\text{bought popcorn} | \text{upperclassman}) = \frac{42}{60}\]
To simplify this fraction, we find the greatest common divisor of $42$ and $60$. Both numbers are divisible by $6$:
\[42 \div 6 = 7\]
\[60 \div 6 = 10\]
So, the fraction simplifies to $\frac{7}{10}$.

Examples

This type of conditional probability is extremely useful in real-world scenarios such as school or business analytics. For example, if a school administrator wants to know if upperclassmen are more likely to purchase snacks than underclassmen, they can compare the conditional probabilities $P(\text{popcorn} | \text{upperclassman}) = \frac{42}{60} = 0.7$ and $P(\text{popcorn} | \text{underclassman}) = \frac{54}{90} = 0.6$. Since $0.7 > 0.6$, the administrator can conclude that upperclassmen have a higher likelihood of buying popcorn at the event.