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?

Question

0.2
0.2

0.5
0.5

0.8
0.8

0.74

Answer 1

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

Let:

• \( A \) be the set of student athletes
• \( L \) be the set of left-handed students

From the problem, we know:

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

Using the inclusion-exclusion principle:

\[ P(A \cup L) = P(A) + P(L) - P(A \cap L) \]

First, we need to calculate the probabilities:

• \( P(A) = \frac{|A|}{\text{Total number of students}} = \frac{35}{50} = 0.7 \)

• \( P(L) = \frac{|L|}{\text{Total number of students}} = \frac{5}{50} = 0.1 \)

• \( P(A \cap L) = \frac{|A \cap L|}{\text{Total number of students}} = \frac{3}{50} = 0.06 \)

Now we can calculate \( P(A \cup L) \):

\[ P(A \cup L) = P(A) + P(L) - P(A \cap L) \] \[ P(A \cup L) = 0.7 + 0.1 - 0.06 = 0.7 + 0.1 - 0.06 = 0.74 \]

Thus, the probability that one of the students on the trip is either an athlete or left-handed is \( \boxed{0.74} \).