Question

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?
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 \( \boxed{0.74} \).