Of 50 students going on a class trip, 35 are student athletes and 5 are left-handed. Of the student athletes, 3 are left-handed. Which is the probability that one of the students on the trip is an athlete or is left-handed?

Answer 1

To find the probability that a randomly selected student is either an athlete or left-handed, we can use the principle of inclusion-exclusion.

Let:

• \( A \) = the event that a student is a student athlete
• \( L \) = the event that a student is left-handed

We know:

• \( |A| = 35 \) (the number of student athletes)
• \( |L| = 5 \) (the number of left-handed students)
• \( |A \cap L| = 3 \) (the number of student athletes who are left-handed)

Using the formula for the union of two sets: \[ |A \cup L| = |A| + |L| - |A \cap L| \]

Substituting the known values: \[ |A \cup L| = 35 + 5 - 3 = 37 \]

Now, the total number of students is 50. The probability that a randomly selected student is either an athlete or left-handed is given by the ratio of the number of students who are either athletes or left-handed to the total number of students:

\[ P(A \cup L) = \frac{|A \cup L|}{\text{Total number of students}} = \frac{37}{50} \]

Thus, the probability that a randomly selected student is either a student athlete or left-handed is: \[ \boxed{\frac{37}{50}} \]